Dr. Ruth Ballard
BIO 184
Page 1
Fall 2003
LINKAGE MAPPING PRACTICE PROBLEMS
KEY
1
a.
b.
c.
d.
+,co+ and gl, +,str
P = +,co and gl,+; R= +,+ and gl,co
(27 + 11 + 8 + 29)/1000 x 100 = 7.5 cM
P = co,+ and +,str; R = +,+ and co,str
(27 + 29 + 0 + 1)/1000 x 100 = 5.7 cM
P = +,+ and gl,str; R = +,str and gl,+
(11 + 8 + 0 + 1)/1000 x 100 = 2.0 cM
e.
gl ------------str------------------------------------------------co
2.0 cM
5.7 cM
2.
f.
(2.0 cM + 5.7 cM) = 7.7 cM
g.
The gl,+,+ and +,co,str classes represent the double crossovers. The smaller
the distance between the two outside genes in a trihybrid testcross mapping
problem, the fewer double crossover events will be seen. The chance of a
double crossover in this particular problem is only (0.02 x 0.057) = 0.00114.
Among 1,000 offspring, one would thus expect to see (1000 x 0.00114) = 1.14
or less than 2 offspring in the recombinant classes. This expectation is
consistent with the data from the cross and simple sampling error explains
why there was one offspring in one of the double crossover class but not in
the other.
a.
White flower color, normal-sized flowers, and tall plants. This can be seen
in the F1 generation, which are trihybrids and express the phenotype
associated with each dominant allele.
If you choose “R” for white and “r” for red flowers, “P for normal-sized
flowers and “p” for peloria flowers, and “D” for tall plants and “d” for dwarf
plants, the cross was: RRPPDD x rrppdd.
dwarf, peloria, wild (d,p,+) and wild, wild, white (+,+,R)
P = d,p and +,+; R = d,+ and +,p
(51 + 43 + 6 + 5)/543 x 100 = 19.3 cM
P = p,+ and +,R; R = +,+ and p,R
(56 + 48 + 6 + 5)/543 x 100 = 21.2 cM
P = d,+ and +,R; R = d,R and +,+
(56 + 48 + 51 + 43)/543 x 100 = 36.5 cM
b.
c.
d.
e.
f.
g.
d-----------------------------p---------------------------------R
19.3 cM
21.2 cM
Dr. Ruth Ballard
BIO 184
3.
Page 2
h.
19.3 cM + 21.2 cM = 40.5 cM
a.
b.
c,wx,Sh and C, Wx,,sh
P = c,wx and C,Wx; R= +,wx and c,+
(84 + 974 + 951 + 99)/6708 x 100 = 31.4 cM
P = wx, Sh and Wx,sh; R = wx,sh and Wx,Sh
(974 + 20 + 951 + 15)/6708 x 100 = 29.2 cM
P = c,Sh and C,sh; R = c,sh and C,Sh
(84 + 20 + 99 + 15)/6708 x 100 = 3.2 cM
c.
d.
Fall 2003
e.
c ---------------------------------------------------------sh----------wx
29.2 cM
3.2 cM
4.
f.
29.2 cM + 3.2 cM = 32.4 cM
a.
There are eight possible phenotypes:
hairless, wrinkled, no scales
hairy, smooth, scales
hairless, smooth, no scales
hairy, wrinkled, scales
hairless, wrinkled, scales
hairy, smooth, no scales
hairless, smooth, scales
hairy, wrinkled, no scales
b/c. Since this is a trihybrid cross among linked genes, the situation is more
complicated than if it was a test cross. However, we do know that each
dihybrid parent will produce gametes as follows:
h, wr, Sc and H, Wr, sc
Equal numbers and greatest proportion
Rationale: These are the two parentals. We know this because of the way the
cross was set up originally between the two pure-breeders.
h, Wr, sc and H, wr, Sc
Equal numbers and smallest proportion
Rationale: We are given the information that Wr is the middle gene. Double
crossover gametes always switch the alleles of the middle gene with respect
to the parental combinations; therefore, these two classes of gametes
represent the rare double crossovers.
Dr. Ruth Ballard
BIO 184
h, Wr, Sc
H, wr, Sc
Page 3
and
and
H, wr, sc
h, Wr,sc
Fall 2003
Equal numbers, intermediate proportion
Equal numbers and intermediate proportion
Now let’s try two scenarios: one in which the genes are extremely tightly
linked and one with very loose linkage. If the same class(es) of offspring are
in the highest and lowest proportions in each case, then it will always be so
for any degree of linkage in between.
Scenario 1: Extremely tight linkage; NO RECOMBINANTS
In this situation, the three genes will remain linked to one another through
every meiosis in both parents and the only gametes produced by the parents
will be parentals: h, wr, Sc and H, Wr, sc.
The Punnett Square can thus be drawn as follows:
h, wr, Sc (1/2)
H, Wr, sc (1/2)
¼
1/2
¼
h, wr, Sc (1/2)
h/h, wr/wr, Sc/Sc (1/4)
H/h, Wr/wr, Sc/sc (1/4)
H, Wr, sc (1/2)
H/h, Wr/wr, Sc/sc (1/4)
H/H, Wr/Wr, sc/sc (1/4)
Hairless, wrinkled, scales
Hairy, smooth, scales
Hairy, smooth, no scales
Scenario 2: Extremely LOOSE linkage; All gametes produced in equal
numbers.
In this case, we get the phenotypic distribution characteristic of a Mendelian
trihybrid cross via Independent Assortment of the alleles of all three genes in
both parents:
Hairy, smooth, scales
27/64
Hairy, smooth, no scales
9/64
Hairy, wrinkled, scales
9/64
Hairy, wrinkled, no scales
9/64
Hairless, smooth, scales
9/64
Hairless, wrinkled, scales
3/64
Hairless, smooth, no scales
3/64
Hairless, wrinkled, no scales 1/64
In each case, the largest class is hairy, smooth, scales and the smallest class is
hairless, wrinkled, no scales. (In fact, if the genes are completely linked, there will
Dr. Ruth Ballard
BIO 184
Page 4
Fall 2003
be no offspring in this category.) This is certainly a different result than if we had
done a test cross, where the largest phenotypic classes would have been the two
parental classes and the smallest classes would have been the double crossovers.
Here, the class containing all the dominant alleles has the most offspring and the
class with all the recessive alleles has the least, regardless of the level of linkage.
5.
Yes. If the genes were not linked, then the offspring in a testcross would be spread
across eight different phenotypes equally. Since 1,033 offspring were produced,
each phenotypic class should contain about 129 offspring. If any one class has 289,
this indicates a very uneven distribution and suggests linkage.