Genetics Question 1 8 point question Assume that Mendel self

Genetics

Question 1 (8 point question) Assume that Mendel selfed a bushy pea plants that was grown from yellow seed (an F1 hybrid). This hybrid had one parent that was spindly and grown from a green seed; these were the wildtype phenotypes and that parent was true-breeding. He then collected 400 seed from the self of the bushy pea plant and recorded the seed color as well as the bushyness of the plant that grew from that seed.

A) (1 points) Given the standard notation rules for naming genes that we discussed in class what would be the most logical name for the two genes being investigated in this cross?

I think this is bushy/spindly trait and also yellow/green?

B) (2 points) For each gene, is the mutation dominant or recessive? Explain your reasoning.

C) (2 points) What ratio of yellow to green seed color would you expect among the 400 F2 offspring? Explain your reasoning using genetic notation (show genotype of parent and progeny).

D) (2 points) Assuming that the two genes show independent assortment, how many of the green peas would you also expect to grow into spindly pea plants?

Solution

A) Gene for growth habit (bushy or spindly)
Gene for seed color (yellow or green)

B) Since the wild type traits are spindly and green seed color which are expressed in true breeding genotype only and the F1 hybrid has bushy habit and yellow seeds, the mutation for both genes is dominant.

C) F1 hybrid YyBb x YyBb
Since Yy x Yy = 1 YY : 2 Yy: 1 yy i.e. 3 yellow: 1 green
The ratio of yellow to green seed color in F2 progeny is 3:1
Yellow seeds: 3/4 x 400= 300
Green seeds: 1/4 x 400= 100

D) If the two genes show independent assortment; the phenotypic ratio for F2 generation would be
9 Yellow seeds with bushy habit: 3 Yellow seeds with spindly habit : 3 green seed with bushy habit: 1 Green seed with spindly habit
So 1/16 x 400= 25 greed seeds will grow into spindly habit.