Answers to Problem Set 1 |
1. This is a case of incomplete dominance shown by both the 1:2:1 ratio and the presence of an intermediate phenotype that is displayed in neither parent.
B = black Bw = black and white splashed
The black chicken is homozygous for the black color allele (B B) and the slate blue chicken is heterozygous (B Bw). In a cross of these two birds we would expect to see a 1:1 ration of black and slate blue birds.
| B | Bw | |
| B | BB Black |
BBw Slate Blue |
| B | BB Black |
BBw Slate Blue |
2a. Four phenotypes are observed in the F2 progeny and the ration of 860:320:330:90 is very close to 9:3:3:1 which would mean this is a dihybrid cross. Both parents are heterozygous for 2 different genes with black dominant to green and yellow spots dominant to red spots.
B = black; b = green; Y = yellow spots; y = red spots
b.
| P | BByy | X | bbYY | ||
| F1 | BbYy | ||||
| F2 | B-Y- 9 : Black with yellow spots: | B-yy 3 : black with red spots: | bbY- 3 : green with yellow spots: | bbyy 1 green with red spots: |
This is just one option for naming the alleles; any letter system is ok as long as it shows the dominance structure.
c.
Black body with yellow spots = BBYY, BBYy, BbYY, and BbYy
Black body with red spots = BByy and Bbyy
Green body with yellow spots = bbYY and bbYy
Green body with red spots = bbyy
3. If the dog is a pure breed it will be homozygous for both traits dark (DD) and short hair (SS). Here are some examples of crosses that could be performed to test this:
(dark = DD and Dd, albino = dd, short hair = SS and Ss, long hair = ss)
a. A test cross: The dog in question is crossed with a dog that is albino and long haired therefore homozygous recessive at both alleles (ddss). If the progeny are all dark and short haired, then you can assume that the dog is homozygous at both genes. If the progeny contain any long haired or albino dogs then the dog being tested is heterozygous in that gene:
This cross will require more progeny to be examined in order to establish the ratios; there are many more opportunities to get dominant phenotypes that are uninformative.
4. One explanation is that there are 3 genes that determine horn formation and a dominant allele of each is needed for a horn. Homozygous recessive in 1, 2 or all 3 genes will give the hornless phenotype (37/64). All others that have at least one dominant horned allele in each of the 3 genes will have a horn (27/64).
| 3/4 A- x3/4 B- x3/4 C- | = 27/64 | A-B-C- |
| 3/4 A- x1/4 bb x1/4 C- | = 9/64 | A-bbC- |
| 3/4 A- x1/4 bb x1/4 cc | = 3/64 | A-bbcc |
| 3/4 A- x3/4 B- x1/4 cc | = 9/64 | A-B-cc |
| 1/4 aa x3/4 B- x3/4 C- | = 9/64 | aaB-C- |
| 1/4 aa x1/4 bb x3/4 C- | = 3/64 | aabbC- |
| 1/4 aa x3/4 B- x1/4 cc | = 3/64> | aaB-cc> |
| 1/4 aa x1/4bb x1/4 cc | = 1/64 | aabbcc |
| 37/64 |
5. Alleles: H = good student, h = poor student, E = on time for tests, e = late for tests
a) Hh X Hh -- 3 Good student (HH&Hh) : 1 Poor student (hh)
b) HhEe X HhEe -- 9 Good students (H-E-) : 3 Poor students (hhE-) : 3 no shows (H-ee) : 1 no show (hhee)
The students that are late for tests fail because the late for tests gene (ee) masks the THP phenotype (H- or hh). This gives us the altered ratio of 9 good students: 3 poor students: 4 no shows. In terms of test performance this is 9 Pass tests : 7 Fail tests.
6.
| III-1 | III-2 | |
| (2/3x1/2) | (2/3x1/2)1/4 | = 4/144 = 1/36 |
P = unaffected, p = phenylketonuria
In the II generation there is an affected individual which tells us that the two parents (I generation) are both heterozygous for the phenylketonuria (Pp). The unaffected offspring of that mating each have a 2/3 chance of carrying the recessive (p) allele. There is a 1/2 chance that II-2 and II-4 will pass on the (p) allele if they carry it. Finally, if both of these individuals are heterozygous they have a 1/4 chance of passing on both recessive alleles (pp) to the 4th generation. Because all of these things must happen in order to get the disease passed on, the individual probabilities must be multiplied.