Key for First Exam, 1996

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[10 points] Some human populations show a very slight sex bias, with probability 0.53 of a child being a boy and 0.47 of being a girl. Assume these probabilities in your answers for this question.
[5 points] Suppose individuals with genotypes AA and Aa show the "dentist" phenotype, while aa individuals show the "chiropractor" phenotype. When an individual showing the unfortunate "dentist" phenotype is crossed to a "chiropractor" individual, all of their five offspring show the dentist phenotype. What is the probability of this if the "dentist" parent is really Aa?
[5 points] Recall that squirrel monkeys have trichromatic females when both copies of the X chromosome contain different color vision alleles (i.e., heteroyzgous females are trichromatic). Suppose there are three such alleles, with frequencies 1/4, 1/4, and 1/2. What is the probability of a trichromatic female?
[25 Points] 2000 Gametes from an AaBbCc individual are scored, with the following numbers seen:

ABC 0 aBC 28

ABc 17 aBc 945

AbC 955 abC 23

Abc 32 abc 0
[5 Points] Why is it the case that human males are more likely to be deficient in either red or green color vision rather than blue color vision?
Red & Green genes are on the X chromosomes (and hence mutants are expressed in all males), blue on the autosomes.
[5 Points] Contrast chromosomal sex determination in flies with that in humans.
Flies: X-autosome ratio (X:A = 1:1 = females, X:A = 1:2 = male)
humans: presence/absence of Y chromosome.
[5 Points] Give two different mechanisms (and an example of a species in which each occurs) in which dosage compensation is achieved.
Humans -- inactivate all but one X
Flies: Double leveles of gene expression on single X in males
[5 Points] Define both Penetrance and expressivity .
Penetrance: Fraction of a genotype that show the disease (or trait)
Expressivity: Extent to which a trait (disease) shows variablity of expression when present
[5 Points] Looking at paired chromosomes in meiosis, how would one distinguish between an individuals heterozygous for an inversion and an individual heterozygous for a translocation? You can use a drawing, if you wish.
Inversions: appear as chromosome loops
Translocations: appear cruciform structures involving mutliple chromosomes

[20 Points] Consider the following three coat-color genes, which have the following phenotypes: D- = full color; dd = dilute color; A-B- = Gray, A-bb = Black, aaB- = brown, aabb = cinnamon. D is unlinked to A and B, while A and B are linked at recombination frequency c=0.2. What are the expected frequencies of the eight possible coat phenotypes in the offspring of a cross of ABD/abd to abd/abd ?
Here, the second parent only gives abd gametes
For the first parent, consider A-B first. The parental gametes AB, ab each have frequency (1/2)(1-c) = 0.8/2 = 0.4, while the recombinant gametes Ab, Ab each have frequency (c/2) = 0.2/2 = 0.1. Since D assorts independently, each gamete contains D or d with frequency (1/2). Putting all these together gives

ABD (0.4)(1/2) = 0.2 Full color gray

ABd (0.4)(1/2) = 0.2 dilute gray

AbD (0.1)(1/2) = 0.05 full color black

Abd (0.1)(1/2) = 0.05 dilute black

aBD (0.1)(1/2) = 0.05 brown

aBd (0.1)(1/2) = 0.05 dilute brown

abD (0.4)(1/2) = 0.2 cinnamon

abd (0.4)(1/2) = 0.2 dilute cinnamon

[5 Points] The following 3 (unordered) octads types are seen on an AaBb parent:

Type 1 (AB, AB, AB, AB, ab, ab, ab, ab) Frequency = 56%

Type 2 2 (AB, AB, ab, ab, Ab, Ab, aB, aB) Frequency = 26%

Type 3 2 (Ab, Ab, Ab, Ab, aB, aB, aB, aB) Frequency = 18%

What is the A-B recombination frequency?
c = Freq(NPD) + (1/2) Freq(T) = .18 + (1/2)(0.26) = 0.31

[5 Points] In a study to estimate mutation rate, 1000 flasks, each containing 10⁷ cells are examined. If 125 of the flasks are mutant-free, what is the estimated mutation rate?
u = [ - ln (125/1000) ] / 10⁷ = 2.08 x 10 ^-7

Key for First Exam, 1996

0.470.470.53 = 0.117

There are three different birth orders to consider: BGG, GBG, GGB, each of which has prob. 0.117, hence 3*0.117 = 0.35

(1/2)⁵ = 1/32 = 0.03

Label these alleles 1, 2, and 3. Can compute this as either
1- Prob(1 hom) - Prob(2 hom) - Prob (3 hom) = 1 - (1/4)² - (1/4)² - (1/2)² = 5/8 = 0.625
or as Prob(12 het) + Prob (13 het) + Prob (23 het) = 2(1/4)(1/4) + 2(1/4)(1/2) + 2(1/4)(1/2) = 5/8 = 0.625

It is not, as there is complete interference, while Haldane's functions assumes no interference.

Red & Green genes are on the X chromosomes (and hence mutants are expressed in all males), blue on the autosomes.

Flies: X-autosome ratio (X:A = 1:1 = females, X:A = 1:2 = male)
humans: presence/absence of Y chromosome.

Humans -- inactivate all but one X
Flies: Double leveles of gene expression on single X in males

Penetrance: Fraction of a genotype that show the disease (or trait)
Expressivity: Extent to which a trait (disease) shows variablity of expression when present

Inversions: appear as chromosome loops
Translocations: appear cruciform structures involving mutliple chromosomes

c = Freq(NPD) + (1/2) Freq(T) = .18 + (1/2)(0.26) = 0.31

u = [ - ln (125/1000) ] / 10⁷ = 2.08 x 10 ^-7

ABC	0	aBC	28
ABc	17	aBc	945
AbC	955	abC	23
Abc	32	abc	0

ABD	(0.4)(1/2) = 0.2	Full color gray
ABd	(0.4)(1/2) = 0.2	dilute gray
AbD	(0.1)(1/2) = 0.05	full color black
Abd	(0.1)(1/2) = 0.05	dilute black
aBD	(0.1)(1/2) = 0.05	brown
aBd	(0.1)(1/2) = 0.05	dilute brown
abD	(0.4)(1/2) = 0.2	cinnamon
abd	(0.4)(1/2) = 0.2	dilute cinnamon

Type 1	(AB, AB, AB, AB, ab, ab, ab, ab)	Frequency = 56%
Type 2	2 (AB, AB, ab, ab, Ab, Ab, aB, aB)	Frequency = 26%
Type 3	2 (Ab, Ab, Ab, Ab, aB, aB, aB, aB)	Frequency = 18%

Key for First Exam, 1996

0.47*0.47*0.53 = 0.117

There are three different birth orders to consider: BGG, GBG, GGB, each of which has prob. 0.117, hence 3*0.117 = 0.35

(1/2)5 = 1/32 = 0.03

Label these alleles 1, 2, and 3. Can compute this as either 1- Prob(1 hom) - Prob(2 hom) - Prob (3 hom) = 1 - (1/4)2 - (1/4)2 - (1/2)2 = 5/8 = 0.625 or as Prob(12 het) + Prob (13 het) + Prob (23 het) = 2(1/4)(1/4) + 2(1/4)(1/2) + 2(1/4)(1/2) = 5/8 = 0.625

It is not, as there is complete interference, while Haldane's functions assumes no interference.

Red & Green genes are on the X chromosomes (and hence mutants are expressed in all males), blue on the autosomes.

Flies: X-autosome ratio (X:A = 1:1 = females, X:A = 1:2 = male) humans: presence/absence of Y chromosome.

Humans -- inactivate all but one X Flies: Double leveles of gene expression on single X in males

Penetrance: Fraction of a genotype that show the disease (or trait) Expressivity: Extent to which a trait (disease) shows variablity of expression when present

Inversions: appear as chromosome loops Translocations: appear cruciform structures involving mutliple chromosomes

c = Freq(NPD) + (1/2) Freq(T) = .18 + (1/2)(0.26) = 0.31

u = [ - ln (125/1000) ] / 107 = 2.08 x 10 -7

0.470.470.53 = 0.117

(1/2)⁵ = 1/32 = 0.03

Label these alleles 1, 2, and 3. Can compute this as either
1- Prob(1 hom) - Prob(2 hom) - Prob (3 hom) = 1 - (1/4)² - (1/4)² - (1/2)² = 5/8 = 0.625
or as Prob(12 het) + Prob (13 het) + Prob (23 het) = 2(1/4)(1/4) + 2(1/4)(1/2) + 2(1/4)(1/2) = 5/8 = 0.625

Flies: X-autosome ratio (X:A = 1:1 = females, X:A = 1:2 = male)
humans: presence/absence of Y chromosome.

Humans -- inactivate all but one X
Flies: Double leveles of gene expression on single X in males

Penetrance: Fraction of a genotype that show the disease (or trait)
Expressivity: Extent to which a trait (disease) shows variablity of expression when present

Inversions: appear as chromosome loops
Translocations: appear cruciform structures involving mutliple chromosomes

u = [ - ln (125/1000) ] / 10⁷ = 2.08 x 10 ^-7