How to do genetics problems 2

Chapter 14 Problems:
1. A rooster with gray feathers is mated with a hen of the same phenotype. Among their offspring 15 chicks are gray, 6 are black, and 8 are white. What is the simplest explanation for the inheritance of these colors in chickens? What offspring would you predict from the mating of a gray rooster and a black hen?

There are three phenotypes, one being intermediate to the other two. Right away suspect incomplete dominance.
Assume gray is the heterozygote since it has a phenotype between the black and the white rooster.
Assign designations for your genotypes:

F^BF^B - black
F^WF^W - white
F^BF^W - gray

Now answer part two. Set up a mating of a gray rooster (F^BF^W) and a black hen (F^BF^B)

	F^B	F^W
F^B	F^BF^B	F^BF^W
F^B	F^BF^B	F^BF^W

There would be equal numbers of black and gray offspring.

4. A black guinea pig crossed with an albino guinea pig produced 12 black offspring. When the albino was crossed with a second black one, 7 blacks and 5 albinos were obtained. What is the best explanation for this genetic situation? Write genotypes for the parents, gametes, and offspring.

Black appears dominant to albino

B - black (both BB and Bb genotypes)
b - albino (bb genotype)

This is a problem showing that 1 phenotype (black color) can be the result of two different genotypes (BB and Bb). These two crosses would be the equivalent of test crosses to determine genotype.
A black guinea pig crossed with an albino guinea pig produced 12 black offspring. This is the result (all black offspring) you would expect when breeding two true breeders (Two homozygotes BB x bb).

	B	B
b	Bb	Bb
b	Bb	Bb

When the albino was crossed with a second black one, 7 blacks and 5 albinos were obtained. This looks like the results you would expect between a heterozygote and an albino (Bb x bb).

	B	b
b	Bb	bb
b	Bb	bb

5. In sesame plants, the one-pod condition (P) is dominant to the three-pod condition (p), and normal leaf (L) is dominant to wrinkled leaf (l). Pod type and leaf type are inherited independently. Determine the genotypes for the two parents for all possible matings producing the following offspring:

P - one pod (PP or Pp) L - normal leaf (LL or Ll)
p - three-pod (pp) l - wrinkled leaf (ll)

318 one-pod normal, 98 one-pod wrinkled.

If one parent is homozygous dominant for the one-pod trait (PP), then mating with any of the other possible genotypes for pod number will give all one-pod phenotypes (PP, Pp, or pp).
This appears to be a 3:1 phenotype for the leaf shape locus. This phenotype is the result of two heterozygotes mating (Ll x Ll)
PPLl x PPLl, PpLl or ppLl

323 three-pod normal, 106 three-pod wrinkled

All offspring are three-pod, and it is a recessive trait. Therefore, both parents must be homozygous recessive for this gene (pp x pp).
This appears to be a 3:1 phenotype for the leaf shape locus. This phenotype suggests a cross between two heterozygotes (Ll x Ll)
ppLl x ppLl

401 one-pod normal

If one parent is homozygous dominant for the pod size gene (PP) and the leaf shape (LL), then mating with any of the other 9 possible genotypes will result in all one-pod normal.

150 one-pod normal, 147 one-pod wrinkled, 51 three-pod normal, 48 three-pod wrinkled

This is a 3:1 phenotype for pod size, suggesting a cross between two heterozygotes (Pp x Pp)
It is a 1:1 phenotype for leaf shape. This suggests one homozygous recessive and one heterozygote (Ll x ll).
PpLl x Ppll

223 one-pod normal, 72 one-pod wrinkled, 76 three-pod normal, 27 three-pod wrinkled.

This is a 3:1 phenotype for pod size, suggesting a cross between two heterozygotes (Pp x Pp)
This is a 3:1 phenotype for leaf shape, suggesting a cross between two heterozygotes (Ll x Ll)
PpLl x PpLl

6. A man with group A blood marries a woman with group B blood. Their child has group O blood. What are the genotypes of these individuals? What other genotypes, and in what frequencies, would you expect in offspring from this marriage?

Type O blood is the result of two recessive alleles (ii). A child needs to receive one recessive allele from each parent. Therefore both parents must be heterozygotes.

man I^Ai He would make two types of gametes I^Aandi
woman I^Bi She would make two types of gametes I^Bandi

Now set up a Punnett Square to determine possible offspring.

	I^A	i
I^B	I^AI^B	I^Bi
i	I^Ai	ii

They can have children with any blood type, in equal proportions.

1/4 I^AI^B - Phenotype AB
1/4 I^Bi - Phenotype B
1/4 I^Ai - Phenotype A
1/4 ii - Phenotype O

8. Phenylketonuria (PKU) is an inherited disease caused by a recessive allele. If a woman and her husband are both carriers, what is the probability of each of the following?

all three of their children will be of normal phenotype.
one or more of the three children will have the disease.
all three children will have the disease.
at least one child will be phenotypically normal.

If they are both carriers, then PKU must be a recessive inherited disease and the parents must be heterozygotes (Pp).
Pp x Pp

	P	p
P	PP	Pp
p	Pp	pp

The genotypes PP and Pp would be phenotypically normal (3 out of 4). The genotype pp would express the phenotype for the disease PKU (1 out of 4).
The question asks what is the probability of each of the following?

all three of their children will be of normal phenotype

The genotype PP and Pp would present a normal phenotype. There is a 3/4 chance that each children would be normal.
The chance of three children being normal would be the product of each individual birth.
3/4 x 3/4 x 3/4 = 27/64

one or more of the three children will have the disease.

The probabilities of all possible outcomes must add up to 1.
In the above problem, we answered that the probability of all three of their children having a normal phenotype was 27/64.
Therefore, the probability that one or more of the three children will have the disease would be 1-27/64 = 37/64.
27/64 + 37/64 = 1

all three children will have the disease.

There is a 1/4 chance each birth that a child would have the disease.
The chance of three children being affected would be the product of each individual birth.
1/4 x 1/4 x 1/4 = 1/64

at least one child will be phenotypically normal

The probabilities of all possible outcomes must add up to 1.
In the above problem, we answered that the probability of all three of their children having the disease was 1/64.
Therefore, the probability that one or more of the three children will be normal would be 1-1/64 = 63/64.
1/64 + 63/64 = 1

14. In tigers, a recessive allele causes an absence of fur pigmentation (a "white tiger") and a cross-eyed condition. If two phenotypically normal tigers that are heterozygous at this locus are mated, what percentage of their offspring will be cross-eyed? What percentage will be white?

We are told that the tigers are heterozygous at the locus for fur pigmentation which also controls the cross-eyed condition. When one gene controls more than one phenotypic effect it is called pleiotropy. Since neither of the tigers are cross-eyed or white, this condition must be recessive. Therefore their genotype must be Ww.

	W	w
W	WW	Ww
w	Ww	ww

Since these phenotypes are displayed in the homozygous recessive individual, 1 out of 4 will be white and cross-eyed. The white tigers will also be cross-eyed due to the pleiotropic effects of this gene.