2. An Fi plant undergoes offipring. A self-fertilization (AaBbCeDdEeFy X AaBbCeD

Question

2. An Fi plant undergoes offipring. A self-fertilization (AaBbCeDdEeFy X AaBbCeDdEeF) to produce F (MaBbCel ssum e independent assortment of these six genes. (10 points) a) How many different types of gametes would be produced by each F, individual? (1 poim) c b) H ow many different types of phenotypes would appear in the F2 generation? (1 point) c) How many different types of genotypes would appear in the Fz generation? (l point) d) What are the probabilities that an F2 offspring would have the following genotype? (5 points) i. aabbecddeeff i.AaBbCcDdEeFy ii AaBbecDdeeFF iv.AaBBCoddEEf V AABBCCDDEeFF e) For cach pea plants with the following genotypes. List the possible gametes that the plant can make (2 points) i. AABbCc ii) AaBBcc

Explanation / Answer

2. a. For the genotype AaBbCcDdEeFf, the number of possible gametes is the number of different alleles per gene multiplied. Therefore, the number of possible gametes is 2 x 2 x 2 x 2 x 2 x 2 = 64

b. There are two possible phenotypes for each gene, meaning there are a 2 x 2 x 2 x 2 x 2 x 2 = 64 possible phenotypes in the F2 generation

c. Each gene can have one from the three different arrangements of alleles (AA, Aa, aa for the first gene, for example), so the total number of possible genotypes in F2 generation is 3 x 3 x 3 x 3 x 3 x 3 = 729

d. Each heterozygous cross for a single has two chances for a heterozygous offspring, and one each for homozygous dominant and recessive out of four (to form the ratio 1:2:1)

i. aabbccddeeff - (1/4)6

ii. AaBbCcDdEeFf - (2/4)6 = (1/2)6

iii. AaBbccDdeeFF - (1/4)3 x (1/2)3

iv. (1/4)3 x (1/2)3

v. (1/4)5 x (1/2)

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