Problem Set 2- answers

1. 
a. 50% type A; 50% type B
b. all type B
c. 25% type AB; 25% type A; 25% type B; 25% type O
d. all type O

2. The probability that the woman has an i allele to contribute 
to her children is (1/2 x 1/2 =) 1/4.  That is, her type B parent has a 1/2 probability of being IB/i rather than IB/IB.  And the probability that she inherited the i allele from that parent is 1/2.

The probability that her husband with type A blood is IA/i rather than IA/IA is 1/2.

And, if she does have the i allele, and he does have the i allele, 
they will have a type O (i/i) child with probability 1/4.

So the total probability is (1/2 x 1/2) x 1/2 x 1/4 = 1/32

3.
a. all agouti
b. 3/4 agouti, 1/4 chinchilla
c. 3/4 agouti, 1/4 albino
d. 1/2 agouti, 1/2 himalayan
e. 1/2 agouti, 1/2 himalayan
f. 1/2 chinchilla, 1/2 himalayan
g. 1/2 himalayan, 1/2 albino
h. 1/2 himalayan, 1/2 agouti
i. 3/4 agouti, 1/4 chinchilla

4. do a segregation test to determine if the tango allele is at a 
second gene, or is an allele at the same locus.  For instance, 
breed a known waltzing heterozygote (which will have a 
normal phenotype) with a tango.  

If the phenotype is controlled by one gene, the cross would
be:  V/v x vt/vt.  We know that the normal allele is dominant over
both waltz and tango alleles, but we don't know the phenotype 
of a v/vt individual, but we can assume it would not be normal.  If
you did a Punnett Square for this cross, you would find that 1/2 of
the progeny would have the normal phenotype, and 1/2 would have 
some other gait (v/vt).

If the phenotype is controlled by two genes, one that produces the
waltz (V/V and V/v are normal, v/v is waltz), and another gene
that produces tango (T/T and T/t are normal, t/t are tango), then
you assume that a double heterozygote V/vT/t is phenotypically 
normal.  Cross this individual with a tango mouse: VVtt.  Do a 
Punnett Square and find that 1/2 would be normal, 1/4 would be 
tango, and 1/4 would have the whatever genotype is caused by 
vvtt.  

Another way to test this is the complementation test, which we 
did not cover in class, and will not be on the exam.

5. 
a. 1/4
b. 1/2 
c. 1/4
d. 0

6. 	P = AABB x aabb
 	F1 genotype = AaBb, phenotype is all agouti
	F2 genotypic ratios = 
		1/16 AABB
		2/16 AABb
		2/16 AaBB
		4/16 AaBb
		1/16 AAbb
		2/16 Aabb
		1/16 aaBB
		2/16 aaBb
		1/16 aabb
	F2 phenotypic ratios=
		9/16 agouti
		4/16 albino
		3/16 black
genotype of 
	mother 1: 	AaBB
	mother 2:	AABb
	mother 3:	AaBb