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Hi, I come across this question from an online website but I dont understand how to get the answer

Here is the question,
you isolate a mutant strain of yeast that cannot grow on medium lacking leucine.
This strain contains a single mutation you call leu1–. The leu1– mutation is near to
drk1– on the same chromosome. When the leu1– mutant is mated to wild-type yeast,
the resulting diploids cannot grow on medium lacking leucine.
The “dark tan” phenotype of the haploid cells you are working with is caused by
two different mutations in the same strain. The two mutations are designated drk1– and
drk2–. Mating of the drk1– drk2– double mutant to wild-type yeast produces
diploids that are white.
You mate leu1– yeast to drk1– yeast and sporulate the resulting diploid.
You grow the resulting spores on medium containing leucine. You then test for growth
on medium lacking leucine. It is apparent that you have isolated only two types of
tetrads, 10 tetrads of Type A and 10 tetrads of Type B. On medium lacking leucine,
only two spores from each Type A tetrad can grow; both are light tan in color. Complete
the chart below so as to indicate: How many spores from each Type B tetrad can grow
on medium lacking leucine, and what color is each spore that can grow?
Anyone knows how to solve this?

http://ocw.mit.edu/courses/biology/7-03-genetics-fall-2004/exams/exam1_2004.pdf

 

Here is the link to the question (Question 3)

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