1. Black pansies must be homozygous recessive for two genes. You can infer this because intercrossing the F1's resulted in 4 different phenotypes, of which black was the rarest. This suggests the 9:3:3:1 phenotypic ratio of a dihybrid intercross. Following this back, the homozygous parents must each have been dominant for one gene and recessive for the other, with the F1 purple being heterozygous for both genes. If you had tried to make this problem work with only a singe gene, you would not have been able to come up with enough phenotypes to cover all the classes (say if AA = blue and aa = magenta, then Aa = purple, but you can get at most 3 phenotypes if you intercross the Aa purple and none of them would be black because you've already defined all the genotypes).

2. The genotype of the full-color parent was Cc^ch. The chinchilla progeny are the key. The only place the c^ch allele could have been hidden was in the full-color parent; it could not have been in the Himalayan parent because chinchilla is dominant over Himalayan.

3. The trait is sex-linked dominant since the father passes it to all his daughters and none of his sons. Therefore, the affected father must be X^AY. His daughters are all X^AX^a, and as such, they have a 1/2 chance of passing the trait on to either their daughters or their sons equally.

4.

White-eyed, normal-winged female: XwXw cc

red-eyed, curly-winged male XWY CC

F1 females: curly-winged XwXW Cc

F1 males: white-eyed and curly- winged XWYCc

F2: 3/8 white-eyed and curly-winged

3/8 red-eyed and curly-winged

1/8 white-eyed and normal-winged

1/8 red-eyed and normal-winged

(Same ratio for females and males )

5. The mRNA sequence is U U A U G C G U A A C U G G A G U U G A C. Note the position of the start codon (AUG). Translation will start with the AUG and end at the UGA (a stop). The protein sequence will be: MET-ARG-ASN-TRP-SER. With the mutation, the first G of the pair is changed to an A, leaving that codon as UAG, a stop codon. This will cause translation of an abnormally short protein with only MET-ARG-ASN.

6.a. The original light blue and pink tribbles had genotypes as follows:

light blue: rrBB

pink: RRbb

b. purple F1's = RrBb

c. 3/16 of the progeny would be expected to be light blue

(9:3:3:1 ratio from the dihybrid intercross)

d. A pink tribble produced by testcrossing a purple F1 would be Rrbb.

e. 1/4 of the progeny of such a testcross would be expected to be pink.

7. a. Phenotypes:AaBb: red aaBb: white Aabb: yellow aabb: white

b. Phenotypic ratio = 9 red : 3 yellow : 4 white