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Name: __________________________ Bio AP

Mendelian Genetics & Chromosomal Inheritance

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ESSENTIAL KNOWLEDGE 3.A.3: The chromosomal basis of inheritance provides an understanding of the pattern of passage (transmission) of genes from parent to offspring (Ex: Sickle cell anemia, Tay Sachs disease, Huntington’s disease, X-linked color blindness, Trisomy 21/Downs syndrome, Klinefelter’s syndrome, reproductive issues) 4.C.2: Environmental factors influence the expression of the genotype in an organism. (ex. Height and weight in humans, flower color based on soil pH, Density of plant hairs as a function of herbivory, Darker fur in cooler regions of the body in certain mammals species, Alterations in timing of flowering due to climate changes, presence of the opposite mating type on pheromone production in yeast and other fungi.) 4.C.4: The diversity of a species within an ecosystem may influence tha stability of the ecosystem, 3.A.4: The inheritance pattern of many traits cannot be explained by simple Mendelian genetics. (ex. Sex-linked genes, Mammal Y chromosome is small as does not carry many genes, Mammal females as XX and males are XY so sex linked recessive traits are always expressed in males. Some traits are sex limited and expression depends on the sex of the individual, such as milk production in female mammals and pattern baldness in males)

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I. Mendelian Genetics A. Background

History Genetics Cytology 1860’s: Mendel proposed that discrete inherited factors segregate and assort independently during gamete formation

1875: Cytologists worked out the process of mitosis

1890: Cytologists worked out the process of meiosis

1900: Three botanists independently rediscovered Mendel’s principles of segregation and independent assortment (Correns, de Vries and von Seysenegg)

1902: Cytology and genetics converged as Walter Sutton and others noticed parallels between the behavior of Mendel’s factors and the behavior of chromosomes.

• Chromosomes and genes are paired in diploid cells • Homologous chromosomes separate and allele pairs segregate

during • restores the paired condition for both

chromosomes and genes

Gregor Mendel-Garden Experiments a. Worked with pea plants because they

had a large number of different traits b. Control the mating because male and

female reproductive structures are in the flower

c. Developed true breeding plants through self-pollination

d. Mated plants with different traits through cross-pollination (hybridization)

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e. Examined the results of his experiments

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Vocabulary a. Parental generation (P1) - parents b. First filial generation (F1)- first generation offspring (filial - son) c. Second filial generation (F2)- second generation offspring

(grandchildren) d. Character-heritable feature (ex. Pea shape, flower color) e. -physical appearance f. -genetic make-up (alleles determining a trait) g. Each individual has determining a trait h. (pure) -individual will have two of the same

alleles for the trait i. (hybrid) -individual will have two different

alleles for the trait B. Mendel’s Laws

Law of Dominance

Law of Segregation

When one allele for a trait can the expression of another allele for the same trait • allele-

stronger allele which is expressed

• allele-weaker allele which is masked

from each other in the formation of gametes

Modern genetics-this is explained in the process of

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3. Law of Independent Assortment-two or more characters in the formation of gametes Recall-modern genetics

C. Mendel’s Crosses & Punnett Square

Basic Concepts a. Symbols/Terms

Key –letters are used to represent alleles. • Dominant allele is represented by an letter • Recessive allele is represented by the form

of the same letter Plant Trait Dominant

Trait Recessive

Trait Homozygous Plant (pure)

Heterozygous Plant (hybrid)

Stem Length Tall (T)

Short (t)

Pure tall (TT) Pure short (tt)

Hybrid tall (Tt)

Seed Color Yellow (Y)

Green (y)

Pure yellow (YY) Pure green (yy)

Hybrid yellow (Yy)

Seed Shape Round (R)

Wrinkled (r)

Pure round (RR) Pure Wrinkled (rr)

Hybrid round (Rr)

Flower Position

Axial (A)

Terminal (a)

Pure axial (AA) Pure terminal (aa)

Hybrid axial (Aa)

b. Genotype/Phenotype Genotype Phenotype

Homozygous Dominant ( )

Hybrid ( )

Homozygous Recessive ( )

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c. Punnett Square-is a way of showing all the possible allele combinations in the offspring of a particular cross.

Key: P=purple flower P=white flower Parental cross: Homozygous purple X White PP X pp P P p p

F1 Results: Phenotype: Genotype:

F1 cross: Hybrid purple X Hybrid purple Pp X Pp P p P p

F2 Results: Phenotype: Genotype:

Punnett

Square

P

unnett S

quare

Steps for using Punnett Square 1. State the Key 2. Write the cross and show the gametes 3. Draw the Punnet square and place the egg alleles on one side of the box

and the sperm allele on the other side of the box 4. Fill in the box representing all the alleles combinations possible in the

offspring 5. Write the results, showing the phenotype and genotype ratios

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2. Mendelian Inheritance (crosses) a. (monohybrid)

Problems: 1) Mendel crossed pure pea plants that produced round seeds, with pure breeding pea plants that produced wrinkled seeds. In the F1 all plants were round. He then crossed the F1 plants. Determine the expected genotypic and phenotypic ratios in the F2 generation.

Key: Parental cross: F1 cross: Pure round X Pure wrinkled Hybrid round X Hybrid round

F1 Results: F2 Results: Phenotype: Phenotype: Genotype: Genotype:

2) When pea plants with axial flowers are crossed with pea plants having terminal flowers, all of the offspring have axial flowers. What were the genotypes of the parent plants?

Punnett

Square

Punnett

Square

Key

Cross: Axial X Terminal

All offspring must have an allele to be axial. The unknown allele must be

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b.

How can you determine if a purple flowered plant is pure or hybrid? • Used to determine the

genotype of an unknown individual Ex. Organisms exhibiting the dominant phenotype can be

• Cross the unknown organism with a and examine the offspring

• The appearance of individual trait will determine that the unknown was

Problems: 1) How can a breeder determine whether a tall pea plant is pure or hybrid?

Unknown Tall plant X plant

If the unknown is pure If the unknown is hybrid (TT) then (Tt) Then

Results: Results: Phenotype: Phenotype:

T

T

t

t

Tt Tt

Tt Tt

T

t

t t

Tt Tt

tt tt

Key T = tall t= short

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Let’s Practice: 1. In watermelons, solid green color is dominant over the stripped pattern and short shape is dominant over long shape. Using the Punnett square method, show the offspring of (1)a cross of a pure green watermelon and a pure stripped watermelon and (2) a cross between a hybrid short melon and a long melon . (a)

(b)

2. In fruit flies, gray body is dominant over ebony body. Two gray fruit flies are mated and produce 186 gray bodied and 58 ebony bodied offspring. Determine the parental genotype.

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3. In humans, albinism is recessive to normal skin pigmentation. If an albino female marries a heterozygous male, what percentage of their offspring would most likely be normal?

4. In guinea pigs, black fur is dominant over white fur. How can a

breeder determine whether a black guinea pig is homozygous or heterozygous black?

5. In fruit flies, dumpy wings are shorter and broader than normal wings. The allele for normal wings (D) is dominant to the allele for dumpy wings (d). Two normal-winged flies were mated and produced 300 normal-winged flies and 100 dumpy-winged flies. The parents were probably (1) DD and DD (2) DD and Dd (3) Dd and Dd (4) Dd and dd (5) dd and dd

6. Suppose that in sheep, a dominant allele (B) produces black hair and a recessive allele produced white hair. If you saw a black sheep, you would be able to identify (1) its phenotype for hair color (2) its genotype for hair color (3) the genotypes for only one of its parents (4) the genotypes for both of its parents (5) the phenotypes for both of its parents.

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c. Inheritance of genes on chromosomes (dihybrid cross) • Genes on separate chromosomes will and follow

the law of

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• Punnett Square needs to be expanded to account for the formation of more gamete combinations and more combinations in the offspring

Cross: Parent Generation Yellow Round X Green Wrinkled F1 Generation 100% hybrid Yellow Round F2 Generation Results:

Key Y = Yellow y = green R=round r=wrinkled

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Lets Practice: 1) Determine the results of a cross between two hybrid tall-yellow pea plants. Results: 2) In humans brown eyes is determined by a dominant allele B and blue eyes is determined by a recessive allele b; free earlobes are determined by a dominant allele F and attached earlobes by a recessive allele f. A brown-eyed man with attached earlobes (whose mother was blue-eyed) marries a blue-eyed woman with free ear-lobes (whose father had attached earlobes. What phenotypes may be expected among their children?

Key Cross:

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3) In cocker spaniels, black color is dominant over red color, and solid color is dominant over white spotting. A solid red male was mated to a black and white spotted female. They had five puppies: one black, one red, one black and white, and two red and white. What were the genotypes of the parents?

d. Rule of Probability The Law of Segregation and the Law of Independent Assortment are “chance” events and will therefore follow rules of probability;

• The chance that alleles from an egg and alleles from a sperm will join together is a zygote is the product of their individual probabilities

Ex. The chance that an egg from a Pp mother will receive a “p” is The chance that a sperm from a Pp father will receive a “p” is The chance that the ova and sperm will fertilize is (Rule of Multiplication)

Problem: 1) What is the probability that a dihybrid cross between two plants with the genotypes TtYy and TtYy will produce offspring with the genotype Ttyy? Since the traits are on separate chromosomes and are inherited independently, set up two crosses and multiply the probabilities.

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Cross 1 Cross 2 Tt X Tt Yy X Yy Chance of Tt is Chance of yy is Chance of the offspring having both combinations is

2) What is the probability that a trihybrid cross between two organisms with the genotypes AaBbCc and AaBbCc will produce an offspring with the genotype aabbcc?

3) What fraction of the offspring of parents each with the genotypes

Kk Ll Mm will be kk LL Mm?

T

t

T t

TT Tt

Tt tt Y

y

Y y

YY Yy

Yy yy

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Complete Dominance (A is dominant)

Codominance (no dominance)

Incomplete Dominance (A is incompletely dominant)

AA and Aa have the phenotype

Aa =Intermediate phenotype (Phenotype is the two homozygotes)

Aa = Both alleles are in the phenotype

Summary of Mendelian Genetics:

II. Extending Mendelian Genetics

A. Incomplete Dominance Cross: Japanese Four O’clock Flowers Key: =red =white P1 Red Flower X White Flower x F1 100% Pink Flowers ( ) Pink Flowers X Pink Flower CRCW x CRCW

B. Codominance

Cross Ratio Monohybrid 3 expressing dominant phenotype:1 expressing recessive

phenotype Dihybrid 9 expressing BOTH dominant phenotypes

3 expressing one dominant and one recessive phenotype 3 expressing one recessive and one dominant phenotype 1 expressing BOTH recessive phenotypes

Test Cross Used to determine an unknown genotype by mating with the PURE recessive

CWCW

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Codominance- the inheritance is characterized by the full in the heterozygote

• MN blood Group the locus codes for the production of surface glycoprotein on red blood cells.

• The MN blood type produces both surface glycoproteins. alleles are fully expressed in the heterozygote

C. Multiple Alleles

• alleles determine a phenotype • Any one individual will possess only of the multiple alleles • Key will use a letter and the allele will be expressed as a superscript

Example: Human Blood Type ( A blood, B blood, AB blood, O blood) There are THREE alleles for Human blood type: Key: =Dominant allele = Recessive allele

Blood Type Possible Genotypes Protein Antigen A

glycoprotein

B

glycoprotein

AB

and glycoprotein

O glycoprotein

Genotype Blood Type MM M (M glycoprotein) NN N (N glycoprotein) MN MN (M & N glycoprotein)

= A antigen (protein) = B antigen (protein) = no protein

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• Blood Testing:

• Blood Transfusions:

√ Universal Donor: √ Universal Recipient:

Let’s Practice: 1) Determine all the possible genotypes of the children of a mother

with type O blood and a father with type AB blood.

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2) Show how two parents who do not have blood type O can produce a child with type O blood.

3) Mrs. Frank and Mrs. Jones both had babies the same day in the same hospital. Mrs. Frank took home her son Bobby and Mrs. Jones took home her son Gene. Mrs. Jones began to suspect that her son had been switched. Blood tests were made: Mr. Frank was type A, Mrs. Frank was type B, Mr. Jones was type A, and Mrs. Jones was type A. Bobby had type O blood and Gene has type B blood. Has a mix-up occurred? Explain.

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D. Pleiotropy • One gene influences

• Example: Sickle cell anemia

√ N=normal red blood cell √ n=sickle red blood cell

E. Gene Interactions: More than is determining a trait

Epistasis •

the phenotypic expression of another GENE

• is the stronger gene

• Example Coat Color in Labrador Retrieves

√ Epistatic gene determines the production of pigment

√ Second gene determines the color of the pigment

Key E=pigment deposited in hair e=no pigment in hair B= black pigment b = brown pigment A 9:3:3:1 ration was

NOT obtained

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Collaboration • Two genes interact to form a phenotype that neither gene can

express alone • Example: Comb Type in chickens

Polygenic Inheritance (Quantitative Characters)

• effect of two or more genes on a single phenotype character

• Example: √ Three genes

determining pigment production

√ Each dominant allele gives one “unit” of skin pigmentation

√ Note the bell shaped curve

Single

Pea

Rose

Walnut

Key R= rose r=single P=pea p=single = walnut

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Gene • Gene which can alter the expression

of another gene • Example: Human eye color

√ Gene -Pigment Production B=brown b=blue

√ Modifier Gene- Pigment distribution √ Modifier Gene-Pigment Tone

• Beagle Spots

F. Environmental Influence

Degree to which the can influence the expression of a gene

Examples: Color in a Himalayan rabbit

Experiment

Expression of the black pigmentation was influenced by Color of hydrangea flowers

Humans: Nature verses

Nurture Sun tanning & skin color Nutrition & height Note: The expressions of many traits are -involving genetic and environmental factors

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III. Pedigree Charts

• Pedigree =family tree that diagrams the relationships among parents and children across generations and that shows the inheritance pattern of a particular phenotypic characteristic.

• Symbols √ Square= √ Circle= √ Horizontal line connecting male and female= has

occurred; are listed below in birth order √ symbols indicate individuals showing the

trait being traced • Tracing Pedigrees

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• Example-Determine all possible phenotypes Key: E=Attached ears e=unattached ears

SUMMARY:

Parent 1= 2= F1 1= 2= 3= 4= F2 1= 2= 3= 4= 5= 6= 7= 8= 9= 10= 11= 12= F3

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IV. Chromosomal Basis of Inheritance

A. Background • Gene map of Drosophila melanogaster (alias-the

Fruit Fly)

• Genes can be on the same chromosome

• Linked genes will during Independent Assortment

• Linked genes can only be separated by

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B. Sex Linked Inheritance 1. Morgan studies:

a. Why did Morgan choose the Fruit Fly? • They are easy to culture in the lab • They are prolific breeders • They have a short generation time • They have only four pairs of chromosomes which

are easily seen under the microscope.

b. A Special Cross

2. Sex Chromosomes

All F1 offspring have eyes

F2 offspring have a 3:1 ratio BUT ALL white eyed flies are

Is eye color linked to the sex chromosome?

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3. Sex Linked Inheritance • Traits that are linked (coded for) on the . • Human sex linked traits

√ √ √ √ √

Sex Determination Male Female Human Birds XX XY Bees XO XX

4. Sex Linked Traits

• Shape-the X chromosome is than the Y chromosome

• Males have “two bottoms” but only one “top” • Sex linked trait are those that have their on

the “top” of the X chromosomes. Males will have only allele for the trait.(hemizygous)

• Males need only recessive allele at that locus to express the trait.

• Hybrid female is a “ ”. She will not express the trait but can pass it on to her male offspring

5. Human Barr Body • Females have “ doses” of alleles on the X chromosome • “ ” is achieved by inactivating one of the X

chromosomes • The inactive X is

and is attached to the nuclear membrane

• Determine the number of Barr bodies in the following:

Genotype Sex Number of Barr Bodies XX XY XO XXX XXY

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• Inactivation of the X chromosome is random and occurs independently in embryonic cells giving rise to a (having clusters of cells derived from the active X inherited from the father and the active X inherited from the mother. Note: traits are associated with the Y chromosome

Barr Body

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6. Sex Linked Inheritance Key: XN= normal allele Xn=recessive allele Y Example: Determine the phenotypes of a cross between a normal male and female carrier for color blindness

KEY: 25% normal female

25% carrier female 25% normal male 25% colorblind male

Let’s Practice:

1. Determine all the possible genotypes of the children of a mother who is a carrier for hemophilia and a normal father.

XNXN=Normal Female XNXn=carrier female XNY=normal male XnY=male expressing the recessive allele

XN=normal female Xn=colorblind Y= male

Normal Male X Carrier Female XNY X XNXn

XN

Y

XN Xn

XNXN

XNXn

XNY

XnY

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2. Determine the percentage of colorblind females which could result from the cross between a colorblind male and a carrier

C. Linked Gene Inheritance 1. History

• Punnett and Baleson Key: P=purple p=white

L=long L=short

Cross: Parent Generation Homozygous purple long X Homozygous white short PPLL ppll F1 Generation PpLl (purple long) 100% hybrid purple long PpLl X PpLl F2 Generation Results: Expected (9:3:3:1)

PL

Pl

PL Pl

PPLL PPLl

PPLl PPll

PpLL Ppll

PpLl Ppll

PpLL PpLl

PpLl Ppll

ppLL ppLl

ppLl ppll

pL pl

pL

pl

A 9:3:3:1 ratio was obtained

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To try to explain the results they performed a test cross Hybrid purple long X Homozygous white short

Results: Expected (4:4:4:4) Actual: 7:1:1:7 The majority looked Purple long and white short

Explanation:

√ The gene for color and the gene for length are on the same chromosome.

√ They will follow the Law of Independent Assortment. √ They can only be separated by

√ The recombinant allele combinations do not occur as frequently

D. Recombination Frequency

• The greater the distance between linked genes, the the chance (percentage) that crossover will occur.

• Percent Recombination=

• One percent crossover or will equal one

pl

pl

PL Pl

PpLp Ppll

PpLp

Ppll

ppLl ppll

ppLl

ppll

PpLp

Ppll

PpLp

Ppll

ppLl

ppll

ppLl

ppll

pL pl

pl

pl

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E. Chromosome Mapping

• The percent recombination can lead to on a chromosome • One map unit= 1% recombination frequency • Procedure

√ Establish the relative distances between those genes farthest apart or with the highest recombination frequency

√ Determine the recombination frequency between the third gene and the first. Consider all possibilities

√ Determine the recombination frequency between the third gene and the second.

√ Eliminate the incorrect sequence Example: (1) b--------------------------------v

Loci Recombination Frequency

Map Units

b v 17.0% 17 c b 9.0% 9 c v 9.5% 9.5

The genes are apart

Possibilities: c---------b---------------------v

or b---------c-----------v

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(2) c--------------- b (3) c----------------v Let’s Practice 1. A tomato breeder wants to determine if the genes for plant height and plant color are on the same chromosome. The allele for tall (T) is dominant over dwarf (t), and the allele for green (G) is dominant over light green(g). A homozygous tall green plant was crossed with a homozygous recessive plant and the F1 progeny were then test crossed. The test cross results are summarized in the table below. Determine if the genes are linked, and if so, how many map units apart are they? Phenotype Number of test crossed progeny Tall-green 434 Dwarf-light

green 466

Tall-light green

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Dwarf-green

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2. Three genes have the following percent recombination. Determine the order of the genes. A-B=20% B-C=10% A-C=30 %

Correct Sequence: b---------c-----------v

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NOTES: