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1. Both loci are showing incomplete dominance: if two heterozygotes are crossed, then
the odds of being red (RR) are ¼, and the odds of being oval (Ll) are ½. If the loci are
independently assorting, they would follow the product rule: Ans: 1/8 (b).
The next two problems outline a situation similar to comb shape in chickens, where 4
different phenotypes can result between the interactions of two independently assorting
gene loci. Let’s define T_ yy as tan; ttY_ as gray, T_Y_ as brown, and ttyy as green in
phenotype.
2. This cross would be TtYy x ttYY. The expected results would be:
½ TtY_ or brown and ½ ttY_ or gray. Ans: e) none of the above.
3. Among the F2’s, only those self-fertilization events where heterozygotes are found
would be able to produce all phenotypic classes, since green, tan, and gray require two
recessive alleles for their phenotypes. ½ of the F2s would be Tt and ½ of the F2s would
be Yy. The answer would be that ¼ would be expected to produce all colors if selfed.
Ans: c)
4. The cross would be: AAYCc x AAYcc
All AYAY will not survive. Therefore, 2/3 of the surviving progeny will inherit AAY and
1/3 will inherit AA. But the homozygous cc genotype shows recessive epistasis:
½ Cc
2/6 (yellow)
2/3 AAY
½ cc
2/6 (albino)
½ Cc
1/6 (agouti)
1/3 AA
½ cc
1/6 (albino)
so the ratio would be 3 albino: 2 yellow: 1 agouti. Ans: (d)
5. The cross would be a modified dihybrid cross: AaBb x AaBb. Let us suppose,
however, that the bb genotype will suppress the dominant phenotype A_. Again, we
expect to recover the following classes:
9/16 A_B_ (dominant mutant);
3/16 aaB_ (wild type)
3/16 A_bb (wild-type, suppressed by bb)
1/16 aabb (wild type)
or 9/16 mutant: 7/16 wild type. Ans: (b).
For questions 6-8, the white trait is dominant to the wild type phenotype. The
organization of the alleles in the heterozygote is as follows:
W++
x +pd
+ pd
+pd
6. True, the white flower color is dominant over wild type in this cross.
7. The distance between the white and pelora gene loci would be:
W p d 56
+++
48
Wp+
5
+ + d 6
total
115
115/543 = 0.212 = 21.2 map units Ans: (d)
8. The distance between the pelora and dwarf gene loci would be:
W + d 51
+p+
43
Wp+
5
+ + d
6
total
105
105/543 = 0.193 = 19.3 map units
(0.212)(0.193) 543 = 22.2 expected double cross over’s
Coefficient of Coincidence = observed DCO/expected DCO = 11/22.2 = 0.495 Ans: (d)
Questions 9 and 10 are from the solved problems at the end of chapter 5 in the Hartwell
text. The father would pass the Xg- allele and the sts allele to his daughter, who receives
the Xg+ and Sts alleles from the mother:
XgXg+
sts
Sts
If 10% of the meiotic products were recombinants (i.e. the genes are separated by 10 map
units) then the father’s chromosome would be expected to be inherited in 45% of the nonrecombinant gametes. If a cross-over occurs between the two loci in question, then 5%
of the recombinants would now be Xg+ and carry the recessive gene for ichtyosis.
9. 45% or none of the above; Ans: (e).
10. 5%; Ans: (a).