P/p----------------15 m.u.------------------A/a----------------20 m.u.--------------------------R/r
1. Probability of P a r (Vulcan ears, earth adrenals, earth heart):
Total recombinant frequency = frequency of single crossovers + frequency of double
cross overs.
0.15 = SCO + DCO, or
0.15 – DCO = SCO
If no interference, then DCO = (0.15)(0.20) = (0.03)
0.15 – 0.03 = 0.12
SCOs = P a r and p A R
Probability of P a r = 0.12/2 = 0.06 Ans. (d)
2. P (PaR/pAr) = 0.03
P(Par/pAR) = 0.12
P(PAr/paR) = .20-.03 = .17
P(PAR/par) = 1- (probability of SCOs + DCOs) = 0.68
P(PAR) = 0.68/2 = 0.34 Ans. (d)
Note that the progeny classes that are missing are wild type, and pallid dwarp, i.e. no
recombination are observed between the pallid and dwarp genes relative to their
orientation in the heterozygote. In addition, the two smallest classes are pallid and
dwarp, raven, rumpled. Therefore raven has to be the middle gene.
(+ dwp)
(pld +)
+
rv
rmp
+
3. Since no recombinants are observed between pallid and dwarp, they are closest
together. Ans: (b)
4. The map distance between rumpled and raven is:
SCO:
dwp
47
pld rv rmp
48
DCO:
dwp rv rmp
2
pld
3
total
100 100/1000 = 0.1 = 10 map units. Ans: (c)
5. size of genome ≈ 3 x 109
Approximate number of genes ≈ 25,000
3 x 109/ 2.5 x 104 = 1.2 x 105 ≈ 100 kb Ans: (e)
Create the two different linkage maps and compare.
a b
a 0 12
b
0
c
d
e
f
a b
a 0 12
b
0
c
d
e
f
Canada:
Spain:
The two maps indicate an inversion with
breakpoints near the b and e genes:
c
14
2
0
d
23
35
37
0
e
3
15
17
20
0
c
14
2
0
d
23
35
37
0
e
3
15
17
20
0
f
29
17
15
50
32
0
f
29
17
15
50
32
0
Canada:


d---------------20-----------------e--3-a--------12------------b-2-c------15--------------------f
Spain:
f--------15--------------------c---4----e---3---a----------12-------b-----------------18-------------d
e
e
a
a
a
b
d
d
b
c
f
c
f
6. True, in both strains c is between f and a. Ans: (a).
7. True, crossing a trihybrid heterozygote to a homozygous recessive individual
represents a three point testcross. Ans: (a).
8. Due to the inversion loop and the distance between the b-c gene loci, we would
expect this to the loop to inhibit crossing over, and prevent recombination in this region
of the chromosome. Ans: zero, (a).
9.
on
(omn)
→
ot
ton
(mhr)
→ tonm
→ tonmh
→ tonmhr
Ans: (c).
hm
10. Both G. herbaceum and G. thurberi are diploid species, with a haploid value of n =
13. Hybridizations between the two yield two sets of chromosome that do not pair
(formation of 13 large and small univalents). G. hirsutum has a haploid value of n = 26,
and when it hybridizes with G. thurberi, it forms a pair of small bivalents. This suggests
that G. hirsutum is an amphidiploid, containing a doubled chromosome set. It is also an
allotetraploid, containing a chromosome set from both G. herbaceum and G. thurberi,
with G. thurberi representing the smaller chromosome set. Ans: all of the above, (e).