Question

1. Given the following multi-gene cross, you’ll need to determine the likelihood/odds that an individual has for having a particular genotype and/or phenotype. Make sure you show all the work that could possibly be done (work is always worth a non-insignificant portion of your grade on this stuff.) parents: a/a; B/B; C/c; d/d; E/E x A/a; B/b; C/c; D/d; e/e a) What is the likelihood that an offspring would be heterozygous for all five genes? b) What is the likelihood that an offspring would display the recessive phenotype for all five genes?

Answer 1

a/a B/B C/c d/d E/E X A/a B/b C/c D/d e/e

a/a x A/a

1/2 Aa; 1/2 aa

B/B x B/b

1/2 BB; 1/2 Bb

C/c x C/c

1/2 CC; 1/2 cc

d/d x D/d

1/2 Dd; 1/2 dd

E/E x e/e

1 Ee

a) likelihood of offspring would be heterozygous AaBbCcDdEe

1/2 x 1/2 x 1/2 x 1/2 x 1= 1/16

b) Likelihood of offspring would be homozygous aabbccddee: The cross between parents with genotypes a/a B/B C/c d/d E/E X A/a B/b C/c D/d e/e cannot produce offspring with genotype of aabbccddee. The parent a/a B/B C/c d/d E/E cannot contribute a b allele and e allele. Therefore their offspring cannot be homozygous for the bb and ee gene loci.