1A. The gene for hair color in rabbits has two alleles Q and q. Q is dominant and codes for brown hair. q is recessive and codes for white hair. Write out all the possible genotypes and phenotypes.
There are three possible genotypes: QQ, Qq, qq
There are two possible phenotypes: Brown and white
1B. Using the above example, fill in the Punnett's Square of offspring genotypes if one parent is heterozygous and the other is white haired. If the pair of rabbits have a litter of 24 babies, write out the expected number of each genotype and phenotype in the table below.
heterozygous parent 

Q 
q 

whitehaired parent 
q  Qq 
qq 
q 
Qq 
qq 
Genotype 
Phenotype 
Expected Number 
Qq 
Brown 
12 
qq 
White 
12 
tall parent 

T_{1} 
T_{1} 

short parent 
T_{2} 
T_{1}T_{2} 
T_{1}T_{2} 
T_{2} 
T_{1}T_{2} 
T_{1}T_{2} 
Genotype: T_{1}T_{2}
Phenotype: Medium Height (100%)
2B. Take any two of the seedlings from part 2A and cross them. Fill in the results below.
T_{1} 
T_{2} 

T_{1} 
T_{1}T_{1} 
T_{1}T_{2} 
T_{2} 
T_{1}T_{2} 
T_{2}T_{2} 
Genotype 
Phenotype 
Expected Number 
T_{1}T_{1} 
Tall 
25% 
T_{1}T_{2} 
Medium 
50% 
T_{2}T_{2} 
Short 
25% 
3A. Sex expression in mammals (including humans) is controlled by the X and Y chromosomes. Females are XX and males are XY. Since cells of the body contain 46 chromosomes, mom must give 23 to baby and dad must give 23 as well. Mom gives 1 sex chromosome to each of her eggs along with 22 body chromosomes. Dad gives 1 sex chromosome to each of his sperm along with 22 body chromosomes. Fill in the possible sex chromosomes contributed to sperm and egg in the table below. If mom and dad have 8 kids, show the expected number of boys and girls below:
Dad 

X 
Y 

Mom  X 
XX 
XY 
X 
XX 
XY 
Genotype 
Phenotype 
Expected Number 
XX 
Female 
4/8 (50%) 
XY 
Male 
4/8 (50%) 
3B. Why do some families end up with unequal sex ratios (more boys or girls)?
With small sample sizes, you can get greater deviations from the expected probabilities.
While the population of the city is likely to have a sex ratio very close to 50:50, a particular family might not.
3C. Colorblindness is a recessive trait caused by an error on the X chromosome. X_{A}=Normal Vision and X_{a}=Colorblind. If mom is normal (not a carrier) and dad is colorblind, fill in the table below:
Dad 

X_{a} 
Y 

Mom  X_{A} 
X_{A}X_{a} 
X_{A}Y 
X_{A} 
X_{A}X_{a} 
X_{A}Y 
Genotype 
Phenotype 
Expected Number 
X_{A}X_{a} 
Female, Normal 
2/4 (50%) 
X_{A}Y_{ } 
Male, Normal 
2/4 (50%) 
None of the children will be colorblind, but the girls are carriers and can pass it down to half of their children.
3D. Let's do the same problem again, but this time with a carrier mom and normal dad. Neither parent is colorblind. X_{A}=Normal Vision and X_{a}=Colorblind.
Dad 

X_{A} 
Y 

Mom  X_{A} 
X_{A}X_{A} 
X_{A}Y 
X_{a} 
X_{A}X_{a} 
X_{a}Y 
Genotype 
Phenotype 
Expected Number 
X_{A}X_{A} 
Female, Normal 
1/2 (50%) 
X_{A}X_{a} 

X_{A}Y 
Male, Normal 
1/4 (25%) 
X_{a}Y 
Male, Colorblind 
1/4 (25%) 
The girls are all normal, but half of them are carriers. Half of the boys are normal and half are colorblind.
3E. Why are more males in the population colorblind than females?
Because colorblindness is carried on the X chromosome and males have only one X, they only need one copy of the defective X in order to be colorblind. Females have two X chromosomes, so they would need two copies of the defective X in order to be colorblind. There are only two genotypes for males (X_{A}Y and X_{a}Y) but there are three genotypes for females (X_{A}X_{A}, X_{A}X_{a}, X_{a}X_{a}). Males cannot be carriers; they either have it or they don't.
4A. In guinea pigs, two different genes affect the coat. One gene codes for coat color and there are 2 codominant alleles C_{1}=Brown and C_{2}=White. The heterozygous form is tan colored. The second gene codes for presence of hair with H=hairy (dominant) and h=hairless (recessive). If mom is C_{1}C_{2}hh and dad is brown and heterozygous for hairiness, fill in the table below.
Dad 

C_{1}H 
C_{1}h 
C_{1}H 
C_{1}h 

Mom  C_{1}h 
C_{1}C_{1}Hh 
C_{1}C_{1}hh 
C_{1}C_{1}Hh 
C_{1}C_{1}hh 
C_{1}h

C_{1}C_{1}Hh 
C_{1}C_{1}hh 
C_{1}C_{1}Hh 
C_{1}C_{1}hh 

C_{2}h 
C_{1}C_{2}Hh 
C_{1}C_{2}hh 
C_{1}C_{2}Hh 
C_{1}C_{2}hh 

C_{2}h 
C_{1}C_{2}Hh 
C_{1}C_{2}hh 
C_{1}C_{2}Hh 
C_{1}C_{2}hh 
Genotypes  phenotypes  fraction produced 

C_{1}C_{1}Hh 
Brown, Hairy 
4/16 (25%) 
C_{1}C_{2}Hh 
Tan, Hairy 
4/16 (25%) 
C_{1}C_{1}hh 
Hairless 
8/16 (50%) 
C_{1}C_{2}hh 
4B. If we didn't know the genotypes of the parents, but mom is hairless and dad is tan haired, and the babies produced included brown, tan, white, and hairless, can you guess the genotypes of the parents?
Mom must be hh because she is hairless. If there are both brown and white haired babies, she must be C_{1}C_{2}
If dad is tan, dad must be heterozygous for coat color, so he is C_{1}C_{2}
If some of the babies are hairless but dad is hairy, then he must be heterozygous for hair, Hh
Therefore dad is C_{1}C_{2}Hh