What would be the probability of obtaining 6 offspring, 4 of which were dominant and 2 recessive if it was a test cross (Aa x aa) instead of a cross between two heterozygotes? Give your answer to 4 decimal places.

Answer:0.0156

Explanation:

If the test cross is between one heterozygous (Aa) and one homologous recessive gene (aa). The possible genotype for their offspring will be

Aa aa Aa aa

The probability of dominant gene (Aa) is 2/4

The probability of recessive gene (aa) is 2/4

The probability that in six offspring 4 is dominant and two recessive is

= 2/4*2/4*2/4*2/4*2/4*2/4

=64/4096

=0.015625

=0.0156 to 4 decimal places