1. The pedigree shows only females being born. Assuming the fathers were normal, it
would be exceedingly unlikely that this would be a standard Mendelian inheritance
pattern (even if it were a sex-limited trait). The most likely possibility is a situation
where a factor in the mother’s egg cytoplasm was inherited, and this contributed to the
death of only males. Ans. (e)
2. Once again, none of the answers are likely. The data set shown is for a population,
and the dominant tasters in that population could be either TT of Tt; this would skew any
expected ‘monohybrid’ or ‘testcross’ data towards a larger number of tasters. The last
situation is less straightforward, since nontasters x nontasters should not give rise to
tasters if the gene is recessive. But again, this is a real data set; the odds are that some of
the participant nontasters could have been misidentified; the question of illegitimate birth
also must be considered. Ans. (e)
3. Both the team of Creighton and McClintock in corn, and Curt Stern in Drosophila,
performed experiments that demonstrated this. See pp. 131-132.
Parents
orange x blue
purple x blue
purple x orange
orange x tan
tan x blue
F1 phenotypes
All orange
All purple
All sienna
All orange
All purple
F2 phenotypes
203 orange, 67 blue
152 purple, 52 blue
86 sienna, 41 orange, 43 purple
130 orange, 60 tan, 44 blue
181 purple, 79 tan, 60 blue
In the above situation, there are two gene loci interacting. At the first locus, there are
three different alleles which show a hierarchy of dominance relationships. Orange (AO) is
dominant to blue (AB) and purple is dominant to blue (AB). Heterozygotes of (AO AP) are
sienna. Individuals who are heterozygous at a second gene locus for a recessive allele
will have a tan phenotype (tt). The crosses are:
1) AOAO x ABAB → AOAB (selfed) → ¾ AO_: ¼ APAP
2) APAP x ABAB → APAB (selfed) → ¾ AP_: ¼ ABAB
3) AOAO x APAP → AOAP (selfed) → ¼ AOAO: ½ AOAP : ¼ APAP
Where all of the above strains are all homozygous for the dominant T allele. The other
crosses are:
4) AOAOTT x ABABtt →AOABTt (selfed) → 9/16 AO_T_: 3/16 AO_tt: 3/16 ABABT_: 1/16 ABABtt
5) APAPtt x ABABTT →APABTt (selfed) → 9/16 AP_T_: 3/16 AP_tt: 3/16 ABABT_: 1/16 ABABtt
4. This cross is an example of recessive epistasis. Ans: two genes are involved (b)
5. The Blue F2 offspring from cross number 4 are either ABABTT or ABABTt. If they are
crossed to true breeding orange offspring, they must be AOAOTT. The progeny of this
cross would therefore inherit at least one T allele and at least one AO allele, and would
therefore be all orange. Ans: (a).
Phenotype
+bc
a+c
abc
++c
a++
ab+
+++
+b+
Number of progeny
436
91
6
92
440
87
8
98
6. The parental classes are + b c and a + +; the DCO
classes are a b c and + + +. The middle gene is therefore a
and the genes are in the following configuration in the
heterozygous parent:
b
+
c
+
a
+
Crossovers in b----a interval:
87
ba+
92
++c
6
bac
8
+++
193, out of 1258 total progeny 193/1258 = 0.153 = 15.3 map units
Crossovers in a------c interval:
98
b++
91
+ac
6
bac
8
+++
203, out of 1258 total progeny; 203/1258 = 0.161 = 16.1 map units
6. The distance between the a and b genes is 15.3 map units; Ans: (c)
7. The number of expected double crossovers is: 30.9. We see 14 double cross overs,
therefore our coefficient of coincidence is 14/30.9 = 0.45 We are seeing about 45 percent
of what we expect, not 1.1%. Ans: False.
Phenotype
+bc
a+c
abc
++c
a++
ab+
+++
+b+
Number of progeny
228
19
20
234
231
227
21
20
When we look at the categories here, we need to
determine which gene are linked to one another. For
example if the a and b loci were not linked and were
segregating independently, we would expect the 4
possibilities of gametes to segregate independently:
++ = 234 + 21
ab = 20 + 227
+b = 228 + 20
a+ = 19 + 231
The a and b loci, therefore are NOT linked.
When we compare a and c however:
++ = 21 + 20
ac = 19 + 20
a+ = 231 + 227
+c = 228 + 234
The a and c loci ARE linked.
The most common classes are in the trans phasing relationship and therefore represent the
parental combinations:
+
c
a
+
b
+
8. The answer would be true.
9. The b gene is unlinked and segregates independently of the a and c genes. The a and
c genes are separated by 80/1000 = 0.08 or 8 map units. Ans: none of the above (e).
Spore pair
1-2
3-4
5-6
7-8
total
td+
td+
tdtd53
tdtd+
td+
td7
Ascus type
tdtdtd+
td+
47
tdtd+
tdtd+
8
10. The number of second division segregation patterns is 7 + 8 + 5 = 20.
The total number of asci are: 120
RF = 20/120 * ½ = 0.083 = 8.3 map units. Ans: (b).
td+
tdtdtd+
5