1. These questions relate to the Pax6 gene: see p. 7 in text.
Answers II and IV are correct; the human and fly genes are structurally related and
program eye development in both mammals and flies.
ans. (d)
2. Mendel was a monk, published his work in 1866 in a journal our library owns and was
a contemporary of Darwin. His work was not widely recognized in his lifetime.
I, II, and III are correct…ans (b)
3. See p. 22-23.
I and III are correct.
the gene encodes an enzyme that is involved in branched chain starch metabolism
the smooth/wrinkled trait assorts independently of the other 7 examined, so it is not
tightly linked to any others; IV must be false
ans (c)
4. Parents: white, round, yellow x purple,wrinkled, yellow
ppR_Y_ x P_rrY_
Are any white-flower progeny observed?
No, therefore second parent must be PP, otherwise we would expect a 1:1 ratio of purple
to white.
Are any wrinkled seed progeny observed?
Yes, so the first parent must be Rr and we expect to see a 1:1 ratio of round to wrinkled.
Are any green seed progeny observed?
Yes, so both parents must be heterozygous, and we would expect a 3:1 ratio of yellow
seeded to green seeded
ppRrYy x PPrrYy Ans: (d)
5. Both plants are true breeding, so they must be homozygous. You would expect the two
dominant traits to be expressed in the F1 hybrid, so white and tall are the dominant traits.
If W = white (dominant phenotype) and T = tall (dominant phenotype), the genotypes of
the parents can be assumed to be wwTT x WWtt and the F1 hybrid WwTt. If the plants
are independently assorting, the rarest class should be the class carrying both recessives,
which would be expected to be inherited in ¼ x ¼ = 1/16 of the selfed progeny. The
rarest class would be blue, short. Ans: (c)
AaBbccddEe x AaBbCcDdEe
6. P(A_B_ccddE_) = ¾ x ¾ x ½ x ½ x ¾ =
27/256 Ans: (d)
7. P(A_B_ccddE_ or A_B_C_D_E_)
P(all dominant phenotype) =
¾ x ¾ x ½ x ½ x ¾ = 27/256
27/256 + 27/256 = 54/256 = 27/128
Ans: (c)
8. P (genotype of either parent) = P (genotype of A or genotype of B)
P(AaBbccddEe) = ½ x ½ x ½ x ½ x ½ = 1/32
P (AaBbCcDdEe) = ½ x ½ x ½ x ½ x ½ = 1/32
P (genotype of either parent) = 1/32 + 1/32 = 2/32 = 1/16
P(genotype of neither parent) = 1- P(of either parent)
1-1/16 = 15/16
Ans: (c)
The characteristics of the pedigree are most consistent with autosomal recessive
inheritance; the disorder skips generations, and males and females are both affected. If
the disease gene is very rare, and both I-1 and V-2 are affected, then individuals IV-3 and
IV-4 are obligate heterozygotes, implying that the gene has to be inherited from their
parents within the family, who must also be heterozygotes (III-3 and III-6). Individuals
III-1, IV-1, IV-2 therefore each have a ½ chance of inheriting the disease gene from their
parents. If individuals IV-1 and IV-2 are heterozygotes, they have a ¼ chance of having
an affected child.
AA
aa
AA
Aa
AA
Aa
½ Aa
Aa
Aa
AA AA
Aa
Aa
½ Aa
AA
½ Aa
?
¼ aa
aa
9. P(V-1 will be affected) = ½ x ½ x ½ x ¼ = 1/32 Ans: (d)
10. IV-3 and IV-4 are obligate heterozygotes, and each subsequent birth has a ¼ chance
of producing an affected child. Ans: (b)