biol 1114 unit 2 notes
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BIOL 1114 Unit 2 Notes
Genetics
Lecture 13
The Central Dogma
Accessing Genetic Information Transcription; Protein Synthesis
6 seconds (60m): ATP reserves depleted
15 sec (150m): Creatine phosphate reserves depleted
Next 45 sec (200m): Fermentation pathways of ATP production
o Lactic acid buildup
o Oxygen debt
o High Altitude Training = more red blood cells/more oxygen
[7.3AB, 7.4AB]
Overview of Protein Synthesis
Amino Acids = monomers of proteins
DNA Codes for proteins
o Transcription:transfer of info from DNAto mRNA
In nucleus
o Translation: transfer of info from mRNAinto a polypeptide
In cytoplasmribosomes
DNA (transcription) -> mRNA (Translation) -> Ribosomes (protein synthesis) ->
RNA Polymerase (-ase = protein)
G1 and G2 in cell cycle
Transcription
ALWAYS read the Template Strand
RNA Polymerase reads the Template Strand
Promoter: [TATA Box], TATA Binding Protein, Transcription Factors, RNA Polymerase
Starts Transcription^
Promoter Region:
o [TATA Box]
o [TATA Binding Protein]: recognizes the TATA box and binds to it
o [Enhancers]
o [Transcription Factors] bind to the enhancers
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o [DNA bends] so the transcription factors connect to the TATA binding
proteins and RNA Polymerase
o [RNA Polymerase]: connects and starts transcription
Always add to the 3 End
DNA template 3 to 5 DNA coding 5 to 3
RNA 5 to 3
[Terminator Region]:
o End of the gene
o Stops the coding of genes
Messenger RNA Processing
Exons (A, B, C)
Introns (1, 2, 3)o Get snipped out
o Non-coding
mRNA cap copied + Poly A tail (AAA)
Splicing
o mRNA contains only Exons (A, B, C)
o mRNA cap --- 5 *Exon A, Exon B, Exon C,+ 3 --- Poly A Tail
o Directed to ribosomes
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Translation
20 Amino Acids; 4 nucleotides (ATCG , AUCG)
o 1:1 code: only 4 amino acids
o Doublet code: only 16 amino acids
o Triplet code: 64 amino acids
mRNA = 3 letter codes for codons
AUG is always the code at the start
Codes for M, or signals the beginning of a protein
UAA, UAG, UGA is always the end
Different from the terminator Signals the end of a protein; release factor connects to
dissemble elements
Peptide bonds are covalent
mRNA = codons (used to read code) AAG = Lysine
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mRNA: language translated to polypeptide language
tRNA is the translator:
o picks up amino acids
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recognizes codons on mRNA (contains anticodons)
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Many ribosomes translate mRNA
Its kinda like a library of books being read by someone
Library = all the cells DNABook = one gene
Reader = RNA Polymerase
Transcription = transposing wiring music from one key to another
Translation = reading music from a page to play music
AUC CUU GCA GUG GAA UGG GAG
I LAV EWE
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Gene Regulation
Mutations
Specialization:cells express different genes in same DNA
(How does a cell know what genes to turn on?)
Ex. Cell type; pancreas eye lens nerveGlycolysis enzyme: on on on
Crystalline gene: off on off
Insulin gene: on off off
Hemoglobin gene: off off off
Regulation of gene expression
Enzyme inductionthe lac operon(in bacteria)
E. coli, exposed to glucose, fat (donut for breakfast)
o Make enzymes that break down glucose and lipids
Repressor protein:binds to DNA and blocks transcription Lactose is not present
E. coli, exposed to lactose, cellulose (salad and milkshake)
o After lactose becomes present, it
o binds to the repressor and pulls it off of the DNA, to continue
transcription;
o enzymes that break down lactose are translated,
o lactose becomes digested, repressor protein binds back to DNA
Gene Regulation in Eukaryotes
Transcription factors are the switches that turn genes on and off
Ovaries: estrogen + progesterone (steroids, lipids made out of cholesterol)
o Lipid-soluble hormones are secreted (non-polar molecules)
Hormone passes through cell membranes and binds to interior
receptor
o Makes mRNA
mRNA Attaches to ribosome
Certain genes activated leading to production of new
proteins
o Hormones change shape by receptor proteins (transcription factors)
o
DNA bends to attach to RNA polymerase and TATA binding proteino Transcribe and translate the genes to make proteins needed to build up
the endometrium
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Mutations
Change in DNA sequence
o Substitution
o Insertion
o
Deletion Base Substitution
o Sickle-cell disease [7.22]
Normal:
DNA: CTC
mRNAGAG
Amino acid: Glutamic Acid
Sickled:
DNA: CAC
mRNA: GUG
Amino acid: Valineo Not Always noticeable (Silent)
Normal:
CTC
GAG
Glutamic Acid
Sickled:
CTT
GAA
Glutamic Acid
Insert/Delete (one or more bases) Ser Gly Leu
o Addition of a letter: Ser Pro Val Ile
o Deletion of a letter: Ser Gly Leu
TAC TAT GTA CGT CAG CTC TCA CGT TCG ACT
AUG AUA CAU GCA GUC GAG AGU GCA AGC UGA
M I HAVE SAS
M I HAVE GAS
Base insertion or deletion
o
What happens to downstream codons?
o Change the whole reading frame
o Near the promotor would change more than by the terminator
o Theyll probably change
o Usually ruins protein shape; fatal if protein is essential
o What is # of bases is a multiple of 3?
o Adds an amino acid to the original coding
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o AT have more chance of mutation than CG
Point: change in a single DNA base
Missense: changes in a codon to code for different Amino Acids
o Sickle-cell
Nonsense: changes a codon to code for stop
Frameshift: add or delete anything exceptmultiples of 3
Figure 7.23 and Table 7.2
Where do these mutations come from?
o You inherit some from your parents
o Some happen at randomin your cells
Physical or chemical mutagensdamage DNA
E.g. radiation, many chemicals, smoking, etc.
Errors in DNA replication
Errors in production of sex cells (meiosis)o Think on this: Your DNA mutates throughout your life
Which mutations can you pass on to your children
Cancer
Two major classes of cancer causing genes
o Oncogens:normally trigger cell division, (such as in would healing), but
are over-exposed in cancer cells
o Tumor suppressor genes: normally prevent a cell from dividing
Arises from mis-timed or misplaced cell division
o Or absence of normal apoptosis
Since proteins control cell cycle, and proteins are coded by DNA, DNA mutations
can lead to inappropriate cell division
10% of human genes code for transcription factors (turn genes on and off)
Meiosis [9.1-9.6]
Figure 9.5
Animal body -> Meiosis (diploid) -> gametes (eggs and sperm) haploid -> Fertilization
(diploid) -> Mitosis (diploid)
Homologous chromosomes: matched pair Both carry genes controlling the same inherited characteristics
o Example: gene for eye color may be located on this chromosome, in this
spot (might be different versions (alleles))
o Not identical; they just carry the same types of genes
o Of our 23 pairs of chromosomes:
22 pairs = autosomes
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1 pair = sex chromosomes (23
rdchromosome)
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Meiosis Stages
Before meiosis is interphase, then:
Meiosis I = reduction division
o Prophase I
Diploid Doubling of DNA
synapsis: homologs line up next to one another to form tetrads
(early)
crossing over occurs(late)
o Metaphase I
Diploid
Spindle fibers
Side by side, alignment; not single file line across equator
Independent Assortment
o
Anaphase I
Diploid
The centromere for the chromosomes are not broken
o Telophase I and cytokinesis
Diploid
Opposite poles
Nuclear envelope
Meiosis II = equational division
o Prophase II
Haploid
2 separate cells
o
Metaphase II Lined up single file
o Anaphase II
Centromeres are broken
o Telophase II and cytokinesis
Haploid
Nuclear envelope
Four nonidentical haploid daughter cells
Create variety
Comparison of meiosis & mitosisMeiosis I
Prophase
o Synapsis; can be long
o Crossing over
Metaphase I
o Homologous pairs align, side by side
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Anaphase
o Centromeres do not divide
Mitosis
Prophase
o No synapsis
o
No crossing over Metaphase
o Individual chromosomes align, single file
Anaphase
o Centromeres divie
Meiosis II is similar to mitosis: sister chromatids separate
Mitosis starts 2n ends 2n (2 identical cells)
Meiosis starts 2n ends n (4 non-identical cells)
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Gamete Formation[9.8a, 34.2b, 34.3b, 10.1]
The Mitosis Waltz
Variation
Diploid # of 4
Haploid # of 2
2 homologous pairs of chromosomes
3 pairs, 8 possible outcomes
number of combinations = 2# of pairs
Meiosis generates genetic variety:
223
= 8,388,606 (form independent assortment)
223
(sperm) x 223
(egg) = 70,368,744,177,664
(this doesnt count crossing over)
If a normal diploid cell has 8 chromosomes, then
There are 4 homologous pairs of chromosomes per pair
There are 16 chromatids per diploid cell between S and the first division of
meiosis
There are four chromosomes per cell after meiosis
Sperm cells made from this diploid cell would have 8 chromosomes
Gamete formation in Males
Chromosomes in the Nucleus
Acrosome = golgi apparatus
o Has enzymes that help burrow through cells to find the eggs
Spiral mitochondria
Tail: to swim
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Sperm are made in the testes
o Seminiferous tubules: 150 meters (place of meiosis)
The walls, stem cells that keep dividing
Spermatogenesiso 2n (diploid): spermatogonium (spermatogonia)
o Mitosis: process another 2n cell (primary
spermatocyte)
begins Meiosis (primary spermatocyte)
o Morph into secondary spermatocytes (2 cells, n =
haploid)
begins Meiosis II
o Morph into early spermatids
o Morph into late spermatids
o Morph into sperm
Production:
50,000/s
70 days
50-500 million/ejaculation
o Epididymis: collect sperm to queue
Cells in the process of metaphase I in the human male are called:
Primary Spermatocytes
If an intestinal cell in a grasshopper contains 12 chromosomes, a grasshopper secondary
oocyte would contain 6 chromosomes
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Gamete formation in females
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Oogenesis
2n = Oogonium (oogonia)
o Primordial follicle
o Primary oocyte
Before birth Mitosis to generate up to 7 million oogonia (potential eggs)
Primary oocyte arrested in prophase I
At birth, about 2 million remain
Childhood
o Primary oocyte arrested
Puberty
o Few get called up to continue
o Prophase I
~400,000
Each month a few follicles begin to grow Growing (nursing) follicle cells
Day 14 = ovulation = n = haploid
Secondary oocyte
Meiosis I complete
First polar body (nucleus growing on
the side
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Secondary oocyte arrested in
metaphase II
If not fertilized, it will stay in
metaphase II
Is fertilized, (w/ sperm cell) meiosis
II is completedo 3
rdand 4
thpolar bodies will
divide (1 fertilized egg and 3
polar bodies)
o Becomes zygote
First cell division
begins ~24 hours
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Single Gene Inheritance [10.1, 10.2A-D, 10.3AB]
Mendelian Genetics
Introduction to Probability
Any event has a probability of occurring between 0 and 1 (inclusive)
o
0 = no chanceo 1= 100% chance
o 0.5 = 50% chance
Flip a coin
Possibility of tossing heads = = 50%
Tossing heads twice = x = chance = 25%
Tossing heads thrice = x x = 1/8
Roll a 6 sided die
Roll a 3 = 1/6 chance
Roll a 3 six times = 1/6
6= 1/46656
Gregor Mendel & Pea Plants
Why garden peas?
o Many distinct varieties
o Lots of offspring, quickly
o Strict control of mating
o Possess either-or traits
o Can cross-fertilizeeasily
Flower
Male = stamens (sperm)
Female = stigma (egg)
Pollen from stamen travels to stigma and begins fertilization
o [10.2
Tracing the Inheritance of One Gene; Principle of Segregation
What Mendel did..
Started with true-breedingparents
Cross-fertilized two true breeding varieties
o Purple flowers; always breed purple flowers (RR)
o
White flowers; always breed white flowers (rr)
P generation = original parents (true-breeding)
o Purple (RR) x White (rr)
F1 Generation hybrids
o All Purple (Rr)
Purple (Rr) x Purple (Rr)
F2 Generation
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o Mostly Purple (RR, Rr) some white (rr)
F2 generation 929 pea plants
Purple 705
White 224
o About purple, white
o
3:1 ratio of purple to white flowers For each cross, Mendel found a 3:1 phenotypic ratio in the F2 generation
Mendel concluded that:
o Alternate forms of genes (alleles) account for variation
o For each character, an individual has two alleles, one from each parent
o If two alleles, differ, the dominantallele is fully expressed, while the
recessivehas no effect
Mendels Principle of Segregation
o Pairs of elementen (alleles) segregate (separate) during gamete formation
Anaphase I
Homologs separate Alleles are on the homologous chromosomes
Replication in interphase
Meiosis: segregates alleles into gametes
Fertilization: gametes combine
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(TT)
o Genotype: homozygous dominant
o Phenotype: tall
(Tt)
o
Genotype: heterozygouso Phenotype: Tall
(tt)
o Genotype: Homozygous recessive
o Phenotype: short
P generation (true-breeding parents)
Phenotype: Purple Flowers (PP) dominant
Phenotype: White Flowers (pp) recessive
P Generation: PP x pp = P (sperm) p(eggs) = Pp
F1 Generation = Pp (heterozygous) Purple Flowers
Pp x Pp
Pp (sperm) x Pp (eggs) [meiosis]
F2 Generation
o PP (purple), Pp (purple), Pp (purple), pp (white)
o 3:1 phenotype ratio
o 1:2:1 genotype ratio
Punnett Squares
F1: Pp x Pp
PP, Pp, Pp, pp
3:1 phenotype ratio (F2 generation)
1:2:1 genotype ratio (F2 generation)
P generation: PP (tall) x pp (short)
F1 generation: Pp, Pp
F2 generation: PP, Pp, Pp, pp; 3 tall : 1 short
Test CrossUsing a Punnett Square to determine unknown parental genotype
Chance is a male, liver-spotted Dalmatian
Spot color is genetically determined
Chances genotype is bb
Probability has black spots; we dont know for certain what her genotype is
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She could be heterozygous (Bb) or homozygous dominant (BB)
99 puppies
48 black, 51 brown = heterozygous (Bb); about 50% chance
Homozygous dominate
bb x BBo Bb 100% black
Heterozygous
bb x Bb
o Bb 50% black
o bb 50% brown
Could she have a litter of 8 black puppies even if she were heterozygous (50% black)?
x x x x x x x = 1/28= 1/256; slim chance, but yes.
The probability of one of her puppies having black spots is, 1(bb) x (Bb)= (50%black)
P generation: RR (homozygous dominant) x rr (homozygous recessive)
F1 generation: Rr (heterozygous) offspring
Principle of Independent Assortment
Each pair of alleles segregates independently of each other
o Metaphase 1
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Meiosis I
Metaphase I
4 possible gamete types
Tracing two genes
Solid color (HH) is dominant to points (hh)Long hair (LL) is dominant to short hair (ll)
HHLL x hhll
How to determine the gametes:
for one set of genes:
o Genotype: Bb
o Gametes: Bb
For two genes, remember independent assortment
o
Genotype: AaBbo Gametes: AB, Ab, aB, ab
4 sperm, 4 eggs = 16 punnett quare
P Generation:
HHLL x hhll
o HL x hl
F1 Generation:
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HhLl (heterozygous for both traits)
Solid color, long hair
F2 Generation (16 squares)
HL, Hl, hL, hl x HL, Hl, hL, hl
o HHLL HHLl HhLL HhLl
o
HHLl HHll HhLl Hhllo HhLL HhLl hhLL hhLl
o HhLl Hhll hhLl hhll
3:1 ratio still exists
12:4 ratio
Solid Colored, Long Haired (H,L): 9 (dominant, dominant)
Solid Colored, Short Haired (H,ll): 3 (dominant, recessive)
Points, Long Hair (hh,L): 3 (recessive, dominant)
Points, Short Hair (hh,ll): 1 (recessive, recessive)
o 9:3:3:1 (phenotypic ratio)
o
Di-hybrid cross
Tracing Inheritance of Multiple Genes; Product Rule [10.4AB]
Tracing Inheritance of Two Genes
The Product Rule:
3 genes: 8 different gametes; 64 boxes
o AaBbCc
o ABC Abc ABc AbC aBc abC aBC acb
4 genes; 16 different gametes; 256 genes
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The chance that 2 independent events will both occursuch as 2 alleles both
being inheritedequals the product of the individual chances that each event
will occur
o Note: you can use the Product Rule if you see statements like this:
Assume the genes assort independently
The genes are on different homologous pairs of chromosomes The genes are not linked
Cross 2 trihybrid plants that are tall, and have round, yellow, seeds. What is the chance
that the offspring is also a trihybrid. [Assume the genes assort independently]
TtRrYy (heterozygous) x TtRrYy (heterozygous)
TT Tt 2/4 = (Tt)
Tt tt
RR Rr 2/4 = (Rr)
Rr rr
YY Yy 2/4 = (Yy)
Yy yy
(Tt) x (Rr) x (Yy) = 1/8 TtRrYy
Cross 2 trihybrid plants that are tall, and have round, yellow, seeds. What is the chance
that the offspring will be short with wrinkled, green seeds?
(tt) x (rr) x (yy) = 1/32 ttrryy
Cross 2 trihybrid plants that are tall, and have round, yellow, seeds. What is the chance
that the offspring will be short, with round seeds and heterozygous for seed color?
(tt) x (RR or Rr) x (Yy) = 3/32 ttRrYy or ttRRYy
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Cross a fish that is heterozygous for big body (Bb), homozygous for long tail (LL), and has
green scales (rr) with a fish that is heterozygous for big body (Bb), has a short tail (ll),
and is heterozygous for red scales (Rr). Big body size, long tail, and red scales are
dominant alleles for these traits. What proportion of the offspring do you expect to be
heterozygous for all three traits?
BbLLrr x BbllRr
BB Bb 1/2
Bb bb
Ll Ll 1
Ll Ll
Rr rr 1/2
Rr rr
(Bb) x 1 (Ll) x (Rr) = Chance(25%) BbLlRr
----
BbLLrr x BbllRr
bbLLRr
(bb) x 1 (Ll or LL) x (Rr or RR) = 1/8 chance
If 80 fish produced, 10will be bbLLRr
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----
Refer to the table and cross a pea plant that is heterozygous for seed form (Rr),
homozygous recessive for seed color (yy), heterozygous for pod form (Vv), and has
yellow pods (gg) with (x) a pea plant that has wrinkled seeds (rr) and is heterozygous for
seed color (Yy), pod form (Vv), and pod color (Gg). If 100 offspring are produced,
approximately how many do you expect to show the recessive phenotype for all fourtraits? [Assume these genes assert independently]
RryyVvgg x rrYyVvGg
Rr rr
Rr rr
Yy yy
Yy yy
VV Vv
Vv vv
Gg gg
Gg gg
x x = 1/32
1/32 x 100 offspring = 3 offspring rryyvvgg
----
Complications to Mendelian Ratios [10.6AB, 10.9AB]
All of these things complicated matters:
Multiple alleles:more than 2 alternative forms of a gene;
o e.g. human blood type, rabbit coat color
Co-dominance:both alleles are equally expressed; more than one allele may be
expressed
Human blood type
o
3 alleles: IA
, IB
, i I
A: red blood cells coated with carbohydrate A
IB: red blood cells coated with carbohydrate B
I: no carbohydrate
Immune system reacts to carbohydrates not found in your own body [Fig. 29.5]
o 6 genotypes
o 4 phenotypes
o Figure 10.15
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Anti-B
Antibodies
Anti-A
Antibodies
Co-dominant
No antibodies
Universal
Recipient
Both antibodies
(Anti-A + Anti-B)
Universal Donor
Recipients blood in test tube: Type A
Anti-B Antibodies
o
Donor has Type Bo The Anti-Bs attack the Bs from the donors blood
o Clumping= bad
If type a gives to type b, would there be clumping? Yes. Bad
A to AB. No. Good
B to AB. No. Good
O to AB. No Good
B to O. Yes. Bad
A to O. Yes Bad
AB to O. Yes. Bad
AB to A Yes. Bad AB to B. Yes. Bad
O to A. No. Good
O to B. No. Good
Couple 1: O (IAIB) x A (I
AIAor I
Ai) = Type A (I
AIB)
Couple 2: B (IBIBor I
Bi) x AB (I
AIB) = Type AB (I
AI
Aor I
Ai)
Couple 3: B (I
BIBor I
Bi) x B (I
BIBor I
Bi)= Type O (ii)
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Epistasis:one gene masks anothers phenotype
o Gene 1 codes for protein 1
o Gene 2 codes for protein 2
If protein 1 is not present, gene 1 and gene 2 cannot be present
H Gene
Incomplete Dominance:
o Snapdragon flowers: red flowers, white flowers, pink flowers
P generation: red and white r1r1 x r2r2
F1: pink r1r2 x r1r2
F2: 1 red (r1r1), 2 pink (r1r2), 1 white (r2r2)
Lethal alleles:causes development to stop before birth
o Chihuahuas:
HH lethal (stop development)
Hh hairless Hh hairy
Hh x Hh
o HH, Hh, Hh, hh
o Phenotype ratio: 2 hairless : 1hairy
o Genotypic ratio: 2 Hh: 1 hh
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Polygenic Inheritance:skin pigmentation/coloration
o More Dominant genes = darker color
Environmental Effects: Siamese Cats
o Dark points on Siamese cats occur on the extremities of the body
Feet, ears, tail
o Effected by the temperature of the cells that dictate the pigment of the
skin at the time of creation
Chromosomes & Gene Linkage [10.1, 10.5AB]
How to solve genetics problems (p 226-227) + emailed problems
Chromosomal Basis of Inheritance
Linked Genes
Genetic Maps Based on Crossover Data
Thomas Hunt Morgan(early 1900s) Drosophila melanogaster (fruit fly)
Morgan associated a specific gene with a specific chromosome
Fruitfly has 8 chromosomes, 4 pairs; male has hooked chromosome
Prophase 1, Metaphase 1 [independent assortment], (diploid 2n=4)
If genes are on the same homologous chromosome
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Linked Genes:
Located close together on the same chromosome
Tend to be inherited together
o If genes are on the same chromosome, only 2 allele combinations are
expected in the gametes, and the number of phenotypes is reduced
b+b vg
+vg (Female) x bb vgvg (Male)
BbVv (Female) x bbvv (Male)
b: refers to black body (mutant body type)b
+: dominant allele (gray body
b: recessive allele (black body)
vg: refers to vestigial wings (mutant type)
vg+: dominant allele (normal wings)
vg: recessive allele (vestigial wings)
BbVv bbVv Bbvv bbvv
965 185 206 944
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So these genes are linkedon the same chromosome
If the genes are lined, why do all the other combinations occur at all?
Crossing over, Prophase 1
S phase: replication of DNA in chromosomes
Prophase 1: BV BVbv bv
End of Meiosis 1: B
End of Meiosis 2: BV -- BvbV -- bv
Genetic Maps Based on Crossover Data
Recombinants / Total = Recombination Frequency
Total: 965 + 185 + 206 + 944 = 2300
Recombinants: 185 + 206 = 391
391 / 2300 = .17 = 17% = Recombination Frequency
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A.H. Sturtevant (Morgans student)
The probability of crossing over between two genes is directly proportional to
the distance between them
Genes are farther apart, genes are more likely to crossover
o Distance between linked genes determines how often they will be split
up Map unit = 1% recombination frequency
Loci/Locus (address) Recombination Frequency Approximate Map Units
b vg 17% 17 (18.5)
cn b 9.0% 9.0
cn vg 9.5% 9.5
32 DE
18 +22
40/100 = 40% (DE)
23 AD
2+3
5/50 = 10% (AD)
36 AE
15+15
30/100 = 30% (AE)
E A D
Or
D A E
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Sex-linked Traits [10.7A-C, 10.8]
Assume genes A and B are linked and are 40 map units apart. An individual
heterozygous (AaBb) at both loci is crossed with an individual who is homozygous
recessive and (aabb) at both loci. If 100 offspring are produced, how many would be
(aabb)?
A40B
40% AB = 20 + 20 for recombinants
100-40=60 /2 = 30 for parentals
AaBb aaBb Aabb aabb
30 20 20 30
Chromosomes and Sex Determination
Mammals:o Male is heterogametic (XY)
o Female is homogametic (XX)
Birds and many other animals: female is heterogametic
Extremely diverse
SRY specifies maleness in humansproduces protein that directs development
of male structures
In humans, female by default until this SRY gene is activated
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X-linked:genes transmitted on X chromosomeo Red-green colorblindness; males more likely to have it
Y-linked:genes transmitted on Y chromosomeextremely rare in humans
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X Inactivation
Inactivation of one X chromosome (randomly) per nucleus in female humans Barr Body
o All subsequent cells will have the same X chromosomes inactivated
In which cats did X inactivation happen earliestin development?
Large patchesor small patches?
A genetic counselor is asked for a probability that the male offspring of a husband and
wife will be color-blind. The woman has normal vision, but her father is color blind, the
man also has normal vision, but his father is color-blind as well. (note: color-blindness is
sex-linked and recessive)
Half the male offspring will be color-blind, and half will have normal color vision
XX (carrier) * XY
XX XY
XX XY
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Review
Transcription: in the nucleus
Translation: in the cytoplasm
Introns are spliced from sections of mRNA before exiting the nucleus
Which of the following forms hydrogen bonds with the anticodon?
MRNA
In translation. mRNA contains the codon
How many of the following terms apply to events that take place during the processes
of transcription and translation in eukaryotes before the mRNA transcript leaves the
nucleus
Release factor attaches to stop codon TAT binding protein binds to promoter
5 cap added
Peptide bond formed
Small rRNA subunits joins mRNA
tRNA binds opposite codon
Introns removed
Transcription factors bind to enhancer
Which of the following early-in-the-strand DNA template strand substitutions would
have the worst results?
CTT -> CTC
GAA (E) -> GAG (E)
GGA -> GGC
CCU (P) -> CCG (P)
ATG -> ATT
UAC (Y) -> UAA (stop codon)
Given that the haploid number (n) for this species is 2 and the diploid number (2n) is 4
Aligned single file as whole (diploid) chromosomes
Metaphase of mitosis
If an intestinal cell in a grasshopper contains 24 chromosomes, a grasshopper secondary
spermatocyte would contain 12 chromosomes (meiosis II; haploid chromosomes: 24 /2
= 12 chromosomes)
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Meiosis occurs in the walls of the seminiferous tubules of male mammals
What is the chance of flipping 5 coins at the same time and getting the result of heads
on each of them?
1/32 = (1/25)
If a man and a woman have four children (none of them twins), the chance that all four
will be boys is 1 in 16.
1/24
= 1/16 True
Assume free ear lobes are dominant to attached ear lobes. A woman with attached ear
lobes mates with a heterozygous male. Their first two offspring have free ear lobes.
What is the probability that their third offspring will have attached ear lboes
P1: Rr x rr
F1: Rr =50%
rr =50%
A heterozygous individual is more likely to pass on a dominant allele than a recessive
allele
False = equal chance Rr
A female fruitfly that is heterozygous for wing type is mated with a male fruit fly that is
homozygous recessive for wing type as shown by the following drawing
Which of the following sets of numbers would be most likely for the numner of
genotypes represented by their offpring?
Vv x vv
432 Vv + 441 vv
50/50 chance
Assume the allele for long pinkie finger is dominant to short pinkie fingers. If a man who
is heterozygous (Ll) for this allele has a child with a woman who is homozygous recessive
(ll) for this same allele, what are the possible genotypes of this offspring
Ll x ll
Ll ll Ll ll
To perform a test crossfor independently assorting traits, you mate an organism with
an unknown genotype with an organism that is homozygous recessive
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What is the expected ratio of the gametes from the following genotype: AABb (assume
that these genes assort independently)
AABb
1 AB : 1 Ab
In a species known as squeaks white ears are dominant to red ears and short hair is
dominant to long hair. You want to breed a red-eared, long-haired female with a white-
eared, shorthaired male. The female and male are homozygous for both traits. These
genes assort independently. If the mating produces 16 offspring, how many will you
expect to have red ears and long hair?
0
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How many different kinds of gametes would you expect from an individual with the
following genotype? AABbCcDdee
1 (AA) x (Bb) x (Cc) x (Dd) x 1 (ee) = 8
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Cross a true breeding female that is homozygous dominant for both characters with a
true breeding male that is homozygous recessive for both traits. Which of the following
represent the genotypes and phenotypes of their offspring (F1)
P1: BBWW (black eyes, wide gape) x bbww (orange eyes, narrow gape)
F1: BbWw black eyes and wide gape
Now cross two of the F1 individuals is 1000 individuals are produced in the F2
generation, approximately how many do you expect to have black eyes and a narrow
gape?9:3:3:1 ratio
black eyes, narrow gape = 3/16 x 1000 = 187.5
3/16 x 1000 = 188
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A couple has three children, all of whome have brown eyes and blond hair. Both parents
are homozygous for brown eyes (BB), but one is a blond (rr) and the other is a redhead
(Rr). What is the probability that their next child will be a brown-eyed redhead
BBrr x BBRr
BBrr (50) BBRr (50)
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Cross 2 trihybrid plants that are tall and have round yellow seeds (TtRrYy). What is the
chance that the offspring will be tall, with wrinkled, yellow seeds
(TT,Tt) x (rr) x (YY,Yy) = 9/64
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ABO blood types in humans are determined by a single gene having three alleles, IA, IB,
and i. A mother with blood type A has a child with blood type O. What is the mothers
genotype? IAi
Mom: IAi
Dad: i?Baby: ii
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Baby Making couples
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Recombination frequency mapping
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Genes D and F are on the same chromosome. The Dross DdFf x ddff yields 1000
offspring. 150 of which are genotype ddFf. About how many map units apart of D and F
150 x 2 = 300
300/1000 = 30%