general terminology chromosomes gametes somatic cells dna · replication • during ... the base...

Janelle Grass 20/09/2012

GENERAL TERMINOLOGY

Chromosomes

A chromosome is a long strand of DNA that contains certain genes.

Inherited from our parents

Eukaryotic cells have multiple chromosomes, and each of these is part of a pair (one from the mother and the other from the father). These are known as homologous chromosomes.

Prokaryotic cells have a long circular chromosome

Cells in the human body contain twenty-three pairs of chromosomes.

Gametes

• The sex cells, sperm and ova, pollen and ova

• Haploid cells, containing half the number of chromosomes (n)

Somatic cells

• Body cells other than the gametes

• Contain a fixed number of chromosomes characteristic of the species

• Diploid cells, containing two copies of every chromosome – one from each parent (2n)

DNA

DNA (deoxyribonucleic acid) codes for the many physical characteristics of every living creature.

• Nucleic acids are made up of polynucleotide chains,

• Each nucleotide is made up of three parts:

• The phosphate group forms the head, known as the 5’ end, and the sugar group forms the tail or 3’ end.

• The four bases in DNA are adenine (A), thymine (T), cytosine (C) and guanine (G). Thymine bonds with adenine and guanine with cytosine.

PROKARYOTIC DNA

Eukaryotic DNA is linear

Prokaryotes have circular DNA in the form of plasmids and one large, round DNA

Plasmids replicate independently of binary fission (the splitting of a bacterial cell into two identical cells)

The long circular DNA replicates only during binary fission

REPLICATION

• During mitosis and meiosis cells make copies of their chromosomes.

• Each of the single strands acts as a template for building a new complementary strand.

• Nucleotide building blocks come into position where they are complementary to the bases on the DNA strand.

• DNA replication results in the formation of two double stranded molecules of DNA, each of which is identical to the original double stranded DNA molecule.

• The two DNA strands produced contain one old strand from the original molecule and one new strand.

http://library.thinkquest.org/20830/Textbook/Glossary/main.htm#Genes�

http://library.thinkquest.org/20830/Textbook/Glossary/main.htm#DeoxyribonucleicAcid�

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Enzymes used in the process

• DNA helicase unwinds the long DNA strand.

• DNA polymerase then travels up from the 3' end attaching complimenting bases to the unzipped strand. Then it travels down toward the 5' end connecting complimenting bases to this side of the double helix.

TRANSCRIPTION

• DNA carries instructions about what proteins to make.

• The instruction is copied or transcribed from the DNA in the nucleus. This copy is in the form of ribonucleic acid (RNA) in the form of messenger RNA (mRNA), which carries a copy of the genetic instruction from the nucleus to the ribosome.

DNA and RNA are both nucleic acids, but they have some differences.

RNA is single stranded

The sugar in RNA is a ribose sugar

In RNA, there is no thymine base. It is replaced by uracil.

DNA base

Complimentary RNA base

A U

T A

C G

G C

DNA template sequence: A T G C C T G A A T

mRNA transcript (copy):

The steps in transcription are:

• DNA unzips and RNA polymerase enzyme binds to one strand of DNA. The RNA polymerase binds at a specific site called the “promoter region”.

• RNA polymerase makes a chain of RNA nucleotides, each complementary to the DNA template strand, and are added one by one into the growing strand, producing a single stranded molecule called pre-mRNA.

• Sections of the pre-mRNA strand, known as introns, are cut out producing a shorter mRNA strand known as final mRNA.

• The sections of DNA that are coded into a protein are known as exons and contain the information for production of the proteins required by the cell.

• The completed RNA strand is released from RNA polymerase and the DNA template and moves from the nucleus to the cytoplasm. This strand of RNA is complementary to the DNA template strand.

TRANSLATION

• Proteins are made on a special organelle in the cell called a ribosome.

• The mRNA moves out of the nucleus, through the cytoplasm and attaches to the ribosome.

• The amino acids required to build a protein chain are floating in the cytoplasm of the cell.

• The amino acid to be added to a growing chain is brought to the ribosome by a carrier molecule called transfer RNA (tRNA). Each tRNA molecule has a set of three bases that are known as an anti-codon. At the other end of the molecule is a region that attaches to one specific amino acid.

http://library.thinkquest.org/20830/Textbook/Glossary/main.htm#DNAHelicase�

http://library.thinkquest.org/20830/Textbook/Glossary/main.htm#DNAPolymerase�


• The information in mRNA is present as sets of three bases or triplets known as codons. Codons contain the information to add one specific amino acid to a protein chain. One codon is a start codon (AUG) and three different codons (UAA, UAG and UGA) are stop codons. The instructions in an mRNA molecule are decoded three bases at a time.

• As each codon is translated, the tRNA molecule with the complementary anti-codon pairs briefly with the mRNA molecule. The base pairing of mRNA and tRNA is as follows:

Codons GAC UCC CGG UAU

Anti-codons

QUESTION 1 (2011)

REVISION: QUESTIONS 1 AND 2

GENES AND ALLELES

Genes are portions of the DNA strand and tell the body what proteins to make. Every cell in your body has the same genes, and DNA.

• An allele is an alternative form of a gene.

• Every gene has at least two alleles, found in the same position on the maternal and paternal chromosomes and codes for the same characteristic as a gene on the other chromosome in the pair. They are usually denoted by a capital letter (T, R, C – dominant trait) or a lower case letter (t, r, c – recessive trait).

• Genes are regulated – they are transcribed and translated as their protein products are required or at different stages of development

MUTATIONS

Genetic material can change and may result in a form of a gene that was never present in a family appearing in the phenotype. Mutations change the instructions that are encoded in genes by changing the sequence of bases in DNA.

Substances that can cause changes in DNA are known as mutagens. If an amino acid in a protein is incorrect, the entire protein may be biologically useless. However, not all mutations may result in altered proteins. Mutations occur at the nucleotide level where individual codons are affected.

Substitution –

Addition –

Deletion –

Single base additions or deletions have a major effect on the genes involved because they alter not only the triplet at the point of the mutation, but also every triplet following that point. These are known as frame-shifts.

Mutations and the Next Generation

If a mutation occurs in the body cells of an organism it is known as a somatic mutation. Only that cell and its daughter cells produced by mitosis will have the mutation. They are not passed onto the next generation. However, if a mutation occurs in the cells that

http://library.thinkquest.org/20830/Textbook/Glossary/main.htm#Proteins�


produce gametes, it is known as a germ line mutation. These can be passed onto the next generation.

Question 5: The Dwarfism Gene

Achondroplasia is the most common form of dwarfism in humans and is inherited as an autosomal dominant condition. The locus for Achondroplasia is on chromosome number 4. Eighty to ninety percent of cases of Achondroplasia are the result of a new mutation which occurs at nucleotide 1138 of the DNA sequence and is a single base substitution resulting in a change in the amino acid sequence of the polypeptide produced by this gene. The frequency of this base change means it is the most commonly mutated nucleotide in the human genome reported so far.

Part of the DNA sequence of the normal allele for the Achondroplasia gene is shown below. Nucleotide 1138 is shown in bold type:

3’,,,A G C T A C G G G T G G…5’

a. What are the names of the bases of the DNA represented by the letters A, G, T, C?

b. What two components, other than the base, are found in a nucleotide of DNA? DNA is used as a template for the production of messenger RNA (mRNA). What is the name given to the process whereby mRNA is produced?

c. What is the base sequence of the mRNA that would result from the piece of DNA shown above?

DNA Sequence shown

A

G C T A G G G G T G G

MRNA

d. What is the name given to the three bases in mRNA that code for an amino acid?

e. Using the genetic code, show the amino acid sequence that results from the DNA sequence shown.

f. At what organelle of a cell would this sequence of amino acids be assembled?

g. The base change which occurs in the DNA of the normal Achondroplasia allele at nucleotide 1138 is a change from G to A. What is the corresponding amino acid sequence that results from the DNA of this mutant Achondroplasia allele?

QUESTION 2 (2011) REVISION: QUESTIONS 3 AND 4

MITOSIS

• Reproduction of eukaryotic cells resulting in the production of two daughter cells which are identical to the original parent cell.

• Mitosis is part of the process of cell multiplication that occurs when:

o

o

o

The process of mitosis involves five main phases:

• Interphase – the DNA in the nucleus is copied. Each duplicated chromosome consists of two strands of genetic information called sister chromatids.

• Prophase – each chromosome is seen as double stranded, held together by a centromere. The nuclear membrane has disappeared.


• Metaphase – the double stranded chromosomes line up around the equator of the cell, forming a line across the middle.

• Anaphase – the centromere divides so that the single stranded copies of each chromosome move to opposite ends of the cell. One copy from each chromosome pair is in each group.

• Telophase – a new nuclear membrane forms around each group of chromosomes. The cytoplasm divides and new cell membranes form. If the cell is a plant cell, a cell wall also forms.

The resulting daughter cells are genetically identical to the parent cell.

http://mitosismeiosis6th.wikispaces.com/

THE CELL CYCLE

The time taken for a newly formed cell to mature and then divide into daughter cells is called the cell cycle. It usually lasts about 18 – 24 hours in animals. The various phases of the cell cycle are shown below:

The phase between successive mitoses is called interphase and is termed S (synthesis).

The S period is between two G (gap) phases during which cell growth occurs. The G phases are periods when it is thought that DNA is checked for errors. Gap 1 exams DNA for errors that might have occurred during cell replication. Gap 2 checks for mistakes that may have occurred during the synthesis of DNA in the S phase

During G1 the cell contains single copies of each chromosome. After replication of DNA in the S phase, the chromosomes duplicate.

Chromosome replication (S)

G2 G1

Mitosis



MEIOSIS

Meiosis creates the cells necessary for sexual reproduction to occur, the gametes, or sperm and egg cells. These have half the number of chromosomes compared to the somatic cells.

Stages of Meiosis:

Meiosis involves two cell divisions. Before meiosis actually starts, a diploid cell goes through interphase forming double-stranded chromosomes.

• The nuclear membrane that surrounds the nucleus of the cell disappears.

Prophase I

• The chromosomes become visible

• Centrioles move to opposite poles of the nucleus

• The chromosomes line up to form homologous pairs of chromosomes

• Chromosome pairs line up along the center of the cell.

Metaphase I

• The process of the chromosomes pairing up is called synapsis. The two sets of chromosomes come together to form a tetrad, consisting of four chromatids. It is during this stage that crossing over of the chromosomes may occur. During this process, segments of chromosomes may be exchanged between homologous chromosomes.

• Each chromosome within the tetrad separates and moves to opposite poles of the cell towards the centrioles. The chromosomes do not separate at the centromere. This is known as disjunction of each homologous pair and is independent of any other chromosomes.

Anaphase I

• The cell begins to split as a nuclear membrane forms around each set of chromosomes. The cells then split in two, producing two daughter cells which are diploid.

Telophase I

• The chromosomes condense and become visible

Prophase II

• The chromosomes move towards the centre of each cell, lining up in single file instead of pairs.

Metaphase II

• The chromatids of each pair split at the centromere and move towards each end of the cells. This is known as disjunction of the single stranded copies of each chromosome and each moves in opposite directions. Again, the separation of the single stranded copies of each chromosome is independent of other chromosomes.

Anaphase II

• A nuclear membrane forms around each set of chromosomes forming four haploid cells.

Telophase II

MEIOSIS AND GENETICS

• Meiosis is the source of variation in sexually reproducing organisms. This variation arises from random assortment of homologous chromosomes at metaphase 1 and through crossing over and recombination of parental alleles.


COMPARISON BETWEEN MITOSIS AND MEIOSIS:

Single gene inheritance

A. Body cells from individual with genetic makeup Aa

Two gene inheritance

Body cell from individual with the genotype ‘AaBb’


MISTAKES IN MEIOSIS

Sometimes during meiosis a pair of chromosomes fails to disjoin so that two copies of a chromosome rather than the usual one are present in a gamete. This failure to disjoin is known as non-disjunction and can happen at anaphase I or anaphase II.

Anaphase I Anaphase II

TRANSLOCATIONS A chromosome translocation is a chromosome abnormality caused by rearrangement of parts between non-homologous chromosomes.

http://laikaspoetnik.wordpress.com/tag/chromosomal-translocation/ HYBRID ORGANISMS Most hybrid animals cannot reproduce because they have different chromosome numbers. Horses have 64 chromosomes and donkeys have 62. When each provides a gamete with its haploid number to make a mule, the mule ends up with 63. Meiosis is impossible in the resulting mule because the 32 horse chromosomes don't pair up easily with the 31 donkey chromosomes. The chromosomes are not homologous, making the mule sterile. QUESTION 4 (2011)

REVISION: 7, 8, 9, 10

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GENETICS AND HEREDITY

• Cells carry pairs of chromosomes (homologous), each chromosome in a pair containing the same genes with one of two alleles

• Phenotype - the way an organisms genes express themselves; either short, tall, green, etc (physical type). The characteristics of an organism

• Genotype - the gene type of an organism; the alleles of a certain characteristic: TT, Tt, tt.

• Allele - A gene on a chromosome that codes for the same characteristic as a gene on the other chromosome in the pair. Each allele can code for a dominant or recessive phenotype and are usually denoted by a capital letter (T, R, C - dominant) or a lower case letter (t, r, c - recessive).

TT - means that the chromosomes that carry the alleles for a certain characteristic both have a dominant phenotype.

Tt - means that one of the chromosomes of a homologous pair has a dominant trait while the other has an allele for the recessive trait for a particular characteristic.

tt - means that both of the homologous chromosomes carry the alleles for a particular recessive trait.

• Homozygous/pure breeding - If an organism's genotype is homozygous it has on both chromosomes, either the allele for the dominant trait or the allele for the recessive characteristic, i.e. TT or tt.

• Heterozygous/hybrid/carrier - If an organism's genotype is heterozygous it has one allele for the dominant trait and one allele for the recessive trait, i.e. Tt.

• Punnet Squares

We use punnet squares to help us determine the genotype of offspring and the probability of a certain genotype of the offspring of organisms. For example:

Cross a heterozygous dominant pea plant (Tt) with a heterozygous pea plant (Tt).

Gametes T t

T

t

Approximately:

• 25% of the offspring will have the genotype of TT.

• 50% of the offspring will have the genotype of Tt.

• 25% of the offspring will have the genotype of tt.

As a result of the dominant T overriding the recessive t when deciding the characteristic, three out of every four plants produced form this cross will be tall.

• 75% of the offspring will have a tall phenotype.

• 25% of the offspring will have a short phenotype.

REVISION: QUESTION 11

INCOMPLETE DOMINANCE:

A trait is incompletely dominant when the heterozygote has a phenotype that is an intermediate of those observed in the two homozygous parents, having different phenotypes.

In snapdragons, there is an allele for white petals (R2R2) and red petals (R1R1). When a red flowered plant is crossed with a white flowered plant the offspring are pink.

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Gametes R2 R2

R1

R1

Both genes express themselves equally so you get a pink colour in the petals. However, by crossing two pink flowers together there is a different outcome.

Gametes R1 R2

R1

R2

There is a possibility of obtaining one white, two pink, or one red flower. The phenotypic ratio is 1:2:1 and the genotypic ratio is 1:2:1.

X – LINKED INHERITANCE

In mammals, genes for some characteristics are carried on the X chromosome only. Sex linked conditions carried on the X chromosome occur more frequently in males than in females.

Sex Determination:

1. In humans, haemophilia is a rare condition where the blood fails to clot. It is inherited as a sex – linked recessive trait. That is, the genes for the condition are carried on the X chromosome. Consider a normal man who marries a carrier female.

a. Work out the percentage chance that the children will have the condition.

b. Work out the percentage chance that the sons will have the condition.

c. Work out the percentage that the daughters will have the condition.

DIHYBRID CROSS:

A Dihybrid Cross is where two characteristics are taken into consideration). Independent assortment occurs because in meiosis the segregation of one pair of homologous chromosomes and the alleles they carry does not influence the segregation of other pairs of chromosomes.

A cross between two plants heterozygous for red bloom and is carried out. Tall is dominant to short and red blood is dominant to white:

TtRr X TtRr

TESTCROSSES:

If we have an individual with a dominant phenotype, how do we find out how many genes control the phenotype?

This is done by carrying out a testcross where the individual in question is crossed with the parental strain showing the recessive phenotypes.

REVISION: QUESTIONS 12, 13, 14, 15 AND 16

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Crossing Over

• Genes that are found on the same chromosome are linked and are inherited together. Genes that are on separate chromosomes are inherited independently and this is known as a dihybrid cross.

• The further apart the genes, the greater the chance that crossing over will occur.

1. At what stage of meiosis does crossing over occur?

2. What is the term used to describe homologous pairs of chromosomes meeting at the equator of a cell?

FAMILY PEDIGREES

1. Autosomal Dominant

http://www.doe.mass.edu/mcas/student/2009/question.aspx?GradeID=100&SubjectCode=bio_hs&QuestionTypeName=&QuestionID=6742

The pattern of inheritance of an autosomal dominant trait includes the following features:

• Both males and females can be affected

• All affected individuals have at least one affected parent

• Transmission can be from fathers to daughters and sons, or from mothers to daughters and sons

• Once the trait disappears from a branch of the pedigree, it does not reappear

• In a large sample, approximately equal numbers of males and females will be affected

2. Autosomal Recessive Pattern

The pattern of inheritance of an autosomal recessive trait includes the following features:

• Both males and females can be affected

• Two unaffected parents can have an affected child

• All children of two affected parents must also show the condition

• The trait may disappear from the pedigree but reappear in later generations


• In a large sample, approximately equal numbers of males and females are affected

3. X – linked Dominant Pattern

• A mother who is heterozygous will have the condition

• If a father is affected but the mother isn’t, all female offspring will show the trait, all males will be unaffected

4. X – linked Recessive Pattern

• Heterozygous mothers will pass the condition on to half of her sons

QUESTION 5 (2011)

THE INTERACTION BETWEEN ENVIRONMENT AND GENOTYPES

An individual’s phenotype is predetermined by their genotype. However, they may or may not reach their phenotypic potential due to the effect of the environment to which they are exposed.

Genotype + environment → phenotype

Identical twins can have identical genotypes, but different phenotypes. One twin may grow taller than the other or be more intelligent.

INHERITED VARIATIONS DUE TO POLYGENES

Some inherited variations in phenotype, such as widow’s peak in humans, are due to the action of a single gene. These are known as monogenic traits.

Monogenic means “one gene”. When the gene has two alleles, with one being dominant to the other in its affect on the phenotype, the trait concerned shows two variations. For example, a person will either have a widow’s peak or they don’t. When variation involves a gene with three alleles, the number of variations is greater.

Variations seen in traits such as human height, skin pigmentation, fat content of cow’s milk, seed mass in bean plants is due to the action of many genes. These traits are polygenic, meaning “many genes”. If we were to graph the height of human populations, we would see a continuous variation.

MANIPULATING GENES – BIOTECHNOLOGY

GENE SEQUENCING

A gene is said to have been sequenced when the order of the nucleotides in a gene is identified. DNA sequencing uses a technique called gel electrophoresis. In the process:

DNA is cut into fragments of slightly different lengths using enzymes

These enzymes make breaks on one side of a particular base eg thymine

The ends of the fragments are labelled with radioactive nucleotides (markers) that latch onto the end of the fragment


The fragments are then sorted according to size by electrophoresis

Each of the four lanes in the gel represents a separate application of the same sample of DNA, but labelled with different radioactive markers (C, G, T and A)

The sequence in the gel represents that of the synthesised strand (ie. the markers) and must be converted into the template strand.

DNA FINGERPRINTING

• In some regions of chromosomes, the DNA contains repetitive sequences. The same sequence in different people have the same core sequence of 10 – 15 bases.

• After this, the patterns vary considerably from person to person. This idea has been used to develop DNA fingerprinting, sometimes known as DNA profiling.

• It has been calculated in humans that only one person in every million is likely to

have identical DNA fingerprints. It is therefore a useful tool in forensic investigations. The process involves the following steps:

The sample is collected from tissue of living or dead organism

The DNA is extracted from the sample, then separated and purified

The DNA is cut into fragments using enzymes that recognise certain base sequences

The fragments are separated by length using electrophoresis

In the process,

• The fragments are placed at one end of a gel

• The gel is then exposed to an electric field, with the positive pole at the far end and the negative pole at the starting point.

• DNA is negatively charged. DNA fragments move towards the positive pole at different rates according to their size. Large fragments move slowly, while small fragments move quickly. Fragments of the same size move at the same rate.

• A series of parallel bands are produced at different distances down the gel.

• The size of the unknown bands can be identified by comparing the distance moved against fragments of a known size.


PCR

• Polymerase Chain Reaction (PCR) is a technique used to create many copies of DNA identical to a trace sample. DNA can be replicated and formed at an exponential rate, resulting in a million-fold amplification of the original minute sample of DNA. The DNA may be a taken from a tiny sample of blood stain or semen sample.

• Temperature cycling is essential to PCR:

o Denaturation is the first step and occurs by heating the reaction mixture to 94oC. This step breaks the hydrogen bonds that maintain the double stranded configuration. The template strand separates into its single-stranded components.

o Annealing is the second step in the PCR cycle. The temperature is lowered so that the primers can form hydrogen bonds with complementary bases on the template DNA. Temperatures of 45 – 72oC are frequently used.

o Elongation completes the PCR cycle. The temperature is raised to 72oC, not hot enough to denature the primer tightly bound to its complement, but hot enough to denature the primer molecules loosely bound with mismatches and thereby prevent any additional non-specific priming.

• One cycle of PCR gives two copies of the desired DNA

• The process is repeated from step 2 onwards for many cycles.

Advantages of PCR:

• Only requires a few cells to gain DNA

• DNA does not have to maintain its integrity – if the DNA source is degraded, contaminated with other chemicals or found within clothing or other tissue the PCR process ignores everything except the desired DNA sequence.


• Useful in diagnosing infectious diseases, especially viruses, by searching for viral genes

REVISION: 17, 18 AND 19

GENETIC ENGINEERING

Plants or animals which have a new gene inserted into them are called transgenic. Plants or animals that have had the sequence of their genes changed are called genetically modified.

Manipulation that crosses the species barrier – a gene can be taken from a plant or animal cell and inserted into another plant or animal cell to give it a desired characteristic.

Through recombinant DNA, proteins that are rare and produce very desirable effects can be mass produced and made available in greater quantities. For example, many vaccinations are produced by recombinant DNA technology.

These days, gene manipulation involves:

• Cutting genetic material of any species into pieces

• Locating a particular piece

• Joining pieces together

• Making many copies of a particular piece

• Inserting a piece of DNA from one organism into the DNA of an organism belonging to the same or a different species.

The process involves:

Method 1: Cutting and Isolating the Gene

• Chromosomal DNA is extracted from cells of an organism, cut into shorter pieces and the required piece of DNA is picked out of the mixture using a gene probe is used in this process.

Method 2: Making a Gene By Chemical Synthesis

This is only possible if the gene sequence of a gene is known, or can be inferred from knowledge of the order of amino acids in the protein required.

Method 3: Making a Copy of a Gene Using an mRNA Template

• An mRNA molecule can act as a template for the formation of a single-stranded DNA molecule, called copy DNA (cDNA)

• The sequence of bases in a cDNA molecule is complimentary to the mRNA

• Reverse transcriptase is the enzyme that catalyses the joining of the nucleotide subunits to form cDNA

• DNA polymerase is the enzyme that makes the cDNA into a double-stranded molecule (ds)

Genetic Engineering Tools

1. Cutting DNA into fragments:

This is done by restriction enzymes that are found in bacteria.


• Restriction enzymes cut DNA at a specific recognition sequence or cutting site, which has a specific order of nucleotides.

• Some enzymes cut the two strands of a DNA molecule at points directly opposite each other, producing ends that are ‘blunt’. Others cut one strand at a point, but cut the second strand at a point that is not directly opposite. The overhanging cut ends are called ‘sticky’ ends.

• When pieces of cut DNA with sticky ends made by using the same restriction enzyme are placed together in solution, these ends will pair up. These sticky ends are complementary.

• Each restriction enzyme has its own name. The first part of the name is made up of three letters derived from the micro-organism from which it was obtained. The second part of the name, the Roman numerals, represents the enzymes obtained from the same organism. ‘I’ represents the first enzyme, ‘II’ the second, and so on. Sometimes a letter appears before the Roman numerals. This represents the strain of the organism.

Enzyme Site Resulting cut

EcoR1

5’ GAATTC 3’

3’CTTAAG 5’

SmaI

5’ CCCGGG 3’

3’ GGGCCC 5’

2. Electrophoresis

Electrophoresis sorts fragments of DNA produced when cut by restriction enzymes. DNA fragments are sorted into groups depending on their lengths.

3. Joining Fragments of DNA

To carry out genetic modification, strands of DNA must be joined using an enzyme, ligase, which joins pieces of double stranded DNA. Thus one longer piece of DNA or a circular molecule of DNA may result if sticky ends are produced. The fragments are permanently joined.

4. Placing DNA in Cells

• Vectors are special carriers that can transport DNA fragments into cells

• The DNA being transported is called passenger DNA

• A plasmid is one kind of vector

• The plasmid DNA is cut at one point using a cutting enzyme that produces sticky ends.

• The DNA fragments are prepared using the same cutting enzyme.

• These DNA fragments and the plasmid DNA are mixed. Their sticky ends pair up.

• Ligase makes this join permanent.

Copying the Gene

• Gene cloning involves the formation of many copies of a particular piece of DNA

• One technique involves inserting a particular gene into a plasmid

• The plasmid with its foreign gene is introduced into a bacterial cell

• The plasmid multiplies within the bacteria and reproduces as the bacteria reproduces



Detecting Huntington’s Disease

• HD is a dominant condition found on chromosome number 4.

• A parent with HD has a 50% chance of passing the allele to their offspring. CVS can be performed at 10 weeks gestation. The DNA from these cells is analysed and if it shows the presence of the H allele, the parents may choose to abort the foetus.

• The H allele can be distinguished from the h allele by the presence of 40 or more copies of a trinucleotide repeat, CAG, in the coding region of the gene. People who are not at risk of HD have repeats in the range of 6 to 30.

• People can be tested to determine the risk of inheriting an H allele from an affected parent. This test determines the number of CAG repeats in their alleles. PCR is used to amplify the region of the HD gene that includes the repeats. The amplified DNA from each person is separated by electrophoresis.

Detecting Haemophilia

• The gene for haemophilia is located on the X chromosome. A carrier female has a 50% chance of transmitting the Xc allele. If she has a son, he will have haemophilia.

• The sex of a foetus must therefore be determined and, if a male, the cells from chorionic villus sampling (CVS) are examined to determine if they are carrying the defective allele.

• Cells from the mother and an affected brother are treated to remove their DNA.

• The restriction enzyme Bcl I cuts DNA within the F8C gene. Depending on the specific alleles, either two or three cut points occur. The C allele has two cut points and a DNA length of 1165bp. The c allele has three cut points and two DNA fragments are produced. One has a length of 879 bp and is detected via the probe.

• The DNA segments are separated by electrophoresis

REVISION: QUESTIONS 21, 22 AND 23

CLONING ORGANISMS

The process of cloning an organism is shown below:


• a somatic cell from the organism to be cloned is removed • the nucleus, containing a diploid number of chromosomes, is removed • an egg cell from a surrogate mother is removed and the haploid nucleus

extracted. This ensures that none of the DNA from the surrogate mother is incorporated into the clone and that the clone will only contain the correct number of DNA molecules from the organism to be cloned

• the nucleus removed from the organism to be cloned is fused with the egg cell, which now has a diploid number of chromosomes taken only from this organism

QUESTION 3 (2011) REVISION: QUESTION 24

EVOLUTION and VARIATION

A population is a group of the same species living in the same region at a given time. The individuals in a population may have variation in many traits.

Change in Populations:

A gene pool in a population is the range of genes and their alleles that are present. Selection pressures result in certain phenotypes being favoured over others and the genes that control the organism are selected. As a result, the gene pool of the population changes and evolution occurs.

The gene pool can be altered by:

NATURAL SELECTION

• Within a population living with certain environmental conditions, different phenotypes may have different survival rates.

• A phenotype that is passed on to the next generation has a higher fitness value and is at a selective advantage.

• A phenotype that has a lesser contribution is less fit, and is selected against.

• The environmental change or agent that is selecting a particular phenotype is called the selective agent.

If a tree were to be sprayed with a pesticide, a small number of beetles living in the tree may have resistance to the pesticide. These beetles are at a selective advantage and will survive to reproduce. Their offspring will also have the gene for resistance to the particular pesticide and will also survive to reproduce resistant offspring. After several generations, the population changes so that the majority of the inhabitants are resistant. Those that are not resistant do not survive to pass on their genes.

Natural selection occurs when:

There is variation in the original population

There is a new selective pressure

Those individuals that carry genes that enable them to survive this change in their environment are able to survive and reproduce.

After many generations, the gene that is fittest or selected for increases in frequency in the population.



ARTIFICIAL SELECTION

This is the deliberate selection of particular individuals from a population to be the parents of the next generation.

Eg. Sheep and cattle breeding, dogs

REVISION: QUESTION 26 AND 27

MUTATION:

The origin of all variation is mutation. Mutations can be harmful or they can be beneficial. If the environment changes, a mutation might provide a phenotype that is better suited to the changed environment and the genes producing the mutation become a part of the gene pool. The mutation must occur in the gametes for it to be passed on. The rate of accumulation of mutations is very slow.

LOSS OF VARIATION:

• Loss of variation – more susceptible to extinction

• Natural populations have a high degree of genetic variation – some individuals are fitter than others and will survive a change in their environment

MIGRATION:

• Also known as gene flow – changes to allele frequency occur quickly

• Emigration

GENETIC DRIFT:

• Small shifts in the allele frequency of a population

• Change is unpredictable due to chance events

• If only a few individuals reproduce the number of alleles passed on to the next generation may be less than the entire gene pool

• Varies from one population to the next

• This change is known as genetic drift

• Smaller populations are more affected by genetic drift

FOUNDER POPULATION:

• The effect of genetic drift when a few individuals carrying a small proportion of the original gene pool are isolated from a population and interbreed

• A small group of organisms, such as a mating pair or a pregnant female, becomes the start of a new population.

• The gene pool of the original members may not be representative of the larger population

BOTTLENECK EFFECT:

A loss of variation and possibly a change in allele frequencies can occur when a population is dramatically reduced by some catastrophic event. The remaining individuals are left to reproduce.

EVOLUTION WITHIN A SPECIES:

Causes of change in allele frequencies include:

• Selection

• Genetic drift

• Migration

• Immigration

• Emigration

• Mutation


SPECIATION:

When members of one population are separated into isolated populations over time, separate races or sub-species may develop. A species is a group of organisms that are able to interbreed producing viable offspring.

• Isolated populations may be exposed to different selection pressures due to different environmental conditions

• Genetic drift can produce different changes in each population by chance

• New alleles may be introduced by mutation

• Subspecies from different populations retain their ability to interbreed and produce viable offspring

• If populations remain separated and continue to evolve in different directions, a new species may arise. If brought together, they are not able to produce fertile offspring. Two new species have formed

REVISION: QUESTIONS 28, 29 AND 30

EVIDENCE FOR EVOLUTION:

1. Fossils

An organisms structure, diet, mode of movement and many other factors can be determined from an analysis of its preserved remains. When organisms die, their chance of becoming fossilised depends on:

• Compression under layers of sediment – tissues are replaced with a carbon film or minerals

• Moulds of organisms form when all or part of the dead organism is covered by sediments that form sandstone or mudstone

• If the cavity within a mould is later filled by other material, a three dimensional cast is formed

• Death in areas where oxygen levels are low – bacterial decomposition is reduced

• Quick burial

• Few scavengers

• Wind erosion is low

Fossils are evidence of organisms that lived in the past. They may include:

• Bones

• Teeth

• Leaves

Dating fossils

• Stratigraphy – the relative positions of the fossils in rock strata, the oldest fossils being those at the bottom of the strata

• Absolute dating depends on the decay of radioactive isotopes present in minerals in rocks

o Potassium – argon dating

o Carbon dating – only useful for rocks that are less than 50,000 years old

• Index fossils are any animal or plant preserved in the rock that is characteristic of a particular geologic time or environment. They must be distinctive or easily recognizable, abundant, and have a wide geographic distribution and a short range through time. Index fossils are the basis for defining boundaries in the geologic time scale and for the correlation of strata. They help in dating other fossils found in the same sedimentary layer. If you find a fossil from an unknown


era near a fossil from a known time, you can assume that the two species were from about the same time.

Comparative Biochemistry

• DNA hybridisation

– separating the stands of DNA from two different organisms

– the separated strands of DNA are mixed, one from each species

– the strands pair up

– organisms that are genetically similar will show a large degree of hybridisation

– the similarity of genomes gives an estimate of the genetic distance between them, i.e. the time elapsed since the two species shared a common ancestor

• Comparing gene sequences

– comparing DNA sequences of genes in different species can indicate an evolutionary relationship

– humans and chimps have the greatest similarity in the sequences of a haemoglobin gene

• Comparing proteins

– amino acid sequence of enzymes is compared

– the longer the time elapsed since two species diverged from a common ancestor, the more time has been available for changes to occur in the amino acid sequence

• Comparative anatomy

– Homologous structures – structures with a similar structure but a different function

– Analogous structures – structures that carry out similar functions but have different basic structure. Analogous structures do not indicate a common ancestor

– Vestigial organs - functionless or reduced remnants of organs that were present in ancestral organisms

– Embryology – all terrestrial vertebrates have gill slits in early embryos, as are tails. This indicates that vertebrates share a common ancestor

– Mitochondrial DNA – mitochondria have a circular molecule of double stranded DNA. It is passed down the maternal line and does not undergo recombination due to the lack of homologous chromosomes in the mitochondria. The coding region of mtDNA mutates very slowly but the non-coding regions have a high mutation rate. The longer two populations have been separated, the greater the differences in their mtDNA


DIVERGENT EVOLUTION

One ancestral species changes over time to give rise to many new species. Each of these species has a different habitat and ecological niche, and has unique features that enable it to survive its particular selection pressures. Such organisms may display homologous structures.

CONVERGENT EVOLUTION

Natural selection acts on distantly related species to produce similar features that are not due to sharing a recent common ancestor, but indicate that the species have similar selection pressures. They have a similar appearance due to the fact that they


share similar environments, not a recent common ancestor. Organisms that have converged may show analogous structures.

PARALLEL EVOLUTION

Where groups of organisms exist in a habitat where no new selection pressures exist, there may be no change in phenotype frequency.


HOMINID EVOLUTION

The evolution of our own species, Homo sapiens, is believed to have occurred through the process of natural selection. However, many of the details of our evolution is the subject of much debate. It is generally accepted that humans share a common ancestor with the other primates.

Features of primates:

• Rounded faces with reduced snouts

• Eyes protected by bony ridges

• Relatively large brains and a large brain case

• Eyes point forward

• Variation in tooth size and shape

• Collarbones for brachiation (swinging by arms)

• Five digits on hands and feet

• Opposable thumbs

• Some bipedal, many quadrupedal

• Internal fertilisation, long gestation, long period of parental care

• Social groups with hierarchies

These similarities cannot be accounted for by convergent evolution. Biochemical evidence based on similar DNA sequences indicates that these animals are closely related.

• The evolution of primates is an example of divergent evolution from a common ancestor.

• Mitochondrial DNA evidence suggests that humans show greatest similarity to chimpanzees and gorillas, which show less similarity to the great apes

The first hominid ancestor appeared in the fossil record in Africa around 4 million years ago. Hominids changed over time from this first ancestor:

• Body size tended to increase

• Thick skeletons – robust, fine boned – gracile

• Skull capacity increased

• Skull bones and teeth were more like apes than modern humans

• Pelvis and thigh bones indicated bipedal gait

• Ability to use tools


There is much evidence that many groups of early Homo sapiens migrated out of Africa about 200,000 years ago. They evolved differently due to different selection pressures in their different habitats.

Species Appearance Features

Homo habilis

2.0 – 1.5 mya Permanent dwellings, stone tools

Homo erectus 1.7 mya – 200,000BP Used fire

Homo sapiens neanderthalis

90,000 – 30,000 BP Ceremonial burials

Homo sapiens sapiens 30,000BP – present Bone needles, art

CULTURAL EVOLUTION

Large brains and highly developed language favour a social life style. This allows for collaborative hunting, sharing of information and transmission of information from generation to generation.

‘Culture’ refers to all the behavioural patterns of a group which is passed on to the next generation by learning and behaviour.

An opposable thumb has allowed allows for high levels of manipulation and this depends on large brains and sensitive fingertips. Combined with language, these factors have provided the skills necessary for technological advancement.

QUESTION 7 (2011) REVISION: QUESTIONS 34, 35, 36, 37, 38, 39, 40, 41, 42 PAST EXAM QUESTIONS

Question 1 (2010)

The following diagram outlines processes that occur in living cells:

a. Name the process represented at X.

b. Describe the sequence of events that occur during the process at X.

c. Name the process represented at Y.

d. Describe the sequence of events that occur during the process at Y.

Question 2 (2002)

The figure below represents a piece of DNA from a gene. The template strand for a section of the gene is shown.

Part of the template strand A A A G T A C T G C G C

Complementary strand


a. In the space provided in the figure above write in the complementary bases for the other strand of DNA.

b. What does the A represent in this DNA sequence?

c. The first step in the process of gene expression is transcription. What is produced during transcription?

d. The compound produced during transcription attaches to an organelle in the cell.

i. What is the name of this organelle?

ii. What is the name of the step in gene expression which follows transcription and occurs at this organelle?

iii. What is produced during this process mentioned in part ii.?

In the section of the gene sequence shown in the above figure a base substitution mutation occurred. The 9th base from the left, Guanine, was replaced by Thymine. The genetic code is provided below:

e. What effect will this mutation have on the sequence of amino acids in the

polypeptide?

Question 3 (2007)

a. Explain what is meant by gene regulation.

Bacteria require amino acids to produce proteins. For example, bacteria in a human intestine may absorb amino acids from digested food, but at times there may be a deficiency of a particular amino acid. If this is the case, the bacteria will produce the necessary amino acid themselves.

The diagram below is a regulation system in a bacterial cell involving the production of the amino acid tryptophan. Note that there are two pathways (X and Y). Tryptophan is the regulatory compound in these two pathways and acts as a repressor in both.

b. Describe the immediate outcome when tryptophan activates pathway X.

c. Describe the immediate outcome when tryptophan activates pathway Y.


d. Suggest how the action of tryptophan as a repressor in this system could be of selective advantage to a bacterial cell in the digestive tract.

QUESTION 4 (2009)

Diagram W shows part of a metabolic pathway involving the amino acids phenylalanine and tyrosine. One or more enzymes are involved at each step in such a pathway.

Diagram W

Two steps have been highlighted, step P and step A. each is under the control of a single gene.

Step P is controlled by gene P which has the alleles

P : phenylalanine hydroxylase produced

p : no enzyme

Step A is controlled by gene A which has the alleles

A : tyrosinase produced

a : no enzyme

A couple, each heterozygous at the P and A loci, have a daughter, Emily.

a. With reference to these two genes, what is the chance that Emily has the same genotype as her parents?

The couple also have twin sons, Jack and Tom. Their genotypes are

Jack Pp aa

Tom PPAa

b. Are Jack and Tom identical twins or non-identical twins? Explain.

Another son, max, has the genotype ppAa.

c. With reference to their genotype for the pathway shown in Diagram W, ooutline two phenotypic differences that would exist for Max and Jack.

In corn, brittle stalk (b) is recessive to normal stalk (B) and green (g) is recessive to yellow (G). A claim was made that a particular plant was heterozygous at both loci.

d. Outline a cross you would carry out with the particular plant. Suggest results that would support the claim.

e. Explain whether the results of your cross would enable you to decide if the genes involved were on the same or different chromosomes.

QUESTION 5 (2009)

a. Describe binary fission.

Diagram X outlines a mitotic cell cycle. Image D shows the appearance of a chromosome during one of the cycles.


b. Explain at which labelled point (P, Q, R, V, W) in the cell cycle image D would be

found.

c. Explain why apoptosis sometimes occurs during the cell cycle represented in the above diagram.

d. Translation occurs within the cytosol of a cell. Outline the steps that normally occur in translation. Use specific terms and names of the molecules involved. Name the final product of the process.

QUESTION 6 (2008)

When a cell replicates it goes through a series of events that can be summarised by the following diagram. The cycle moves in a clockwise direction and includes mitosis. Note the four points labelled A, B, C and D.

Given that two cells are formed as a result of replication, a cell must replicate its DNA during the cycle.

a. Using crosses, mark on the graph below the relative amount of DNA present at each of the points B and D in the cycle.

The following photograph shows a group of cells, some of which are replicating by mitosis. The letters P, Q, R and S indicate cells that are at different points in the cell cycle.


b. Starting with the cell closest to the beginning of mitosis, arrange the letters P, Q, R

and S in the order in which they would occur during the cell cycle.

If a serious error has occurred during mitosis, the daughter cells may not survive to complete the cell cycle.

c. Name the process by which such cells are destroyed.

Meiosis is another form of cell division. A student claimed that there was no significant difference between mitosis and meiosis.

d. Identify one significant feature of meiosis and explain how it indicates that it is a different process from meiosis.

Question 7 (2010)

Humans have 46 chromosomes in each of their body cells.

a. How many autosomes are present in a single normal gamete?

During gamete formation, homologous chromosomes pair and exchange genetic material. This process is known as crossing over.

b. What is the advantage of crossing over in gamete formation?

One kind of translocation of genetic material between chromosomes occurs when part of one chromosome exchanges with a part of a different chromosome. If a translocation is present in a fertilised egg, all cells of the individual that has developed will carry the translocation. A phenotypically normal man carried a translocation involving chromosomes 1 and 2.

c. Explain why the male is phenotypically normal in spite of carrying the

translocation shown.

The following diagram outlines one meiotic division that occurred in the male.


d. What stage of meiosis is shown in step R of the diagram above?

The particular makeup of gametes produced by meiosis depends on the orientation of the chromosome pairs at stage N. one example is given above. However, this is not the only orientation possible and, during gamete formation in testes, all possible orientations occur.

e. What is the chance that a sperm from the man will contain a normal chromosome 1 and a normal chromosome 2?

f. Outline your reasoning.

Question 8 (2003)

A cell with a diploid number of 4 underwent meiosis. The following diagrams show different stages throughout the total process of meiosis in this cell:

a. Using the letters under each cell (A – F) put the cells in order commencing with

the earliest stage of meiosis shown.

b. During meiosis crossing over and recombination occur between homologous chromosomes. Describe the outcome of recombination.

c. During meiosis the nucleus undergoes two divisions. Which of the cells (E or F) represents anaphase 1?

d. Explain how you reached your answer.

Question 9 (2007)

a. Name the process by which the gametes are produced in humans.

b. Use the symbols X and Y to complete the diagram below to show how the sex chromosomes behave during meiosis in a human male.

Sometimes a mistake occurs in sperm production resulting in sperm which carry an extra X chromosome or an extra Y chromosome. Part of such a process is shown in the diagram below. Note that only the sex chromosomes are shown.


c. Describe the mistake in the process that led to the production of abnormal

sperm.

d. If sperm T fertilised a normal ovum, what would be the diploid number of the resulting zygote?

e. One kind of error in sperm production has been shown in the diagram above. In another kind of error, sperm are produced carrying XX or YY chromosomes. Explain how this could occur.

QUESTION 10 (2008)

In the late 1950s, a series of nuclear tests involving the detonation of hydrogen bombs was carried out in the mid-Pacific Ocean. Witnesses of these tests included 500 New Zealand sailors. Since the tests, the New Zealand sailors have claimed that their lives have been affected because, as a direct result of radioactivity from the tests, a number of genetic disorders have appeared among them and their offspring.

In the year 2000, the New Zealand veterans commissioned research to assess DNA damage that they may have.

a. Describe three factors that scientists would need to consider in the design of their research.

The chromosomes of men who participated in the study were examined. Each homologous pair of chromosomes was stained with a specific colour. This technique meant that segments of chromosomes that have moved between non-homologous chromosomes can be readily identified. The following diagram shows the result for three homologous pairs of chromosomes from veteran X.

b. Assume that you are veteran X. Use the data to present an argument supporting

your claim that you have genetic defects that adversely impact on your health. c. Explain how the results for veteran X could lead to an increase in genetic

disorders in his children and grandchildren. d. One individual claimed the chromosomal changes observed in veteran X were

mutations involving a single base pair on each chromosome. Explain whether you agree or disagree with this individual.

Question 11 (2003)

The back of the leopard frog can be either patterned or non-patterned offspring. Several patterned frogs were allowed to breed and they produced 75 patterned offspring and 25 non-patterned offspring.

a. i. Which of the phenotypes, patterned or non-patterned, is dominant?


ii. Explain your answer.

b. Using your own allelic notation, show the genotypes with their respective phenotypes for the parents and offspring of the cross between the patterned frogs described above.

Crosses between patterned and non-patterned frogs were performed. Not all crosses produced the same outcome. The results are shown in the table below. For both cross A and cross B there were large numbers of offspring produced.

Parents Offspring

Cross A Patterned Non-patterned All patterned

Cross B Patterned Non-patterned ½ patterned

½ non-patterned

c. The parents in crosses A and B have the same phenotypes. Explain why the outcome of crosses A and B are different.

Question 12 (2003)

Purebreeding guinea pigs with tough textured black coats were crossed to purebreeding guinea pigs with smooth textured white coats. The F1 were all rough textured with black coats. The F1 were allowed to interbreed to produce an F2. The numbers of each phenotype produced in the F2 are shown below:

F2 offspring Number

Rough, black 95

Rough, white 32

Smooth, black 27

Smooth, white 11

Total 165

a. There are two genes involved in these crosses. What information provided in the results confirms there are two genes each with two alleles?

b. Use the allelic symbols R and r for the texture locus, and B and b for the colour locus.

i. What are the genotypes of the purebreeding parental guinea pigs? Indicate in your answer which genotype matches which phenotype.

ii. What is the genotype of the F1 guinea pigs?

iii. Give one genotype for an F2 smooth, black guinea pig.

c. The two pairs of chromosomes on which these two gene loci are located assort independently. What does ‘assort independently’ mean? Include a diagram in your answer.

Question 13 (2002)

Coat colour in cocker spaniel dogs varies. Four of these colours are black, liver, red and lemon. These four colours result from the interaction of two particular autosomal genes. The pedigree below shows the inheritance of coat colour in a group of cocker spaniels.


a. I-1 and I-2 are heterozygous at both the R and B locus. What evidence from the

pedigree supports this conclusion?

b. What is the specific genotype of II-4?

c. What is the specific genotype of III-4?

d. Explain how many different phenotypes could be expected in the offspring of a mating between individuals II-4 and III-4. Show all working.

Question 14 (2010)

Consider the following pedigree:

a. What is the mode of inheritance of the trait?

b. What is the chance that Molly is heterozygous for the trait? Show your working out including the genotype of Molly’s parents.

Eukaryotic chromosomes, prokaryotic chromosomes and plasmids all contain DNA and vary in size.

c. What is one other difference between a:

i. Eukaryotic and a prokaryotic chromosome

ii. Prokaryotic chromosome and a plasmid

Question 15 (2007)

The pedigree below represents a family of rabbits. The shaded rabbits have an inherited disease. The phenotypes of rabbits I-1, I-2, I-3 and I-4 are not known.


On the basis of the offspring produced by II-1 and II-2 it has been suggested that the disease is inherited as an X-linked dominant characteristic.

a. What evidence from generations II and III supports the suggestion made about the mode of inheritance of the disease?

b. If the mode of inheritance suggested is correct, complete the table below to show the possible phenotypes and genotypes of rabbits I-1, I-2, I-3 and I-4. For each rabbit’s phenotype, select from:

• Has the disease

• Does not have the disease

• Impossible to tell from the information given

Use the following symbols XD, Xd and Y

Phenotype Possible genotypes

Rabbit I-1

Rabbit I-2

Rabbit I-3

Rabbit I-4

Assume each female in generation III was mated to a male from the same litter.

c. What genotypic and phenotypic ratios would you expect in the offspring?

Question 16 (2010)

DNA includes sections that are called short tandem repeats (STR). Mutations in STRs occur, on average, every 500 generations. Different numbers of these repeats have no obvious effect on the individual.

a. What is the likely reason for this?

A young man, Ben, wants to find out more about his genetic ancestry. He sends a sample of cells, obtained from a swab of his mouth, to a laboratory. On receipt of the sample, the laboratory treats the cells to release the DNA to enable identification of STR.

b. Name the process used to produce many copies of the STR markers

Each if the STR markers produced is labelled with a dye and subjected to gel electrophoresis. Five of Ben’s STR markers were compared with three family groups who have the same surname as him. The following gels resulted.

c. Explain which family is Ben’s most recent common ancestor.

Question 17 (2002) There are three suspects in an assault case. A forensic scientist found blood, other that the victim’s, at the site. DNA was extracted from five blood samples.

• The victim

• The blood at the site (not the victim’s)


• The three suspects

Polymerase chain reaction (PCR) was used on the extracted DNA.

a. A DNA polymerase enzyme is involved in the PCR process. Explain the role of the polymerase enzyme in PCR.

One of the regions used in the forensic analysis was a short tandem repeat (STR) sequence of three bases, called HUMTHO1. This sequence, located on chromosome 11, has many alleles which differ from each other by the number of times the sequence AATG is repeated. It was this region of chromosome 11 which was amplified using PCR. The amplified samples were loaded onto a gel and electrophoresis was performed to separate the fragments of DNA.

b. Name two properties of the DNA fragments which allow them to be separated from each other during gel electrophoresis.

A diagram of the gel is shown below.

c. Why is there only one band in lane 2 but two bands in lanes 3, 4, 5 and 6?

d. How many different alleles at the HUMTHO1 locus are represented on the gel in individuals 2, 3, 4, 5, 6?

e. Which piece of DNA, A or B, has the greater number of 4 base repeat sequence?

f. Which of the suspects appears to have committed the assault? Explain.

Question 18 (2007)

Victoria police forensic scientists conduct DNA profiling using samples taken from crime scenes. Traces of DNA less than 1 nanogram can be amplified and then profiled.

a. Name the process which is used to amplify DNA.

b. What must be done between stages 1 and 2 to separate the strands of the DNA

molecule?

c. Complete and label the diagram at stage 2.

Small pieces of DNA of differing length can be compared to determine whether or not a sample could have come from a particular person. In a case, samples of DNA from the victim and the crime scene were compared with samples from two suspects.

The DNA samples were treated with restriction enzymes, amplified and run through gel electrophoresis. The results for one gene locus are shown in the diagram below:


d. Draw an arrow on the right hand side of the diagram to indicate the direction of

movement of the DNA fragments.

e. What do the standards consist of, and what is their purpose?

f. From these results, give a conclusion which could be drawn about the sample taken from the crime scene.

g. What further action would you recommend to the forensic scientist investigating this case?

QUESTION 19 RFLP (Restriction Fragment Length Polymorphism) analysis is commonly used to determine genetic variation between individuals. The procedure is summarised below.

In this procedure, scientists select a particular restriction enzyme from an available range.

a. Explain the reason for their choice. Electrophoresis uses electrical current to sort DNA fragments.

b. Describe one characteristic of this sorting process. c. Explain why the DNA of each individual produces a different pattern of

fragments after gel electrophoresis, even when the same restriction enzyme is used.

Examine stages Y and Z. d. Describe, at the molecular level, what is meant by the term hybridised. Why is it

necessary to carry out hybridisation?

Question 20 (2003)

Genes can be transferred from one species to another in different ways. One method is to use plasmids, circular pieces of DNA found in some bacteria. In this method, a plasmid is cut and a foreign piece of DNA inserted. The foreign piece of DNA contains more than one gene. This process is shown below:


Many copies of the new plasmid are then incubated with bacteria.

a. What is the name given to the plasmid that is used to transfer DNA from one organism to another?

b. What is used to cut the DNA of a plasmid?

c. What is used to join the inserted piece of DNA to the plasmid?

One of the foreign genes inserted into the plasmid codes for resistance to a particular antibiotic.

d. Explain why it is important to include a gene for antibiotic resistance in the plasmid produced?

Bacteria containing plasmids can be used in a variety of settings with plants and animals. For example, some plants are resistant to attack by insects. The plants produce a protein that poisons the larval stage of some insects that feed on them. The production of the protein is under genetic control. A particular species of crop was genetically engineered to contain this gene. Such plants are referred to as GM (genetically modified) plants.

e. Explain why a farmer might choose to grow a crop that was genetically engineered to be resistant to insects, rather than spray the crop with insecticide.

Some plants are resistant to particular herbicides, chemicals used to kill plants. This trait is also under genetic control. The gene that confers herbicide resistance has also been incorporated into some GM crops. This enables a farmer to spray his crop with a herbicide that will not harm the GM crop but does kill weed plants growing within the crop.

f. Suggest one advantage for a farmer to be able to spray his crops with a herbicide.

Two farmers have properties nest door to each other. They grow the same cereal crop. Farmer X wishes to grow GM crops that are resistant to the herbicide. Farmer Y wishes to continue to grow non-GM crops. Farmer Y was concerned and suggested to farmer X that pollen from the GM crop could fertilise the non-GM plants.

g. Explain why farmer Y might be concerned about the possibility of his crop being fertilised by pollen from farmer X’s crop.

QUESTION 21 (2008)

Some gene loci have several alternative alleles. With these gene loci, the DNA profile of an individual may be given in terms of the relative sizes of alleles. For example, gene locus THO has three alleles that are called THO7, THO9 THO THO10. In a maternity hospital, a mother claimed that she had accidentally been given the wrong baby boy. A DNA analysis of two gene loci, locus FGA and locus DS, was carried out on members of the family and the results are included in the following pedigree. Assume that no mutation involving these two loci has occurred in the family.

a. Does the data support the mother’s claim that she has been given the wrong

baby? Give reasons to support your answer.

Many trials involving the technique of gene therapy have been carried out in humans, with varying success.

b. What is the aim of gene therapy?

The technique of gene therapy also involves the use of vectors.


c. Name one agent that could act as a vector in gene therapy.

In 1999, during a gene therapy trial for a particular enzyme deficiency, a young man died from multiple organ failure four days after starting treatment. He was in excellent health before participating in the trial.

d. Suggest one event that could have occurred within the young man as a result of him receiving gene therapy that resulted in his death.

QUESTION 22 (2009)

Some people prefer to eat Wagyu cattle because of the high level of marbling (fat) in the meat. Four separate DNA markers are used to test for marbling in an animal. Tested cattle are scored on a scale of zero to eight, eight indicating the highest degree of marbling.

a. What does the use of four markers suggest about the inheritance of this characteristic?

A Wagyu breeder discovered a small number of individuals in her elite herd that were suffering from Chediak-Higashi Syndrome (CHS). CHS is an autosomal recessive condition that can affect species other than cattle. The breeder required further information.

Gene probing was used to target CHS1, the allele responsible for the condition. The genetic probes for the Wagyu CHS locus were derived from human alleles.

b. Given that the gene probe for a human works for the Wagyu, what can you infer about the chemical code for this allele?

The Wagyu CHS1 allele was isolated and given a fluorescent tag. It was introduced into a yeast cell as a large, independent, cytoplasmic chromosomal segment called a Yeast Artificial Chromosome (YAC). In addition to the allelic DNA, a YAC includes a centromere and a replication sequence. The yeast cells are then incubated in the presence of growth stimulants and given time to replicate.

The procedure is similar to genetic engineering of bacterial plasmids, however the YAC is able to contain much larger pieces of DNA than a plasmid.

c. In the bacterial cell above, draw a plasmid in the blank box.

d. Bacterial plasmids lack a centromere. Why are YACs made with a centromere?

e. What term describes the process of copying a gene?

A test was developed to identify each of the normal and mutant alleles. Two cows were chosen for testing.

Cow X – a cow known to have the autosomal recessive disease CHs

Cow Y – a phenotypically normal cow with a family history off CHS

The CHS locus was isolated from each, amplified and then treated with FokI restriction enzyme which recognises the nucleotide sequences

5’ – G G A T G – 3’ and 5’ – C A T C C – 3’

The genotypes at the CHS locus for the two cows are shown in the following figure:


f. Explain whether cow Y is heterozygous of homozygous at the CHS locus.

g. On the electrophoretic gel diagram below, draw in the band(s) that would accurately show a profile for an unaffected cow Z with no history of CHS in the family.

A farmer suspected that one of his cows was a CHS carrier. He sent a sample of the cow’s hair follicles for testing. A technician ran a gel of DNA sequences from the hair follicles and obtained the follwing result.

h. What mistake must the technician have made in his procedures to obtain this result?

QUESTION 23 (2009)

Many techniques in molecular biology require the use of gene probes.

a. What is a gene probe?

DNA microarray technology, also known as DNA chip technology, allows screening to detect mutations. A DNA chip is made of glass and can contain thousands of fields. Each field is like a tiny well in which reactions can occur.

DNA microarray technology has been used to survey the p53 gene because a mutation of this gene is present in about 60% of all cancers. The position of a mutation in the p53 gene of a patient, patient X, who has breast cancer, can be determined.

Steps in the screening procedure are outlined below.

Step 1 Treat a normal allele of p53 to break it down into nucleotide sequences


Step 2 Each segment must be tested, one nucleotide at a time. Tests for the first segment of the allele are outlined below.

Consider nucleotide five (G) in the normal sequence. Sequences of this section are manufactured so that all possible mutants of base five are formed. Each of these sequences is placed in a different field.

Step 3 Two solutions are added to each of the fields. These solutions contain

Solution 1 complementary normal strand, labelled with green fluorescent dye

Solution II Complementary strand from DNA of person with breast cancer, labelled with red fluorescent dye

Step 4 Allow time for hybridisation of strands and then wash the DNA chip to remove excess dyes

Step 5 Examine fields under UV light to distinguish colours remaining. Interpret results.

The results for patient X are shown below:

b. What is the function of each of the two different fluorescent dyes used?

c. What does hybridisation mean?

d. What mutation resulted in patient X having breast cancer?

e. A daughter of patient X was also tested for the first segment of the allele. Would you expect the result of her test to be green, red or black? Explain your answer.

QUESTION 24 (2008)

The following diagram summarises the steps involved in the production of a cloned sheep.


The chromosomes in the cells of the cloned sheep will be identical with those in the cells of

A. Sheep M B. Sheep N C. Sheep P D. Sheep Q

Question 25 (2010)

The blue mussel, Mytilus edulis, lives along the north eastern coastline of the USA. A species of Asian shore crab, Hemigrapsus sanguineus, was accidentally introduced into the area about 15 years ago. As shown below, the Asian shore crab has only migrated to the southern half of the total area inhabited by the blue mussel.

The Asian shore crab feed off the blue mussels. The thinner the mussel shell, the easier it is for the crab to crush and eat the mussel.

In recent times, scientists have observed that the overall population of the southern blue mussel has a thicker shell than that of the northern and southern blue mussel populations.

a. Explain the process of natural selection that has occurred in the population of southern blue mussels over the last 15 years that has resulted in thicker shells.

Assume that the Asian shore crab is unable to migrate past the northern limit line into the northern blue mussel area.

b. What would you expect to happen to the shell thickness of the northern blue mussels over time? Explain your reasoning.

Question 26 (2010)

Long before the development of agricultural crops, hunter gathers in southern Africa would pick the tastiest nutty fruits of the marula tree and scatter them around their camps. These would germinate and grow into fruit-bearing trees. The best seeds would be chosen from these trees and the process would be repeated.

a. Explain how this practice is an example of selective breeding. In your answer, include the selective agent and the phenotypic characteristic being acted on.


Current domestication processed include marcotting. This involves peeling away the bark from a branch, stimulating the branch to produce roots. The branch is then cut and planted in the soil.

b. What can you infer about the genotype of trees propagated through marcotting?

c. Outline one disadvantage of a plantation of marula trees grown through marcotting compared to a natural population of marula trees.

d. Should the fruit from marcotted marula trees be labelled as genetically modified (GM). Explain why.

Question 27 (2003)

Streptococcus pneumonia is a bacterium that causes pneumonia in humans and may show resistance to antibiotics.

a. How would antibiotic resistance have first occurred in the Streptococcus pneumonia population?

The incidence of antibiotic resistant Streptococcus pneumonia has increased in the last 15 years. Approximately 40% of infections in bacterium are resistant to commonly used antibiotics.

b. Explain how the increase in bacteria resistant to antibiotics has occurred.

c. What is the selective agent associated with the increase in antibiotic resistance in Streptococcus pneumonia?

Question 28 (2007)

Eastern tiger snakes (Notechi scutatus) living on desolate islands off mainland Australia have longer jaws than the mainland populations of snakes. The diet of island snakes includes large prey, such as seagull chicks, while the diet of the mainland snakes consists of small prey, such as frogs and mice.

Researchers set up experiments using baby snakes from both locations. Snakes were fed either large or small mice over several months, until they reached maturity. The method and results are indicated in the table below.

Experiment 1 Experiment 2

Group A island snakes

Group B island snakes

Group C mainland snakes

Group D mainland snakes

Length of eastern tiger snakes jaws at birth

Long Long Normal Normal

Type of prey given over several months

Small mice Large mice Small mice Large mice

Length of eastern tiger snakes’ jaws at maturity

Normal Long Normal Normal

a. What were the researchers investigating in these experiments?

b. What was the independent variable in experiment 1?

c. What evidence from the results suggests that the size of eastern tiger snakes’ jaws is:

i. a genetically inherited trait


ii. affected by environmental factors

At present the island and mainland populations are both classified as the same species. It has been proposed that the two populations of snakes may eventually evolve into two separate species.

d. Outline the steps involved in the process of speciation, with particular reference to the snakes in the two populations. You may use labelled diagrams or flow chart to illustrate your answer.

Question 29 (2002)

a. Lack of genetic variation is believed to put a species at greater risk of extinction. Explain why low levels of variation put a species at risk of extinction.

b. Explain what is meant by the founder effect in the context of population genetics.

Question 30 (2002)

The diagram below shows the natural distribution of a mammal, the red-necked wallaby.

a. Give two reasons why populations of this species in Tasmania have not evolved

into a separate species despite being geographically isolated by the waters of Bass Strait.

b. Another mammalian species common in Tasmania is the Eastern Quoll. This species, which is about the size of a domestic cat, was widely distributed in south-eastern mainland Australia until about 50 years ago. It is now believed to be extinct in Victoria and possibly over the rest of its former range in mainland Australia.

Assuming this species is extinct on the mainland of Australia, provide two possible reasons for its extinction.

Question 31 (2010)

The islands of Hawaii in the Pacific Ocean were formed as a result of volcanic action in which small land masses were thrown up by submarine volcanoes. The youngest of the islands lies to the east of the oldest.

A similar pattern of deposition has been found across all islands, shown by the profile below:


a. What assumption is made about the formation of strata when interpreting profiles

such as this?

b. State a hypothesis to account for the disappearance of many of the bird species from the groups of islands

c. Provide evidence to support your hypothesis.

Biologists studied many species of the fruit fly, Drosophila, living on the Hawaiian islands. The species vary widely in appearance, behaviour and habitat. The diversity of Drosophila can be explained by the successive colonisation of newly formed islands by a small number of individuals ‘island hopping’ from the neighbouring westerly island. This is represented in the diagram below:

d. What name is given to this small group of colonising individuals?

e. Explain how the new and old colonies became separate species.

Question 32 (2007)

There are two varieties of lice which live on humans: body lice which only live in clothing but feed on the body, and head lice which only live in hair and feed on the scalp. It is not known when humans began wearing clothes. It is difficult to find evidence of cultural evolution in early humans as changes in behaviour are rarely reflected in physical changes visible in fossils. However, indirect evidence can be found.

A scientist used DNA hybridisation to measure differences between the DNA of head lice and body lice. He estimated that the two groups diverged about 72,000 years ago.

a. Explain how DNA hybridisation can be used to determine evolutionary relationships.

b. A scientist claimed to have found other evidence showing the time at which humans began wearing clothes. What might this evidence be?

c. Explain a possible advantage for lice of living in clothes.

Question 33 (2010)

DNA sequencing is often performed to help produce phylogenetic trees or to classify organisms. Sections of nuclear DNA from similar organisms are sequenced and compared for similarities.


a. Name and describe another DNA method that is used to determine how closely related two species are.

Scientists use a specific mitochondrial gene called cox 1 in comparative studies in fish. The cox 1 gene lacks introns and is only 654bp long, making it economical and easy to sequence. The sequenced gene appears like a barcode found on grocery products, for example:

A worldwide database has over 45,000 cox1 gene sequences obtained from different kinds of fish. Illegal fishing occurs in Australian waters. Often only a small portion, such as a fin, is kept by the fisherman and the remainder is thrown overboard.

b. How would the barcode database be used to identify the fish species that had been caught by the fisherman? Provide an example of a situation (other than legal fishing) when humans would want to identify the fish species.

Scientists use the cox 1 gene have seen that even within the same species of fish, differences within the DNA sequence can occur.

The scientists believe this is often due to the redundancy in the genetic code.

c. Explain what redundancy in the genetic code means

As the database grows it is hoped that at least five samples for each species will be collected and analysed. The species samples will be taken from as many different locations as possible.

d. Why is it important to sequence multiple samples from the same species?

Question 34 (2003)

The table below shows the number of nucleotide differences between a region of mitochondrial DNA in humans, chimpanzees and a Neanderthal.

Human 2 Chimpanzee 1 Chimpanzee 2 Neanderthal

Human 1 15 77 76 20

Human 2 79 80 27

Chimpanzee 1 23 72

Chimpanzee 2 71

Neanderthal

a. Based on the data in the table, which individual is most closely related to the Neanderthal?

The differences between the mitochondrial DNA recorded are the result of base substitutions. There are 77 nucleotide differences between Human 1 and Chimpanzee 1.

b. Explain why 77 nucleotide differences is a minimum number of base substitutions.

The Neanderthal DNA was extracted from a fossil approximately 25000 years old.

c. What other type of information obtained from the fossil could be used to assist in determining the evolutionary relationship of Neanderthals with humans and chimpanzees?

d. What method would be used to estimate the absolute age of the Neanderthal fossil?

e. What method could be used to determine the relative age of the Neanderthal fossil?


QUESTION 35 (2009)

In 1877, German workers found a slab of stone containing the fossil of an ancient bird form. The fossil was called Archaeopteryx.

a. Describe how this fossil could have been formed.

b. Scientists use information gained from sedimentary rock to arrange animal and plant fossils into some kind of evolutionary sequence over time. Explain how such sequencing is possible.

c. Name one isolation barrier involved in allopatric speciation.

d. Explain how isolation may result in speciation.

Question 36 (2003)

Trilobites are an extinct group of marine arthropods. They are very well studied group due to the abundance of fossils. Trilobites had a tough exoskeleton and bodies and legs divided into segments. They were distributed world wide and occupied a range of habitats. They existed for almost 300 million years before becoming extinct around 250 million years ago.

a. What structural feature improves the chances of a trilobite becoming a fossil?

b. Suggest one other reason why trilobite fossils are abundant?

Trilobites are thought to be closely related to three other groups of fossil arthropods; helmetids, tegopeltids and naraoids. The tegopeltids and helmetids are the two most closely related groups. These two groups are more closely related to trilobites than they are to naraoids. The diagram below illustrates the evolutionary relationships between these four groups.

c. Write the names of these four groups in the boxes at the top of the diagram below so that the evolutionary relationships between them are consistent with the information provided.


d. Suggest one reason why all species of trilobites became extinct.

Question 37 (2003)

The following photographs show two fossil hominid skulls.

a. Which of the two skulls is the more ancient fossil?

b. List one characteristic of this skull in support of your choice.

Although extensive searches have occurred, fossils classified in the genus Australopithecus and other early hominid genera have been found only in Africa. Assume that these fossils in fact only exist in Africa.

c. What is a possible explanation for these fossils being limited to Africa alone?

Question 38 (2010)

The following diagram shows two lower jaws, one from Australopithecus and the other from a gorilla (Gorilla gorilla).

a. Outline the differences between two characteristics, other than tooth number,

that enable you to distinguish which jaw is that of Australopithecus.

Molecular techniques allow detailed examinations of DNA sequences of chromosomes and the rearrangements that have taken place during evolution.

Two chromosomes of the last common ancestor that humans had with chimpanzees fused and gave rise to human chromosome number 2. This fusion is represented in the following diagram.

After fusion of the chromosomes, one of the centromeres from the fused chromosomes was inactivated.

b. Why was the inactivation of one of the centromeres a significant step in human evolution?


Question 39 (2007)

A complete skeleton of Thylacoleo carnifix, a huge marsupial ‘lion’, was found in a cave in the Nullarbor Plain. Further exploration uncovered more specimens of Thylacoleo as well as other extinct mammals, some previously unknown to science.

The specimens of Thylacoleo were dated to be between 100,000 and 200,000 years old.

In attempting to date the other mammal specimens, scientists had a range of methods available.

a. Describe a relative dating technique which could be used to date the extinct mammals.

Thylacoleo had powerful jaws with only two types of teeth – one type for killing and one type for slicing meat. It had short, strong limbs and opposable thumbs with prominent claws. It has been compared with large placental carnivores such as lions.

b. What kind of evolution has produced similar characteristics in Thylacoleo and lions.

c. Describe two methods by which Thylacoleo is likely to have obtained its food, and state the evidence that supports your answers.

Method Evidence

QUESTION 40 (2008)

Two paleoanthropologists each used fossil data to draw a model of the human evolutionary tree. The two models they produced are shown below.

a. Explain how it is possible that the paleoanthropologists produced different

models for the human evolutionary tree.


b. State one feature of agreement between the models.

c. State one feature of conflict between the models.

d. Give two structural features that would distinguish between the fossils of Homo erectus and Australopithecus afarensis.

Evidence suggests that Homo sapiens and Homo neanderthalensis were living in the same areas some 30000 years ago, but did not interbreed.

e. Give one reason why interbreeding might not have occurred.

Scientists have recently discovered a tiny fossil skull in Indonesia. It has been named Homo floresiensis (the hobbit) and dated to the time our own ancestors were colonising the world. Some scientists believe Homo floresiensis evolved from Homo erectus. Fossils of Homo erectus have also been discovered in Indonesia.

f. Modify one of the models below to include Homo floresiensis.

There is still some debate about what the hobbit is. Two explanations have been proposed. Explanation 1: the hobbit belongs to a species of small brained dwarf humans Explanation 2: the hobbit is a stone age Homo sapiens with a disease that stunts brain development.

g. Suggest one piece of evidence that would support explanation 1.

h. Suggest one piece of evidence that would support explanation 2.

Question 41 (2009) The press recently reported: ‘Anthropologists have uncovered ancient fossil footprints in Kenya dating back 1.5 million years, the oldest evidence that indicates our ancestors walked like present day humans…’

a. Give one significant feature of the footprints that would have led to anthropologists to this conclusion.

b. According to one interpretation of the hominid fossil record, Homo habilis is thought to have existed about 2 million years ago. What kinds of discoveries have been made at Homo habilis sites that have increased our understanding of the technological evolution of hominids?

‘The rate of technological evolution has increased by cultural evolution of Homo sapiens.

c. Describe on example of the effect that cultural evolution has had on the rate of technological evolution.

Cultural evolution depended on the development of physical capabilities of the Homo genus.

d. What physical feature has played the most important role in this advancement? How has this feature developed over evolutionary time?


Arguably, modern Homo sapiens has taken the manipulation and control of the environment to its highest level in history.

e. Does this mean that our species will no longer physically evolve by the mechanism of natural selection? Justify your answer.

EXPERIMENTAL DESIGN

QUESTION 42 (2009)

An experiment was set up investigate the growth of two varieties of pea plants, Palma and Vaspa. Two hundred plants of each variety were grown from seed for 10 weeks. The histograms below show the heights of the plants (measured to the nearest whole number at the end of the ten weeks.

a. Name the independent variable in this experiment.

b. Equal numbers of each type of seed were grown for the same amount of time. State one environmental variable that should be controlled in this experiment.

c. What difference in height exists between the tallest Palma and tallest Vaspa plants?

d. Is height in pea plants controlled by one or many genes? Explain your answer.

The endangered pygmy possum lives in three restricted alpine areas, Mt Buller, Bogong High Plains and Mt Kosciusko. About 2000 individuals remain in the wild. Studies show that there is a lot of genetic diversity between the three populations. Due to the isolation of these populations, scientists think that each population has a separate gene pool.

e. Explain what is meant by a gene pool. f. Explain how exchange of genetic material may be beneficial in the survival of

endangered species like the pygmy possum.

SUGGESTED ANSWERS Question 1

a. Transcription

b. The DNA template strand is read and a strand of pre-mRNA is produced by RNA polymerase. The pre-mRNA is modified to remove all of the introns and mRNA leaves the nucleus.

c. Translation

d. Ribosomes read the mRNA code. Transfer RNA anticodons attach to the mRNA codons and bring a specific amino acid to the ribosome. A protein is produced.

Question 2

a. Complementary strand - TTTCATGACGCG

b. Adenine

c. mRNA or messenger RNA

d. i. Ribosome

ii. Translation

iii. A protein (polypeptide chain)

e. Aspartic acid is replaced by glutamic acid


Question 3

a. Genes are only activated or transcribed when required. Not transcribing all genes at the same time saves energy for the cell

b. The transcription of all genes is inhibited

c. Enzyme 1 is inhibited and compound 1 is not produced

d. When tryptophan is available, its presence prevents further tryptophan being produced. Thus the cell does not waste energy or resources producing tryptophan

Question 4

a. ¼

b. Non identical twins, they have different genotypes for the given gene loci

c. Max is not able to convert phenylalanine to tyrosine/Jack can convert phenylalanine to tyrosine

Jack is albino and does not produce melanin/Max has pigmentation

Max has higher levels of phenylpyruvic acid than Jack

d. A test cross should be performed where a plant that is homozygous recessive for the phenotypes, bbgg. The cross would be BbGg X bbgg and the offspring produced would be: 1BbGg (normal stalk, green); 1 Bbgg (normal stalk, yellow); 1bbGg (brittle stalk, yellow); 1 bbgg (brittle stalk, green)

e. The cross shown above gives the results for genes located on separate chromosomes. If the offspring was not produced in the same ratio, it could be inferred that the genes were on the same chromosome (linked).

Question 5

a. Cell division occurring in bacteria/prokaryotes producing two daughter cells

b. The chromosomes are visible/it is showing anaphase of mitosis

c. A fault has occurred such as spindles break or chromosomes do not separate/cells produced are unable to function or are not required

d. mRNA travels to the ribosomes where its codons are read

tRNA carries specific amino acids to the ribosomes are complementary base pairing occurs between the codons and anticodons

product: protein or polypeptide

Question 6

a. D has half the amount of DNA compared to other stages

b. Q, S , P, R

c. Apoptosis

d. Meiosis produces gametes whereas mitosis produces somatic cells; homologous pairs of chromosomes line up during metaphase 1 of meiosis in mitosis the homologous pairs line up ‘single file’ during metaphase 1; crossing over or non-disjunction occurs in meiosis but not mitosis.

Question 7

a. 22

b. Provides greater variation in the gametes

c. The same genetic material is present, just in a different location

d. Anaphase II

e. 25%

f. For each homologous pair, there are two ways of lining up during metaphase 1. This gives four different possible combinations, one of which will have both normal chromosomes

Question 8

a. CEDBFA b. Increased variation in gametes or offspring; recombining of paternal and maternal alleles;

exchange of genetic material between homologous chromosomes c. E d. Chromosomes are double stranded in anaphase 1.


Question 9

a. Meiosis

b.

c. The sex chromosomes failed to separate at meiosis 1. (Non-disjunction)

d. 47

e. There was non-disjunction of the chromatids in stage two of meiosis.

Question 10

a. the magnitude of the exposure to the veterans, how common disorders are in the general population, comparison to other populations exposed to the H bombs, compare chromosomes of offspring and parents, prior or subsequent exposure to mutagens.

In the design of the experiment the factors that need to be considered are: select veterans from a cross section of the community (range of socio-economic groups), have a control group of males for comparison with the veterans, ages of control group should be similar to the veterans, similar living conditions between control group and veterans, similar chromosome tests, similar tests for children and grandchildren of vets and controls

b. parts of non-homologous chromosomes have been translocated and the insertion of the section of DNA may have interrupted the function of other genes.

c. Changes occurred in the gonads of the vets and are passed onto the offspring through gametes

d. Disagree because large sections of chromosomes have been translocated, and this has involved many bases, not a single base.

Question 11

a.

i. Patterned phenotype is dominant ii. 2 heterozygous parents produce three patterned to one non-patterned offspring, or two

patterned parents produce non-patterned offspring

and

offspring are produced that have a different phenotype from the parents, thus indicating that the parents are heterozygous. If the patterned trait was recessive, then no patterned offspring would be produced.

b. P = patterned, p = non-patterned

Parent genotypes Pp X Pp

Parent phenotypes patterned X patterned

Offspring genotypes ¼ PP, ½ Pp, ¼ pp

Offspring phenotypes patterned, patterned, non-patterned

c. Patterned frogs can have a homozygous genotype (PP) or heterozygous (Pp). The frogs in cross A are homozygous PP, producing all patterned offspring. The patterned frog in cross B is heterozygous.

Xor XX

Y or YY

X

X

Y

Y


Question 12

a. There are four phenotypes in the F2 generation. This suggests that two genes are involved, one determining coat colour (black or white) and the other controlling coat texture (rough or smooth). Each gene has 2 alleles.

There are four phenotypes in the F2 generation in the ratio 9:3:3:1, thus the parents are heterozygous for the two genes, each with two alleles.

b. i. Rough, black – RRBB; smooth , white – rrbb

ii. RrBb

iii. RrBB or rrBb

c. In anaphase 1, the movement of one pair of homologous chromosomes does not influence the behavious of the other member of the pair.

Question 13

a. In the pedigree, red and liver puppies are present. The dogs that are rrB_ and R_bb show that the parents are heterozygous.

II-1 is homozygous rr and II-3 is homozygous bb.

b. II-4 is RrBb

c. III-4 is rrBb

d. Parental genotypes RrBb X rrBb

RB Rb rB rb

rB RrBB RrBb rrBB rrBb

rb RrBb Rrbb rrBb rrbb

RrBB or RrBb – black (3)

Rrbb – red (1)

rrBB or rrBb - liver (3)

rrbb – lemon (1)

Question 14

g. Autosomal recessive

h. 2/3 – need to show that Molly’s parents are heterozygous

i. Eukaryotic chromosomes are linear and replicate during mitosis and meiosis; prokaryotic are circular and replicate during binary fission

j. Plasmids replicate independently of binary fission/can be used as vectors in delivering genes during genetic transformation

Question 15

a. X linked dominant because in generation II the father is affected and the trait is passed onto all of his daughters in generation III

b.

Phenotype Possible genotypes

Rabbit I-1 Impossible to tell XDXd, XdXd

Rabbit I-2 Does not have the disease XdY

Rabbit I-3 Has the disease XDXd, XDXD

Rabbit I-4 Impossible to tell XDY, XdY

c. 1 XDXd : 1 XDY : 1 XdXd: 1 XdY

1 affected female 1 affected male 1 unaffected female 1 unaffected male


Question 16

a. The repeats are in the non coding section of DNA, or introns

b. PCR

c. Family Y – there is only one difference between Y and Ben; Y has more of the five markers the same as Ben; X and Z have a higher number of differences than Y does to Ben

Question 17

a. The polymerase enzyme catalyses production of a new DNA strand. DNA polymerase replicates DNA by adding complementary bases, using the original strand as a template.

b. DNA fragments move according to their charge and their molecular weight. DNA is negatively charged and moves to the positive pole, while heavier fragments do not travel as far as lighter ones.

c. Individual 2 is homozygous. The others are heterozygous. Individual 2’s fragments are therefore of the same size, having the same number of repeats in the two fragments.

d. 5

e. Piece A has the greatest number of repeat sequences (4). The greater the number of repeats, the heavier the fragment and the smaller the distance it moves.

f. The bands for suspect 5 match the sample of blood found on the victim, but do not belong to the victim.

Question 18

a. Polymerase chain reaction (PCR)

b. Heating to approximately 90oC to separate the DNA strands by breaking the hydrogen bonds between bases

c. Show primers annealing, one per strand, at opposite ends of the gene of interest. Must include one of the following:

i. Primers attach at 3’ end

ii. Temperature reduced to 50oC

iii. DNA/Taq polymerase added

iv. Elongation of the DNA fragment away from the primer

d. Arrow points towards top of page

e. The standard consists of fragments of DNA of a known length and are used to estimate the size of the samples

f. The sample is not from the victim and could be from either suspect

g. Apply the same process to a different region (gene), use a different DNA technique such as sequencing, or use another forensic method such as fingerprinting or blood analysis

Question 19

a. Restriction enzymes cut at a particular base /recognition sequence

b. DNA is negatively charged and travels to the positive electrode; smaller DNA fragments travel further along the gel

c. DNA of different people is unique and therefore is cut at different places and produces fragments of different lengths

d. The RNA probe is complementary to the single stranded DNA sample and these join or hybridise, therefore enabling the DNA to be visualised

Question 20

a. Vector

b. Restriction enzyme

c. DNA ligase

d. The gene for antibiotic resistance is incorporated in the plasmid so that bacteria that have taken up the plasmid can be identified from those that haven’t. Only the bacteria growing on media containing antibiotics have taken up the plasmid.

e. Insecticide may be harmful to humans.

GM crop may be cheaper and easier to grow

Insecticide may kill other useful insects.


f. There may be less competition for resources from other plants, resulting in increased yield. It may also mean the farmer spends less time weeding.

g. Farmer Y may be selling his crop as non-GM, and he cannot be certain if his crops are still non-GM.

Question 21

a. Because the baby has the DS alleles 11 and 15, either of these could have come from the mother but neither could have come from the father

b. The aim of gene therapy is to insert a normal or functioning allele into the cell

c. Virus or liposome

d. The virus may be active and cause illness, or the inserted allele has disrupted the functioning of other genes

Question 22

a. The trait is controlled by many genes and is an example of polygenic inheritance

b. The chemical code is identical

c. Drqw a circle

d. The centromere is needed for spindles to attach/ to allow mitosis to occur

e. Gene cloning, DNA replication or gene replication

f. Heterozygous, has different alleles

g. Two bands – one slightly above 100 and one slightly above 50

h. The electrode had been connected the wrong way around

Question 23

a. A single stranded piece of DNA which is radioactively labelled

b. Green indicates hybridisation with a normal allele, red indicates a specific mutation and hybridisation with that allele have occurred

c. The joining of complementary DNA from different sources

d. G was replaced by A

e. Green if she inherited a normal allele from her mother, red if she inherited the defective allele from her mother

Question 24

C

Question 25

a. There was variation in the mussel population/thick shelled mussels had a selective advantage

The crabs eat more of the thin shelled mussels/crabs are a selective pressure

The thicker shelled mussels survived and reproduced, passing on the alleles for the favourable trait to the next generation

b. No change in variation and the crabs are not present

Variation occurs due to a different selection pressure of gene flow with the southern population

Question 26

a. The humans are the selective agent and the phenotype being selected is tastiness/nuttiness of the fruit

b. The genotype is identical and the plants are clones

c. The lack of variety means that if there is a new selection pressure, they may be all unresisting and become extinct

d. No – there has been no change to the plant’s DNA

Question 27

a. Mutation

b. There was variation in the population before the antibiotics were used. Some had resistance to the antibiotics, others did not. The antibiotics were introduced, and acted as a selective pressure that killed the non-resistant bacteria while those with resistance survived. The resistant bacteria reproduced and passed on their resistance allele to the next generation. Over many generations, the frequency of the resistant bacteria increased.


c. Antibiotics

Question 28

a. The effect of diet on the jaw size of the snakes

b. The size of the prey given

c. i. Mainland snakes had normal jaw length at maturity

ii. Island snakes jaw length in group A and B increased in size when large prey was eater

d. the island and mainland snakes were geographically isolated and had no gene flow. Each island population was subjected to different selection pressures. When reintroduced, the populations were unable to produce fertile offspring

Question 29

a. If there is a change in the selection pressure, individuals with the same phenotype will behave in the same way. Thus all of the population will respond in the same way to the selection pressure, thus many of the individuals may die if they are not resistant to the selection pressure.

b. The founder effect is when a small group of individuals founding a population may not have allele frequencies that are representative of the original population.

Question 30

a. The population of red necked wallabies in Tasmania may not have been isolated long enough from the populations of red necked wallabies on the mainland. Therefore, there has been insufficient time for genetic differences to accumulate.

They live in similar habitats and have similar selection pressures

b. The eastern quoll may have become extinct on the mainland due to a disease spreading through the mainland population, killing all the quolls. There may have been a predator on the mainland that was not present in Tasmania.

The habitat of the quoll may have been destroyed on the mainland.

Question 31

a. strata near the top are the more recent

strata are laid down in chronological order

b. predation by humans/dogs; habitat destruction by humans, dogs, volcanic eruptions

c. dog bones in the strata

pottery indicating human settlement

charred plant remains indicating habitat destruction by humans

charred plants remains indicating human settlement or volcanic eruption

d. founders

e. population separated with no gene flow

natural selection occurs

when brought back together again, they are unable to produce fertile offspring

Question 32

a. DNA hybridisation involves the dissociation of different samples of DNA and their reassociation, providing a measure of similarity. The greater the similarity of DNA, the more closely related the two organisms are. The higher the melting point of the hybrid DNA, the greater the similarity

b. Artefacts associated with clothes, such as bone needles, fossilised remains of clothing, cave paintings showing humans wearing clothes

c. Clothing provides a warm environment, close contact with the body, hiding places and an easy way to be transferred between hosts

Question 33

a. DNA hybridisation – DNA from two different sources is made single stranded and mixed together. The degree of bonding is determined by the melting temperature.

Mitochondrial sequencing – the DNA from different species is extracted, sequenced and compared

b. A sample is compared to the database and used to identify endangered species, identify people selling incorrectly labelled fish, or to monitor fish numbers


c. One amino acid is coded for by more than one codon

d. To ensure accuracy of results or to identify variations that occur on a species

Question 34

a. Human 1

b. Mutations are reversible. Thus some mutations may reverse an earlier mutation.

c. Skeletal structure may be used to determine the relationship between Neanderthals and humans and chimpanzees.

d. Carbon dating.

e. Stratigraphy shows the oldest layers of rock are at the bottom while younger layers are at the top.

Question 35

a. The bird was rapidly covered in sediment, hidden from scavengers, slow decomposition, undisturbed

b. Stratigraphy is where layers of sediment build up over time. The oldest fossils are found in the lowest stratum

c. Mountain range, dry ground, road, etc

d. The two separated populations have different gene pools/ genetic variation/ mutations present

Different selection pressures/environments/natural selection acts on each population

If the two populations, when brought together, do not produce fertile offspring, they are different species

Question 36

a. Tough exoskeleton

b. They were widespread

They were able to survive in many habitats

They were around for a long time (300 million years)

They lived in an area where they were buried quickly

c. Trilobites, helmetids and tegopeltids, naraoids

d. A change in the global climate

Question 37

a. Skull 1

b. Large brow ridge, more prominent jaw

Question 38

a. Jaw X is more parabolic than jaw Y. The canine teeth are larger than jaw Y. The teeth are a more uniform size than in jaw X

b. Inactivation of one centromere enables meiosis and cell replication to occur. Gametes are able to be produced.

Question 39

a. Stratigraphy, where the relative age of the fossil can be determined from its position in the rock stratum

b. Convergent evolution

c.

Method Evidence

Killed and ate fresh meat

Hunted

Killing teeth, slicing teeth, no grinding teeth for crunching old bones

Ambush, not pursue

Short strong limbs suited to grappling, not running fast or climbing

Grappled and killed prey Opposable thumbs for holding

Claws for killing or climbing


Killed by biting Long front teeth

Powerful jaws

Question 40

a. Different paleoanthropologists interpret the same data differently

b. H. afarensis evolved about three million years ago; H. heidelbergensis evolved from H. ergaster about one million years ago

c. H. erectus evolved from A. afarensis in model 1; H. erectus evolved from H. ergaster in model 2; model 2 shows linear evolution from A. afarensis

d. H. erectus had a larger brain case, less prominent brow ridge, a more parabolic jaw, position of foramen magnum more central, size of teeth reduced

e. Different lifestyles or customs prevented interbreeding, different mating behaviour and rituals, mating may not have produced fertile offspring

f. A line coming off H. erectus linking them to H. florensiensis in Model 1 or Model 2

g. If adult and child fossils of H. florensiensis were found that had skulls indicating they all had small brains

h. If fossils were found in the same area and all had normal sized skulls; if adult and child fossils of H. sapiens were found in the same area and had skulls indicating the child brain size was much larger that that of H. florensiensis

Question 41

a. Big toe is parallel to other toes, big toe is not opposable/prints indicate two feet not four, the use of hands and knuckles

b. Evidence of tools and use of fire

c. Articulate speech/writing/painting/ceremonies enabled information to be passed on

d. The brain – an increase in capacity leads to greater processing of information. Precision grip led to the availability to make tools for fine manipulation. Structures involved with speech led to communication through speech.

e. Yes – medical advances or genetic manipulation mean that modern humans are interfering in their own selection to the extent that natural selection no longer operates

No – humans still exist in many different environments and are still subjected to different selective pressures such as disease

Question 42

a. Variety of pea plant

b. Amount of sunlight/availability of water/temperature

c. 9cm

d. Many genes, continuous variation

e. The sum of alleles within a given population

f. Increase in genetic diversity/variation. The greater the variation within a species, the more likely it is able to survive a change in the environment

general terminology chromosomes gametes somatic cells dna · replication • during ... the base...

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