Solutions For An Introduction to Genetic Analysis Chapter 4 Problem 41P

ii) l+

l+

+c

+c

This tetrad is a parental ditype. Due to random alignment and segregation during the process of meiosis I, two linear tetrads are possible. One is shown above while the other is the exact reverse of it.

Hence, the predicted frequency of the given tetrad can be calculated in this way:

iii) lc

l+

++

+c

This tetrad is a tetratype. This tetrad shows one of the two meiosis-I segregation patterns for leu3 and one of the four meiosis-I segregation patterns for cys2.

Hence the predicted frequency for this tetrad would be:

iv) lc

+c

++

l+

In this tetrad, leu3 shows a second-division segregation pattern. Since the question says that it always segregates in the first division, hence the predicted frequency would be 0%.

v) lc

++

++

lc

In this tetrad, the leu3 gene shows a second-division segregation pattern. This does not match with the data given in the question. Hence, the predicted frequency would be 0%.

vi) l+

+c

+c

l+

Since the leu3 gene shows a segregation pattern corresponding to the second division, this goes contrary to the data presented in the question. Hence, the predicted frequency would be 0%.

vii) l+

+c

++

lc

This tetrad also shows a second division segregation pattern for leu3. Since this does not match with the data given in the question, the predicted frequency would be 0%.