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1. RrLl x RrLl
What fraction are purple and long?
(½ Rr)(¼ LL) = ⅛ RrLL
Ans: (b)
2. In this cross, we see that the inheritance pattern in the F1 shows a ‘criss-cross’ similar
to that observed for the white eye phenotype in Drosophila- only the sexes are reversedyellow males give rise only to green females.
Remember, however, that females in birds are the heterogametic sex (ZW), and males
are the homogametic sex (ZZ).
The crosses can therefore be interpreted as:
ZyZy (yellow male)x ZYW (green female), where the green allele (ZY) is dominant over
the yellow allele (Zy)
The F1 progeny would therefore be green males (ZyZY) and yellow females (ZyW).
This is a single gene locus with complete dominance and sex-linked expression- none the
above are true (e).
3. green males (ZyZY) x yellow females (ZyW)
¼ ZyZy (yellow males)
¼ ZyW (yellow females)
¼ ZYZy (green males)
¼ ZYW (green females)
All of the above are true (d)
4. c+ > cch > cc > c
4 homozygotes: c+c+; cchcch; chch; cc;
6 heterozygotes
c+cch; c+ch; c+c; cchch; cchc; chc
Ans: 10 total (c)
+ h ch
5. c c x c c
¼ c+cch (agouti)
¼ c+c (agouti)
¼ chcch (chinchilla)
¼ chc (himalaya)
2 agouti: 1 chinchilla : 1 himalaya
none of the above (e)
6. This is discussed on page 69 of your text.
The myocilin gene is associated with a form of juvenile glaucoma; there are multiple
allelic variants and not all individuals heterozygous for the mutations manifest the
disease.
Ans (b)
The orientation of the alleles in the
heterozygous parent is:
+
b
y
+
Phenotype Progeny Recovered
4
byp
93
by+
54
b+p
349
b++
331
+yp
66
+y+
97
++p
6
+++
Total
1000
+
p
7.
+ + p 97
y b + 93
+++ 6
ybp 4
Total 200
200/1000 = 0.2 = 20 map units between y and b.
+ b p 54
y + + 66
+++ 6
ybp 4
Total 130
130/1000 = 0.13 = 13 map units between b and p.
The distance between y and p would therefore be 20 + 13 = 33 map units
Ans: greater than 30 (e)
8. (0.2)(0.13)(1000) = 26 expected DCOs
COC = observed DCOs/expected DCOs =
10/26 = 0.385
Ans: (c)
Classes
1 and 2
Spore pairs 3 and 4
5 and 6
7 and 8
total
Ditype
Segreg.
1
ag
++
ag
++
1
NPD
2:2
2
a+
++
ag
+g
37
TT
2:1
3
ag
ag
++
++
2
NPD
1:1
4
ag
++
a+
+g
3
TT
2:2
5
a+
+g
a+
+g
3
PD
2:2
6
a+
a+
+g
+g
316
PD
1:1
PD >>NPD (319>>3), so genes are linked
Centromere to a distance: ½ * (1+ 37 + 3 + 3)/437 = 0.05 (5 m.u.)
Centromere to g distance: ½ * (1+ 3 + 3 + 75)/437 = 0.094 (9.4 m.u.)
a to g distance: [½ (37 + 3 + 75)] + 3(1 + 2)/437 = 15.2 map units
9. Genes are on either side of the centromere: Ans: (c)
10. Centromere to a distance is 5 map units Ans: (a)
7
ag
a+
++
+g
75
TT
1:2