Question

·     A strain of yeast requiring both tyrosine (tyr-) and arginine (arg-) is crossed to the wild type. After meiosis, the following 10 asci are dissected. Classify each ascus as to the segregational type (PD, NPD, TT). What is the linkage relationship between these two loci?

1             arg-tyr-                arg+tyr+               arg+tyr+               arg-tyr-

2             arg+tyr+               arg+try+               arg-tyr-                arg-tyr-

3             arg-tyr+               arg-tyr+               arg+tyr-               arg+tyr-

4             arg-tyr-                arg-tyr-                arg+tyr+               arg+tyr+

5             arg-tyr-                arg-tyr+               arg+tyr-               arg+tyr+

6             arg+tyr+               arg+tyr+               arg-tyr-                arg-tyr-

7             arg-tyr-                arg+tyr+               arg-tyr+               arg+tyr-

8             arg+tyr+               arg+tyr+               arg-tyr-                arg-tyr-

9             arg+tyr+               arg-tyr-                arg-tyr-                arg+tyr+

10           arg-tyr-                arg+tyr+               arg+tyr+               arg-tyr-

Answer 1

Given that two yeast strains are crossed. We can show the cross as

arg+tyr+    arg-tyr-

Also given that 10 asci are obtained . We will now classify each tetrad as;

1. Parental ditype (PD). This is because all the four products resemble parental combinatons (arg+tyr+ and arg-tyr-)

2. Parental ditype (PD). This is because all the four products resemble parental combinatons (arg+tyr+ and arg-tyr-)

3. Non-parental ditype(NPD). All the four products are recombinants (arg+tyr- and arg-tyr+) and neither of these resemble parental combinations.

4. Parental ditype (PD). This is because all the four products resemble parental combinatons (arg+tyr+ and arg-tyr-)

5. Tetratype(TT). This is because 50% are parental ( arg+tyr+ and arg-tyr-) and 50% are nonparental (arg-tyr+ and arg+tyr-)

6. Parental ditype (PD). This is because all the four products resemble parental combinatons (arg+tyr+ and arg-tyr-)

7. Tetratype(TT). This is because 50% are parental ( arg+tyr+ and arg-tyr-) and 50% are nonparental (arg-tyr+ and arg+tyr-)

8. Parental ditype (PD). This is because all the four products resemble parental combinatons (arg+tyr+ and arg-tyr-)

9. Parental ditype (PD). This is because all the four products resemble parental combinatons (arg+tyr+ and arg-tyr-)

10. Parental ditype (PD). This is because all the four products resemble parental combinatons (arg+tyr+ and arg-tyr-)

From the above information, we have 1 NPD, 2 TT and 7 PD, a total of 10 tetrads.

It is clear that PD>>>NPD. So it shows that the two genes are linked ( If PD was equal to NPD, then the two loci were not linked.)

We can now calculate the recombination frequency RF between the two loci as

RF = NPD + 1/2TT 100

Total asci

or RF = 1 + 1/2 (2) 100

10

or RF = 20 map units or 20%