Problem 1:
Yeast sporulation can produce tetrads of three varieties:
Parental Ditype (PD), Non-parental ditype (NPD), and Tetratype (T)
Recombination is required to generate tetratype tetrads. Use a diagram to illustrate why this is the case.
For the purpose of the diagram, consider mating haplotype parents Ab with aB. The A and B loci are on
separate chromosomes. Locus A is inseparable from the centromere, while Locus B is far away.
If crossovers could not take place, what types of tetrads would you observe?
Only NPD and PD
Extras - How would this diagram differ if the A and B loci were on the same chromosome (A close to
centromere and B far away)? How are NPD tetrads observed if both loci are on the same chromosome?
NPD only observed if a crossover event occurs in both sister chromatids.
Problem 2:
Dave stated in lecture that the ratio of parental ditypes to non parental ditype to tetratype is 1:1:4 for
genes on different chromosomes. Prove it using this handy worksheet! Let’s say that we are mating AB
X ab and gene “A” and gene “B” are unlinked.
To make calculating the probabilities easier, let’s say that we’ll pre-allocate all of the A’s, and then you
only need to fill in the B’s and determine the probability of each event. You only need to calculate the
probability of the first two spores in each tetrad because the outcome of the second two spores is
dependent on the outcome of the first two.
Tetrad A
A1
A2
A3
A4
A
A_
a_
a_
B1
Tetrad B
B2
B3
B4
A_
A_
a_
a_
C1
Tetrad C
C2
C3
C4
A_
A_
a_
a_
D1
Tetrad D
D2
D3
D4
A_
A_
a_
a_
P(A1)=___
P(B1)=___
P(C1)=___
P(D1)=___
Tetrad type is ___________
P(tetradA) is P(A1)*P(A2) = ____________
P(A2)=___
Tetrad type is ___________
P(tetrad B) is P(B1)*P(B2) = ___________
P(B2)=___
Tetrad type is ___________
P(Tetrad C) is P(C1)*P(C2)= ____________
P(C2)=___
Tetrad type is ___________
P(Tetrad D) is P(D1)*P(D2) = _____________
P(D2)=___
Now add up the totals: PD ____ NPD_____ T_____ =
Tetrad A
A1
A2
A3
A4
AB
AB
ab
ab
B2
B3
B4
Ab
aB
aB
C2
C3
C4
Ab
aB
ab
P(A1)=_1/2__ P(A2)=_1/3_
Tetrad B
B1
Ab
P(B1)=_1/2__ P(B2)=_1/3_
Tetrad C
C1
AB
P(C1)=_1/2__ P(C2)=_2/3
Tetrad D
D1
D2
D3
D4
Ab
AB
ab
aB
P(D1)=1/2___
P(D2)=_2/3__
Tetrad type is PD
P(tetradA) is P(A1)*P(A2) =
(1/2)(1/3) = (1/6)
Tetrad type is NPD
P(tetrad B) is P(B1)*P(B2) =
(1/2)(1/3) = (1/6)
Tetrad type is _T__________
P(Tetrad C) is P(C1)*P(C2)=
(1/2)(2/3)= 1/3
Tetrad type is _T__________
P(Tetrad D) is P(D1)*P(D2) =
(1/2)(2/3) = 1/3
Now add up the totals: PD( 1/6) NPD (1/6) T (2/3) =
Problem 3: You have isolated mutants in genes required for synthesis of adenine and wish to order
them into a pathway. Conveniently, ade2 mutants are red colored while ade4 mutants are white. You
are given the strains Mata, ade2::URA3, ura3, his3 and Matα, ade4::HIS3, ura3, his3.

How would you generate the double mutant?
Cross the two strains, sporulate and plate on media lacking uracil and histidine

The double mutants are white colored. Which gene is epistatic? Can you order them into a
pathway?
Ade4 is epistatic and is upstream of ade2

Suppose instead you were given the strains Mata, ade2-1 and Matα ade4-1 which contain point
mutations in the respective adenine synthesis genes. Could you still carry out the above
experiment?
Not as easily because it’s impossible to distinguish between the ade4 and the ade2, ade4
phenotypes. You could do it by careful tetrad analysis. Such as if we assume ade2 and ade4
are unlinked then 2/3 of the time you expect to get a tetratype with 1 wt, 1 ade2, 1 ade4, and
1 ade2 ade4 spore. The wildtype spore can be identified by replica plating to –adenine plates.
Of the remaining spores 2 are white and one is red implying that the double mutant is white
so ade4 is epistatic. If you get the uninformative parental ditype you will have 2 red spores
and 2 white spores all of which need adenine. If you get the non parental ditype two of the
spores will grow on –adenine media implying that both mutant alleles are in the remaining
two (white) spores again demonstrating that ade4 is epistatic. If ade2 and ade4 are linked
then you will most get parental ditypes. You will then be screening through a lot of tetrads.
Moral being auxotrophic markers are awesome!