1. Affected father transmits disease to all daughters
Unaffected individuals do not pass the trait
Affected females pass trait to ½ of their progeny,
regardless of sex
inherited as an X-linked dominant: ans (e) none of
the above.
2. X+ X+ww x XbYw+w+ 
¼
Xb X+w+w x X+Yw+w
¾ w +_
3/16 solid black female
¼ ww
¾ w+_
1/16 solid white female
3/16 blotchy black male
¼ ww
¾ w+_
1/16 blotchy white male
3/16 solid black female
¼ ww
¾ w+_
1/16 solid white female
3/16 solid black male
¼ ww
1/16 solid white male
XbX+
¼ XbY
¼ X+X+
¼ X+Y
Ans: (e) all of the above.
3. False. There are no variant 9:3:3:1
ratios. All are consistent with
monohybrid ratios with a multiple
allelic series showing different
dominance relationships between
alleles: C>H>A. It is possible that this
data set would be seen with dominant
epistasis- i.e. a dominant allele at the C
locus, masking H being dominant over
A at a second locus. If this were the
case, however, you would also expect
to see some variant 9:3:3:1 ratios arising in the rabbit crosses. For example, one expected
‘C x C’ cross would be CcHA x CcHA, which would give a 12C:3H:1A ratio; similarly a
CcHA x CcAA cross would give a 6:1:1 ratio.
4. cross 7: CA x CA or ¾ chinchilla: ¼ albino; (e) none of the above.
5. The strains are haploid; td;su x td+su+: the two traits assort independently during
meiosis, so all classes would be expected to be recovered in equal frequency:
td+su+ (independent)
tdsu+ (mutant)
td+su (independent)
tdsu (independent)
Ans. (b) 3 independent: 1 dependent
+
s
+
x
+
w
6. Waxy-shrunken interval:
White
626
Waxy, shrunken
601
Wild type
4
Waxy, shrunken, white
2
Total
1233
1233/6708 = 0.184 = 18.4 map units
116
4
2538
601
626
2708
2
113
sw+
+++
s++
s+x
+w+
+wx
swx
++x
Shrunken-white interval:
Shrunken, white
116
Waxy
113
Wild type
4
Waxy, shrunken, white
2
Total
235
235/6708 = 0.035 = 3.5 map units
The distance between the white and waxy genes is 18.4 + 3.5 = 21.9 m.u. (d)
7. Expected crossovers: (0.35)(0.184)(6708) = 43.2
Coefficient of coincidence = 6/43.2 = 0.139
Interference = 1-0.139 = 0.86 (d)
A-------------------------10 m.u.---------------P--------------------8 m. u. -----------R
8. Assuming no interference, the frequency of the double crossover classes should be:
(0.1)(0.08) = .008. Therefore, the frequency of the two double recombinant classes
would be:
Adrenals absent, right heart: 0.004
Pointed ears: 0.004
A single crossover between the A and P loci would be observed (0.1) –(0.008) or in 0.092
of the individuals. The frequency of the two single recombinants in this interval would
be:
Adrenals absent: 0.046
Pointed ears, right heart: 0.046
A single crossover between the P and R loci would be observed (0.08) –(0.008) or in
0.072 of the individuals. The frequency of the two single recombinants in this interval
would be:
Right heart: 0.036
Adrenal absent, Pointed ears: 0.036
The parental classes would be:
1-(.008+.092+.072) = 0.828/2 = 0.414 would have all Vulcan features (c)
9. Adrenals absent/pointed ears would represent 3.6% of the progeny: (e) none of the
above
10. RF = # of second division asci/total
asci * (1/2) =
20/120 * (1/2) = 0.083 = 8.3 map units =
(b)
Ascus type
td+
tdtd+
td+
tdtd+
tdtd53
7
tdtdtd+
td+
47
tdtd+
tdtd+
8
td+
tdtdtd+
5
8.