Question

The genes for Mendel’s characters fall on five chromosomes. Seed color is on I (yellow vs green), flower color is on II (purple vs white), stem length is on III (tall vs dwarf), flower position is on IV (axial vs terminal), and seed shape is on V (round vs wrinkled). One way to test independent assortment would be to set up a five-factor cross. The parent cross would be YYPPTTAARR x yyppttaarr. The F1 cross would be YyPpTtAaRr x YyPpTtAaRr. (a) How many unique gametes would the F1 produce? (b) If you did a Punnett square, how many zygotes or cells would be in the square? Represent it in exponential notation. (c) Set up the first step in the multiplication method for calculating the expected phenotype frequencies but do not do the multiplication.

Answer 1

The F1 cross would be YyPpTtAaRr x YyPpTtAaRr

The number of unique gametes would be - 2^n

where n is the number of heterozygous alleles in the cross.

here n = 5

Therefore, number of unique gametes = 2^5 = 32

With a single trait , Punnet square will have two alleles for each parent and thus, the square has two rows and two columns.

We have been given that the all the 5 traits show independent assortment, therefore, the number of allele combinations which will be produced by the Punnet square is 2 raised to the power (exponential) of the number of traits.

In a given Punnett square, there would be 32 by 32 squares.

we have 5 traits here, all heterozygous therefore, the number of zygote combinations (squares required) produced will be 2^10 = 1024