Question

1. In mice, one of the gene determining coat color is the A gene, which has multple alleles. two of these are an Allele for black hair (a) and allele for yellow hair (A^y₎ . (A^y) is completely dominant to a; however, for reasons that are unclear ( at least to me), A^YA^Y mice always die before embroyic development is completed.

a) two mice with yellow hair mate with each other. what are the genotype of the mice?

b) for this cross, give the expected genotype (s) and phenotype (s) for pups that survive embroyic development. A punnet square may help you to solve this part of the probelm.

2. The lubber grasshopper is very large, and as nymph is black with red and yellow stripes, Assume that individuals with the genotype RR have red stripes; that individuals with phenotype rr have yellow stripes; and that individuals with genotype Rr have both red and yellow stripes.

a) if you cross two grasshoppers that, as nymphs, had both red and yellow stripes, then give the expected genotype (s) and phenotype (s) for the offspring. A punnet square may help you to solve this part of the problem.

b) If you cross a grasshopper with both red and yellow stripes with one that has just red stripes, then give the expected genotype (s) and phenotype (s) for the offspring. Again, a punnet square may help you to solve this part of the problem.

3. The common grackle is a blackbird that is common over most of thge eastern and central united states. Suppose that, in grackles, the alleles for long tail (L) is cpmpletly dominant to the alleles for short tail (l). A female with a short tail mates with a long-tailed male, and that mating produces a brood of four chicks- of which just one chick survives, and that a short tail.

a) what is the genotype of the female parent?

b)what is the genotype of the male parent?

c) what is the genotype of the one survival chick?

d) imagine that, instead of just one chick surviving, all four survived. give the expected genotype (s) and phenotype (s) for the chicks. Apunnet square may help you to solve this part of the problem.

4. in human, tongue rolling is controlled by a single gene (the R gene); for that gene, the rolling alleles (R) is completely dominant and the non-rolling (r) is reccessive. Also in human, curly hair is controlled by a single gene (the H gene); for the gene, the curly-hair allele (H) is completely dominant and the straight-hair allele (h) is recessive.

a) there is a curly-haired non-tongue rolling man, and of the man's parents has straight hair. what is man's genotype?

b) there is a straight-haired tongue-rolling woman,and one of the woma's parents is a non-roller. what is the woman's genotype?

c) the man from part a marries the woman from part b. give the expected genotypoe (s) and phenotype (s) of their children. Be sure to use a punnet square to solve this part of the problem.

5) In the breeding season, male anole lizards court females by bobbing their heads up and down while displaying a colorful throat patch. suppose that the allele for fast head-bobbing (F) is completely dominant to the allele for slow bobbing (f), and that the allele for red throat patch (R) is completely dominant to the allele for a yellow throat patch (r). further, suppose that a female homozygous recessive for both traits mates with a male homozygous dominant for both traits.

a) for the offspring of this pairs- the F₁ generation give the expected genotype (s) and phenotype (s).

b) imagine that one of the F₁ individuals mates with an unrelated individual of identical phenotype. if this happens, then the offspring of this mating would be expected to have what genotype (s) and phenotype (s)? be sure to use a punnet square to slove this part of the problem.

Answer 1

Q1) This is an example of Lethal allele where Ay of yellow coat is a lethal gene.

a)Parent 1 = Aya

Parent 2= Aya

b)

	Ay	a
Ay	AyAy	Aya
a	Aya	aa

Genotype of Progeny that will survive = Aya and aa

Q2) This is an example of Co-dominance.

Red stripe = RR

Yellow stripe = rr

Red-Yellow stripe = Rr

	R	r
R	RR	Rr
r	Rr	rr

Genotypic ratio:1:2:1

Phenotypic ratio: 1:2:1

phenotype = Red:Red-Yellow:Red-Yellow:Yellow

Q3)Since long tail (L) is completely dominant over short tail(l) , henceforth

a) Genotype of female = ll

b) Genotype of male = Ll

	L	l
l	Ll	ll
l	Ll	ll

d) Genotypic ratio: 2:2or1:1

Genotyoe = Ll:ll:Ll:ll

Phenotypic ratio: 2:2

Phenotype = Long tail:short tail:Long tail:short tail

Q4)Dominant tongue rolling = R ; Recessive non rolling tongue = r

Dominant curly hairs = H ; Recessive straight hair = h

a) Curly hair non rolling tongue genotype =Hhrr

b) Straight hair rolling tongue genotype = hhRr

	Hr	hr
hR	HhRr	hhRr
hr	Hhrr	hhrr

Genotype: HhRr:hhRr:Hhrr:hhrr

Phenotype: Curly and rolling tongue:Straight and rolling tongue:Curly and non rolling tongue:Straight and non rolling tongue

Q5) Fast head bobbing = F ; Slow head bobbing = f ; Red throat patch = R ; Yellow throat patch = r

Female = ffrr

Male = FFRR

Male x Female = ffrr x FFRR

F1 progeny genotype = FfRr

F1 progeny phenotype = Fast head bobbing and Red throat patch

b)

	FR	Fr	fR	fr
FR	FFRR	FFRr	FfRR	FfRr
Fr	FFRr	FFrr	FfRr	Ffrr
fR	FfRR	FfRr	ffRR	ffRr
fr	FfRr	Ffrr	ffRr	ffrr

Phenotypic ratio: 9:3:3:1

Phenotype : 9 x Fast head bobbing and red throat patch:3x Fast head bobbing and yeallow patch :3 X Slow head bobbing and red patch : Slow head bobbing and yellow patch.