Answers to Quiz #2:
1. In birds, the female is the heterogametic sex (ZW), and the male is the homogametic
sex (ZZ). In this cross, there are two allelic variants on the Z chromosome that govern
feather color: G = green, is dominant; g = yellow, is recessive.
ZGW ♀ x ZgZg♂
Green females x yellow males
↓
ZgZG♂ and ZgW ♀
Green males and yellow females
↓
g g
F2 males would be ½ Z Z (yellow) and ½ ZgZG (green)
Ans: all of the above (d).
2. C (full color) >ck (sepia) > cd (cream) > ca (albino).
If both of the parentals had an albino parent, their genotypes must be:
ck
ca
ckca x cdca
sepia x cream
↓
d
c
ca
d k
a k
c c (sepia) c c (sepia)
cdca (cream) caca (albino)
Ans: 2 sepia: 1 cream: 1 albino (e) none of the above.
Cross
Parental
F1
F2
1
Green x gray
Green ¾ green: ¼ gray
2
Green x brown Green ¾ green: ¼ brown
3
Gray x brown Green 9/16 green: 3/16 brown 3/16: gray: 1/16 blue
The last cross here shows a 9:3:3:1 relationship, indicative of a dihybrid cross situationtwo separate gene loci are involved. This would imply that at least one dominant allele is
required for green (A_B_) and the blue phenotype is all homozygous recessive (aabb).
Gray and Brown phenotypes would represent situations where only one locus had a
dominant allele and the other was homozygous recessive. (e.g. A_bb = gray; aaB_ =
brown). Since cross #3 is with true-breeding strains, AAbb x aaBB → AaBb x AaBb
3. Ans: false.
4. AABB x aabb would produce green individuals who were homozygous recessive for
both loci, the same as cross # 3. Ans: (c) the same results as cross #3.
5. The male parent (AO) must have inherited an O allele from his mother: otherwise, the
male would be AB. The same logic holds for the female parent (BO); she must have
inherited an O allele from her father. The cross is therefore:
AO x BO→ 1 AO: 1BO: 1AB: 1OO; Ans: (e) all classes would be expected.
Phenotype Number of progeny
Abc
18
aBc
630
abc
148
AB c
132
a BC
22
abC
120
ABC
140
AbC
615
total
1825
Phasing relationships in heterozygous parent:
a
c
B
A
C
b
a-c interval
a+b 120
+c+ 132
a++ 22
+cb 18
Total 292/1825 = 0.16
c-b interval
acb
148
+++ 140
a++ 22
+cb 18
Total 338/1825 = 0.1797
6. Ans: 16 map units (c)
7. (0.16)(0.1797)1825 = 52.48 COC = 40/52.48 = 0.762 Ans: (c)
Phenotype Number of progeny Looking at this data set, we see distributions that
are significantly different than in the data set above;
abC
228
in this example, four classes are found in high
aBc
235
frequency and 4 are found in low frequency. This
aBC
16
suggests that one gene locus may be showing
abc
22
independent assortment, i.e. is not linked to the
AbC
224
other two. To determine this, look at how each of
ABc
233
the two genes segregate relative to one another:
ABC
22
Abc
20
total
1000
AB = 233 + 22
BC = 16 + 22
Ab = 224 + 20
Bc = 235 + 233
aB = 235 + 16
bC = 228 + 224
ab = 228 + 22
bc = 22 + 20
The A and B gene loci are assorting independently- all gametes are found in about
equal frequencies; the B and C gene loci are linked- the nonrecombinant
chromosomes have genes in the trans configuration : Bc/bC are the nonrecombinants,
bc/BC are the recombinants. Recombination between the B and C gene loci here are
80/1000 = 0.08 or 8 map units.
Ans: (d) all of the above
Spore
pairs
total
1
Ab
Ab
aB
aB
127
2
AB
AB
ab
ab
125
3
Ab
AB
ab
aB
100
4
Ab
ab
AB
aB
36
5
Ab
aB
Ab
aB
2
6
AB
ab
AB
ab
4
7
AB
ab
Ab
aB
6
9. To measure the distance between a gene and a centromere, identify number of asci
showing a second division segregation pattern…
36 + 2 + 4 + 6 = 48/400 (1/2) = 0.06
Ans: (a) 6 map units
10. Are the two genes linked? If the genes are linked, then PD >> NPD
PD = 127 + 2 = 129
NPD = 125 + 4 = 129
The genes are NOT linked…
Ans: (e)