Problem Set 5

Problem Set 5
Unordered and Ordered Tetrads and Bacterial Linkage Mapping
1. The yeast Saccharomyces cerevisiae has unordered tetrads. In a cross made to study
the linkage relations between three genes, the following tetrads were obtained. The cross
was between a strain of genotype + b c and one of genotype a + +.
Tetrad type Genotypes of spores in tetrads Number of tetrads
1 a++ a++ +bc +bc 132
2 ab+ ab+ ++c ++c 124
3 a++ a+c +b+ +bc 64
4 ab+ abc+ +++ ++c 80 the extra plus was a typo
Total: 400
a. From these data determine which, if any, of the genes are linked.
b. For any linked genes, determine the map distances.
Ans:
a. To determine the linked genes, first consider each possible set of pairs and classify
each tetrad type as PD, NPD, and TT with respect to the gene pairs. For a and b, the
tetrad types 1 through 4 are PD, NPD, PD, and NPD, respectively, totaling 196 PD and
204 NPD; since PD < NPD, genes a and b are unlinked. For genes b and c, the tetrad
types 1 through 4 are PD, NPD, TT, and TT, respectively, totaling 132 PD, 124 NPD, and
144 TT; since NPD approx equals PD, genes b and c are unlinked. Finally, for genes a
and c, the tetrad types 1 through 4 are PD, PD, TT, and TT, respectively, for a total of
256 PD and 144 TT. In this case, NPD < PD and so genes a and c are linked.
b. Because NPD = 0 for the a and c genes, the map distance between a and c is calculated
as (1/2) × ([TT]/Total) × 100 = (1/2) × 144/400 × 100 = 18.0%.

2. In yeast Saccharomyces cerevisiae the “gender” of the gametes is determined by a
single gene, the mating type (mat) gene. Haploids are either type a or type α (equivalent
of male and female) depending on which allele they have, and diploids are heterozygous
a/α. Therefore in the cross mat-a met2 x mat-α arg7, three genes are heterozygous
(mat, met2 and arg7). Two of these genes are linked and one is unlinked. Determine the
map for the linked genes from the following unordered tetrads:
Tetrad genotypes number of tetrads
a met2 arg+ 119
a met2 arg+
α met+ arg7
α met+ arg7
a met+ arg+ 125
a met+ arg+
α met2 arg7
α met2 arg7
a met2 arg+ 33
a met2 arg7
α met+ arg+
α met+ arg7
a met+ arg+ 27
a met+ arg7
α met2 arg+
α met2 arg7
Answer:
Tetrad mat/met mat/arg met/arg
I = 119 PD PD PD
II = 125 NPD PD NPD
III =33 PD TT TT
IV = 27 NPD TT TT
PD = NPD PD>T>NPD PD=NPD
conclusion: unlinked linked unlinked
mat - arg = ½ TT / Total x 100 = [0.5(33 + 27)/(119 + 125 + 33 +27)] x 100 = 9.87 cM

3. In Neurospora, the following ordered tetrads were obtained for gene A near the
centromere. What is the distance in map units of A from the centromere?
A A a a 131 FDS
a a A A 127 FDS
A a A a 23 SDS
a A a A 19 SDS
Total: 300
The two mitotic daughters are not indicated because they are identical, so the eight spores
would be:
A A A A a a a a, etc.
Distance from centromere is ½ SDS x 100 = ½ (23+19) x100 = 7 cM
Total 300

4. A small portion of the genetic map of Neurospora crassa chromosome VI is illustrated
here. The cys-1 mutation blocks cysteine synthesis, and the pan-2 mutation blocks
pantothenic acid synthesis. In a cross of cys-1 pan-2 ◊ CYS-1 PAN-2, what are the
expected frequencies of the following types of asci, assuming complete chromosome
interference?
a. First-division segregation of cys-1 and first-division segregation of pan-2.
b. First-division segregation of cys-1 and second-division segregation of pan-2.
c. Second-division segregation of cys-1 and first-division segregation of pan-2.
d. Second-division segregation of cys-1 and second-division segregation of pan-2.
e. Parental ditype, tetratype, and nonparental ditype tetrads?
Ans: The gene-centromere map distance equals 1/2 the frequency of second-division
segregation, which also equals the frequency of crossing over in the region. In this
problem it is easiest to answer the problem by taking the cases out of order, considering
the second-division segregations at the beginning.
c. The frequency of second-division segregation of cys-1 must be 14%, since the map
distance is 7 cM. Because of the complete interference, a crossover on one side of the
centromere precludes a crossover on the other side, and so these asci must have firstdivision
segregation for pan-2.
b. Similarly, the frequency of second-division segregation of pan-2 must be 6%, since the
map distance is 3 cM; these asci must have first-division segregation for cys-1.
d. Because of the complete interference, second-division segregation of both markers is
not possible.
a. The only remaining possibility is first-division segregation of both markers, which
must have a frequency of 1 - 0.14 - 0.06 = 80%.
e. First-division segregation of both markers yields a PD tetrad and second-division
segregation for one of the markers yields a TT tetrad. Since there are no double
crossovers, there can be no NPD tetrads. Hence, the frequencies are PD = 80% and TT =
20%.

5. Challenge Problem:
From the map below, predict the type and number of tetrads from the cross a leu1 x a
his4. Assume you recover 900 tetrads.
mat his leu
_|____________|_________________|__
10.5 15.4
The cross is a leu1 his+ x α leu+ his 4:
α his4 leu+
a his+ leu1
region 1 region 2
Four events can occur:
1 no recombination
2 a single recombination event in region 1
3 a single recombination event in region 2
4 two recombination events, one in region 1 and the other in region 2
#1 will produce a parental ditype = Tetrad I
#2 will produce a tetratype in which mat changes linkage = Tetrad II
#3 will produce a tetratype in which leu changes linkage = Tetrad III
#4 will produce a tetratype in which his changes linkage = Tetrad IV
I II III IV
no recomb recomb region 1 recomb region 2 recomb regs 1 & 2
a leu1 his+ a leu1 his+ a leu1 his+ a leu1 his+
a leu1 his+ α leu1 his+ a leu+ his+ α leu1 his4
α leu+ his4 a leu+ his4 α leu1 his4 α leu+ his+
α leu+ his4 α leu+ his4 α leu+ his4 α leu+ his4
The distance between two genes is
cM = 0.5 [ (single XO’s + double XO’s) / total ] x 100
Therefore the total number of recombination events occurring between two loci
whose distance is known is derived by simple algebra from the mapping formula:
Total XO’s = single XO’s + double XO’s = 2 [ (cM x total) / 100 ]
This tells us the total number of recombinants, but to get the number of singles,
we must subtract the doubles. The doubles must be calculated independently.
cM expresses the probability of a recombination event. If two recombinations are
independent of one another, then the number of double recombinants can be
calculated by determining the product of their independent probabilities and
multiplying by the total.

double XO’s = 2 [ (cM reg 1) / 100 ] x [ (cM reg 2) / 100] x total
This will give the number of tetrads showing the IV pattern. (The factor of 2 in the
equation is derived from the 1/2 that we use when dealing with tetrads.) Now, the
total number of single recombinations between two loci can be calculated, giving
the numbers of II and III tetrads. All that is left is to calculate the number of non
recombinants:
Total Tetrads = I + II + III + IV
double XO’s = tetrad IV = 2 [(10.5 / 100)] x [(15.4 / 100)] x 900 = 29
single XO’s in region 1 = tetrad II = 2 [ (10.5 x 900) / 100 ] - 29 = 160
single XO’s in region 2 = tetrad III = 2 [ (15.4 x 900) / 100 ] - 29 = 248
non recombinants = tetrad I = 900 - 160 - 248 - 29 = 463