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 any one of four combinations: FS, Fs, fS, and fs. This sets up a 4x4 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 whole flock of spotted floppy-eared rabbits. The number of individuals and their phenotypes produced were: 30 monocolored straight-eared rabbits; 27 monocolored floppy-eared rabbits; 29 spotted straight-eared rabbits; and 31 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 are usually noticeable deviations from Mendel's predictions, just because of chance.

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 incompletely dominant with a white wattle (H^W) so that an individual with H^RH^W will have a pink wattle.

A. A chicken with a wrinkled red 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. In turkeys, bronze body color (B) is dominant over red (b). Normal feathers (N) are dominant over hairy feathers (n). A bronze male turkey is mated to a bronze female, and some of the poults (baby turkeys) produced by this cross are red. What are the genotypes of the male and female parents and the red poults?

Ignore the hairy vs. normal feathers thing for a moment. In this cross, some of the baby turkeys are red. Since bronze is dominant to red, this means that the red poults must have the genotype bb. This in turn means that the red poults must have received a b allele from Mom and a b allele from Dad. But we know that Mom and Dad Turkey both have bronze feathers, so they must have either the BB or the Bb genotype. The only way that both parents can be bronze, but both can pass the red allele on to some of their babies, is for both parents to have the Bb genotype.

5. Some of the poults from the cross described in #1 above are bronze. What proportion of the poults would you expect to have the BB genotype? What proportion would have the Bb genotype? What proportion would have the bb genotype?

Do the Punnett square for a cross between a Bb parent and another Bb parent -- you'll get 1/4, or 25% of your offspring, having the BB genotype; 1/4, or 25%, with the bb genotype; and 1/2, or 50%, with the Bb genotype.

6. If you were to cross a Bbnn turkey with a bbNn turkey, what proportion of the offspring will have red, normal feathers? What proportion will have bronze hairy feathers? red, hairy feathers? bronze, normal feathers?

The Bbnn turkey passes on one of each pair of alleles to each of his kids. He (let's say it's a he, although it's not important here) can only pass on the n allele for hairy feathers, but he also passes on either the B or b allele. So he can pass on either the combination Bn or the combination bn. The other parent, by the same reasoning, can pass on either bN or bn. So your Punnett square should look like this:

             Bn             bn
        ----------------------------
       |              |             |
       |    BbNn      |    bbNn     |
bN     |              |             |
       |--------------+-------------|
       |              |             |
bn     |    Bbnn      |    bbnn     |
       |              |             |
       |----------------------------|

You'll end up with one-quarter of the offspring with bronze, normal feathers; one-quarter with red, normal feathers; one-quarter with bronze, hairy feathers; and one-quarter with red, hairy feathers.

7. A breeder of hamsters crosses a golden hamster with a black hamster. All the baby hamsters are golden. Which trait is dominant and which is recessive? What are the genotypes of the parents? What are the genotypes of the babies?

Golden fur is dominant to black fur. Let's write the golden fur allele as G and the black fur allele as g. Thus the golden parent must be GG and the black parent must be gg, which means that all the offspring must have the genotype Gg.

8. The breeder takes one of the babies, raises it to maturity, and then mates it to a black hamster. Use a Punnett square to predict the results of this crossbreeding. Predict the genotype(s) and phenotype(s) of the offspring, and the expected ratios.

The baby hamster has the genotype Gg, and the black hamster it's mated with must have the genotype gg. So the Punnett square looks like this:

              G             g
        ----------------------------
       |              |             |
       |      Gg      |     gg      |
 g     |              |             |
       |--------------+-------------|
       |              |             |
 g     |      Gg      |     gg      |
       |              |             |
        ----------------------------

One-half of the offspring will have the Gg genotype and will have golden fur. One-half of the offspring will have the gg genotype and will have black fur.

9. In hamsters, long fur is recessive and short fur is dominant. Suppose the breeder takes a hamster that is heterozygous for both golden fur and short fur, and mates it to a hamster with long black fur. Use a Punnett square to predict the results of this crossbreeding. Predict the genotype(s) and phenotype(s) of the offspring, and the expected ratios.

OK, let's represent the allele for short fur as S and the allele for long fur as s. The hamster with long black fur must have the genotype ggss (any other genotype and he'd have golden and/or short hair). The other parent I told you is heterozygous, which makes her have the genotype GgSs. Now: The longhaired black hamster can only pass on the allele combination gs. The heterozygous parent, however, can pass on any one of four possible combinations: GS, Gs, gS, and gs. That sets up the Punnett square, which should look more or less like this:

             GS            Gs             gS             gs
        ---------------------------------------------------------
       |              |             |              |             |
gs     |    GgSs      |    Ggss     |    ggSs      |    ggss     |
       |              |             |              |             |
       |--------------+-------------+--------------+-------------|
       |              |             |              |             |
gs     |    GgSs      |    Ggss     |    ggSs      |    ggss     |
       |              |             |              |             |
       |--------------+-------------+--------------+-------------|
       |              |             |              |             |
gs     |    GgSs      |    Ggss     |    ggSs      |    ggss     |
       |              |             |              |             |
       |--------------+-------------+--------------+-------------|
       |              |             |              |             |
gs     |    GgSs      |    Ggss     |    ggSs      |    ggss     |
       |              |             |              |             |
        ---------------------------------------------------------

So that's. . . one-quarter of the offspring will have the GgSs genotype and have short gold fur; one-quarter will have Ggss and have long gold fur; one-quarter will have ggSs and have short black fur; and one-quarter will have ggss and have long black fur.

10. 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!

11. 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: You could write this as 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.

12. Budgies (Melanopsittacus undulatus) are attractive birds that come in many colors. Budgies may produce either or both of two types of pigment in their feathers: a blue pigment and a yellow pigment. Whether or not each pigment is produced is governed by one gene with two alleles. Blue pigment (B) is dominant to no blue pigment (b), and yellow pigment (Y) is dominant to no yellow pigment (y). A budgie that produces neither blue nor yellow pigment in its feathers looks white. A budgie that produces both blue and yellow pigment in its feathers looks green.

A. What is/are the possible genotype(s) of a yellow budgie?

Either bbYY or bbYy. Just YY or Yy isn't correct; you have to take both genes into account. (A YY budgie could be either yellow or green, depending on whether or not it also carries B alleles.)

B. What is the phenotype of a budgie with the genotype BbYy?

It's green.

C. Suppose you cross a homozygous blue budgie with a homozygous white budgie. Predict the results of this cross, both genotype and phenotype.

Both budgies are homozygous -- meaning that the two genes in each pair are both of the same allele. A blue budgie also can't produce any yellow pigment -- if it did, it wouldn't be blue, but green. So the blue parent must have the genotype BByy. The white parent, which produces no blue and no yellow, must be bbyy. The blue parent can only pass on the allele combination By to its offspring, the white parent can only pass on the combination by, and the result is that all the bouncing baby budgies will have the genotype Bbyy, and all will have blue feathers.

D. Now suppose you cross one of the offspring from the mating in part C above with a budgie whose genotype is BbYy. Predict the results of your cross. What are the expected genotypes and phenotypes of the offspring from this cross, and in what ratios?

OK, you're crossing a Bbyy budgie with a BbYy budgie. The Bbyy budgie may pass on either of two combinations of alleles to its offspring: either By or by. However, the BbYy budgie may pass on any one of four allele combinations to each of its babies: BY, By, bY, or by. So the Punnett square looks like this:

             By            By             by             by
        ---------------------------------------------------------
       |              |             |              |             |
BY     |    BBYy      |    BBYy     |    BbYy      |    BbYy     |
       |    green     |    green    |    green     |    green    |
       |--------------+-------------+--------------+-------------|
       |              |             |              |             |
By     |    BByy      |    BByy     |    Bbyy      |    Bbyy     |
       |    blue      |    blue     |    blue      |    blue     |
       |--------------+-------------+--------------+-------------|
       |              |             |              |             |
bY     |    BbYy      |    BbYy     |    bbYy      |    bbYy     |
       |    green     |    green    |    yellow    |    yellow   |
       |--------------+-------------+--------------+-------------|
       |              |             |              |             |
by     |    Bbyy      |    Bbyy     |    bbyy      |    bbyy     |
       |    blue      |    blue     |    white     |    white    |
        ---------------------------------------------------------

In other words, out of all the offspring from the cross, three-eights of the babies would be green, three-eighths would be blue, one-eighth would be yellow, and one-eighth would be white.