2. Your son is interested in obtaining a spotted floppy-eared rabbit for entry into the fair. You have a male FfSs and a female ffss, and female rabbits often produce about eight babies per litter. Figure out the ratio of poffspring for each phenotype you can expect from crossing these rabbits, so you can decide if it is likely that your son can enter the fair with a spotted floppy-eared rabbit.

A. Ratio of the offspring with:

Monocolored fur and straight ears: One-fourth, or 25%
Monocolored fur and floppy ears: One-fourth, or 25%
Spotted fur and straight ears: One-fourth, or 25%
Spotted fur and floppy ears: One-fourth, or 25%

Each of the parent rabbits will pass on one of its alleles for spots and one of its alleles for ear shape. The female rabbit can only pass on one combination of these alleles: fs. The male, however, can pass on four combinations: FS, Fs, fS, and fs. This sets up a 4x1 Punnett square. There will be four types of offspring, all equally common: FfSs, Ffss, ffSs, and ffss.

Remember that mono-colored fur (F) is dominant over spotted fur (f), and straight ears (S) is dominant over floppy (s). This means that you'd get 25% straight-eared solid-colored rabbits (FfSs), 25% floppy-eared monocolored rabbits (Ffss), 25% straight-eared spotted rabbits (ffSs), and 25% floppy-eared spotted rabbits (ffss).

B. A male rabbit is mated to a warren of spotted floppy-eared rabbits. The number of individuals and their phenotypes produced were: 30 monocolored straight-eared rabbits; 27 monocolored floppy-eared rabbits; 19 spotted straight-eared rabbits; and 27 spotted floppy-eared rabbits. Using what you know about the genotype of the mothers, decide what the genotype of the father is.

The females are all ffss. For the male to father all these kinds of baby rabbits, he must be FfSs -- this is the same cross as in 2A above. You don't get exactly 25% of each phenotype -- it's fairly close, but unless you have really huge numbers of offspring, there's usually noticeable deviation from Mendel's predictions just because of chance effects.

3. The ruffed chicken has an under-chin wattle. A smooth wattle (S) is dominant over a wrinkled wattle (s). A red wattle (H^R) is codominant with a white wattle (H^W) so that an individual with H^RH^W will have a pink wattle.

A. A chicken with a red wrinkled wattle is mated to a homozygous chicken with a smooth white wattle. What is the genotype and phenotype of the chicks?

Genotype of the parents: ssH^RH^R x SSH^WH^W
Genotype of the chicks: SsH^RH^W
Phenotype of the chicks: smooth pink wattles

4. Humans have four possible blood types (A, B, AB, and O) and these blood types are controlled by three alleles (I^A, I^B, i). The I^A and I^B alleles are codominant (they share expression, thus we have an AB blood type), but they are both dominant over the i allele.

A. List the possible genotypes from the phenotypes:

Blood Type A: I^AI^A, I^Ai
Blood Type B: I^BI^B, I^Bi
Blood Type AB: I^AI^B
Blood Type O: ii

B. A court case has been filed by a mother with type O blood who has a son with type O blood. There are two fathers being accused; one has AB blood and the other A blood. Which one of the men could be the father of the child?

The son, with type O blood, must have the genotype ii. He must have received one i from his mother -- which makes sense, because she has type O blood and thus the the genotype ii as well -- and one i from his father. The accused man with type AB blood cannot be the father, because his genotype must be I^AI^B -- he does not have an i allele to pass on to any of his children. The man with type A blood is presumably the father, since he could have the I^Ai genotype, with an i allele to pass on. (He could also have I^AI^A blood, but that would mean that he's not the father, and we're assuming that one of the suspects is the father.)

C. In another court case there are three possible fathers. The mother has type B blood, and the child has type O blood. One suspect father has type AB blood, another has type A blood (both his parents were AB), and the third father has type A blood (one parent had AB and one parent had A blood). Construct a pedigree and decide who the father is:

Again, the child has type O blood and so has the ii genotype. The mother has type B blood, but she must have given an i allele to her child, so her genotype must be I^Bi. The father must have contributed the other i allele. . .

• Father 1: Blood type AB. He must have the I^AI^B genotype. This means that he does not have an i allele to pass on to any of his children. He cannot be the father.
• Father 2: Blood type A. He could have either the I^AI^A or the I^Ai genotype. Which is it? We know that both his parents had blood type AB. That means that both his parents had the I^AI^B genotype. For Father 2 to have type A blood, he would have to have received an I^A allele from each parent. This makes him I^AI^A. He cannot be the father.
• Father 3: Blood type A. Again, he could have either the I^AI^A or the I^Ai genotype. If he has I^AI^A blood, he's off the hook -- but we are assuming that one of the three suspects is the father, and the other two cannot be the father. So is there any way for Father 3 to have the I^Ai genotype? Yes: One of his parents had type AB blood and so had the I^AI^B genotype. The other had type A blood. If that parent had the I^Ai genotype, then Father 3 could have received I^A from the parent with AB blood, and i from the parent with A blood. That would mean that Father 3 would have an i allele to pass on to the kid. Father 3 is a daddy!

5. Red-green color blindness, in humans, is a sex-linked trait controlled by alleles on the X chromosome. Normal color vision (X⁺) is dominant to colorblindness (X^c). [NOTE: The lab manual has X^C for the normal allele and X^c for the recessive colorblind allele. Unfortunately, capital C and lowercase c are hard to tell apart as superscripts, especially in type. I will use X⁺ -- "X-plus" -- for the normal allele.]

A. If a colorblind man marries a woman with normal vision and they have a colorblind son, what are the genotypes of the individuals?

Remember that men are XY and women are XX. A colorblind man must be X^cY, and his colorblind son must also be X^cY. Men pass on either an X chromosome or a Y chromosome to their children -- and when a man has a son, he passes on his Y chromosome; that's what makes his kid a son and not a daughter. So the son did not get his colorblindness from Dad. This means that Mom, who is female and therefore XX, must have the X⁺X^c genotype, since we know she's not colorblind.

B. If the mother and father were to have more children, what proportion of the girls would be colorblind? Why?

We know that the man is X^cY and the woman is X⁺X^c. You can do the Punnett square, but in brief, the answer is that half the girls would be colorblind. Why? A colorblind woman must have the X^cX^c genotype. Each girl gets one X chromosome from Dad and one from Mom. Each girl automatically gets one X^c allele from Dad (because the only other option is that she gets the Y chromosome from Dad, which would make her a boy and not a girl). There's a fifty-fifty chance that each girl will get Mom's X⁺ allele, in which case she'll have the X⁺X^c genotype and normal vision; and a fifty-fifty chance that she'll receive the X^c allele from Mom, and end up with the X^cX^c genotype -- and be colorblind.