Question

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 1

In a test cross between a heterozygote (Aa) and a homozygote (aa), the probability of obtaining 6 offspring with exactly 4 dominant (Aa) and 2 recessive (aa) is 0.2344.

Step-by-step explanation:

The question involves predicting the genotypic ratio of offspring from a test cross between a heterozygote (Aa) and a homozygote (aa) for a single gene trait. Each offspring has a 50% probability (1/2 chance) of being dominant (Aa) or recessive (aa) in this scenario.

To determine the probability of getting exactly 4 dominant and 2 recessive offspring from 6, we use the binomial probability formula which is given by: P(x) = (n! / (x!(n - x)!)) * p^x * (1-p)^(n-x), where n is the total number of offspring, x is the desired number of dominant offspring, p is the probability of a single offspring being dominant, and (1-p) is the probability of an offspring being recessive.

Plugging in the values (n=6, x=4, p=0.5), we get:

P(4) = (6! / (4!(6 - 4)!)) * (0.5)^4 * (0.5)^(6-4)

Calculating the values gives us:

P(4) = (6 * 5 / (2 * 1)) * 0.0625 * 0.25

P(4) = 15 * 0.0625 * 0.25

P(4) = 0.234375 or 0.2344 to four decimal places

Therefore, the probability of obtaining 6 offspring with 4 being dominant and 2 being recessive in a test cross of Aa x aa is 0.2344.

Answer 2

Answer:0.0156

Step-by-step 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