BIO 340 Problem Set #1 ANSWER KEY

1. a) Using the same convention that we have used in class, let’s use a letter corresponding to the recessive phenotype, and use lower case for the recessive allele. So, S = long, and s = short. Many other schemes would be acceptable.

b) The long-stem allele, S, is dominant because the heterozygotes (F1) all had long stems.

c) This looks like a classic monohybrid inheritance problem. Note that the phenotypic ratio in the F2 is 2.84 long : 1 short. Therefore, it looks like ONE gene controls this trait difference.

d) True breeding parents are SS (long) and ss (short). The F1 is all Ss (long). The expected F2 genotype ratio is 1/4 SS : 2/4 Ss : 1/4 ss; and the expected F2 phenotype ratio is 3/4 long : 1/4 short.

e) The expected number of short F2 plants is (0.25)(1064) = 266.

f) I would feel fairly confident about the model. The observed results differ from expectation by only 11 plants in 1064, or about 1%.

g) The two copies of each gene that are carried by a diploid individual separate (or segregate) during gamete formation, so that each gamete receives only one copy (i.e., is haploid).

h) All of the short-stem F2 plants (1/4 of the total F2) would be homozygous ss according to our model. They would produce only short-tem F3 progeny. 1/4 of the F2 would be expected to be homozygous SS. They would produce only long stem F3. 1/2 of the F2 plants would be expected to be heterozygous Ss. They would be expected to produce 3/4 long stem F3 progeny and 1/4 short stem. So, putting all of this together, we would expect :

{(1/4) (1)} + {(1/2) (1/4)} = 3/8 short-stem F3; where 1/4 in the first curly brackets refers to the proportion of homozygous ss F2 plants and 1 is the proportion of their offspring that will have short stems, and 1/2 in the second curly brackets is the proportion of heterozygous F2 plants and 1/4 is the propostion of their progeny that is expected to have short stems.

{(1/4) (1)} + {(1/2) (3/4)} = 5/8 long stem F3; where 1/4 in the first curly brackets refers to the proportion of homozygous SS F2 plants and 1 is the proportion of their offspring that will have long stems, and 1/2 in the second curly brackets is the proportion of heterozygous F2 plants and 3/4 is the propostion of their progeny that is expected to have long stems.

2. a) Recessive because two normal parents had a PKU child. If it were dominant, then at least one parent would have had to have PKU in order to pass on the allele to the child.

b) Both parents are heterozygous.

c) Best evidence would be to examine a large number of families with at least one PKU child but normal parents. Then if we saw that about 1/4 of the children in such families had PKU, it would confirm the idea that PKU is a genetic disorder due to a recessive allele.

3. a) Yes, because both Jenny and Mark could be heterozygous type B (I^Bi)

b) No, it wouldn’t. Jenny could still be heterozygous type B if her type A parent was also heterozygous (I^Ai).

4. a) Two types: cT and ct.

b) Considering just kernel color, cc x CC -> all Cc. Considering just height: Tt x TT -> 1/2 TT and 1/2 Tt. Assuming independent assortment, and putting both traits together, we get the following genotypic expectation: 1/2 CcTT and 1/2 CcTt. Therefore, there are two genotypes (CcTT and CcTt) in the progeny.

c) From (b), the expected phenotypic ratio is 1/2 purple, tall (CcTT) and 1/2 purple, medium (CcTt). So, there are two phenotypes: (1) purple, tall; and (2) purple, medium.

d) During gamete formation, the alleles of one gene segregate into gametes without regard to the alleles of other genes.

5. a) and b) The simplest explanation would seem to be that there are two genes. One gene controls stippled vs. non-stippled, the other gene controls patternless vs. twin-spot (with crescent being a heterozygous phenotype). In the cross between stippled, patternless and non-stippled twin-spot, all the (F1) progeny were stippled, crescent. Therefore, it seems like stippled is dominant to non-stippled and patternless and twin-spot exhibit incomplete or partial dominance (resulting in the crescent phenotype for heterozygotes). This is borne out by the cross of the F1 fish. Considering just stippled vs. non-stippled, we would expect 3 stippled : 1 non-stippled in the F2. In fact, the ratio is 147 stippled : 48 non-stippled, or 3.06 : 1. Considering the spotting pattern, we would predict the following F2 phenotypic ratio: 1 patternless : 2 crescent : 1 twin-spot. In fact, the ratio is 47 patternless : 110 crescent : 38 twin-spot, or 1.24 : 2.89 : 1. The observed ratio for spotting pattern doesn’t look too great. But let’s go one step farther and calculate the expected numbers of each F2 phenotype (given that there were 195 F2 individuals).

F2 phenotype Expected Proportion Expected Number Observed Number

Crescent, stippled (1/2) (3/4) = 3/8 (3/8) 195 = 73.125 83

Crescent, non-stippled (1/2) (1/4) = 1/8 (1/8) 195 = 24.375 27

Twin-spot, stippled (1/4) (3/4) = 3/16 (3/16) 195 = 36.563 29

Twin-spot, non-stippled (1/4) (1/4) = 1/16 (1/16) 195 = 12.19 9

Patternless, stippled (1/4) (3.4) = 3/16 (3/16) 195 = 36.563 35

Patternless, non-stippled (1/4) (1/4) = 1/16 (1/16) 195 = 12.19 12

Overall, our model looks fairly good given that the experiment size (195 F2 individuals) is not very large.

c) No chance, because crescent is a heterozygous phenotype. Therefore, it cannot breed true. Crosses between crescent individuals will always yield about 1/2 non-crescent offspring.