Map Units
 Snapdragons with genotypes CCDD and ccdd are
crossed. The CcDd progeny were then crossed with
ccdd plants and we obtain 30% CcDd, 20% ccD 20%
Ccdd and 30% ccdd offspring. We conclude
a. Genes C and D are linked, separated by 20 map
units
b. Genes C and D are linked, separated by 30 map
units
c. Genes C and D are linked, separated by 40 map
units
d. Genes C and D are unlinked
Answer:
 CD= 30%
 cd= 30%
 Cd= 20%
 cD= 20%
 20% cD + 20% Cd = 40%
 40% = 40 map units
 Snapdragons with genotypes DDEE and ddee are
crossed. The DdEe progeny are then crossed with ddee
plants and we obtain 45% DdEe, 5 % ddEe, 5% Ddee
and 45% ddee. How far apart are genes D and E?
a. 5 map units
b. 10 map units
c. 45 map units
d. Genes D and E are not linked
Answer:
 DE= 30%
 de= 30%
 De= 5%
 dE= 5%
 5 % dE + 5% De = 10%
 10% = 10 map units
A bit tougher:
 In fruit flies, there are 3 recessive mutations.
 s = Scute (no thorax bristles) S = wild type
 e = Echinus (rough eye surface) E = wild type
 v = vestigial wings V = wild type
 P: We cross homozygous recessive flies (sseevv) with
homozygous dominant wild type flies (SSEEVV)
F1:
 All F1’s have the wild type phenoytype, and genetically
would be:
SsCcVv
To produce the F2:
 Cross the F1 flies (SsCcVv) with recessive flies (ssccvv)
and you get an F2 with 6 different phenotypes:
Out of 1000 flies: + = wild type
Scute
Echinus
Vestigial
Total
scute
echinus
vestigial
420
+
+
+
435
scute
+
+
46
+
echinus
vestigial
44
scute
echinus
+
20
+
+
vestigial
25
scute
+
vestigial
6
+
echinus
+
4
Assignment:
 From the genetic data, determine the
recombination frequencies between each
pair of genes (s/e , e/v, s/ v).
Using these frequencies, construct a
genetic map:
ANSWER:
 s-e recombination
 46+44+6+4= 100/1000= 10%
 s-v recombination
 46+44+20+25= 135/1000= 13.5%
 e-v recombination
 20+25+6+4= 55/1000= 5.5%
s
10 map units
e
v
5.5 map units