Question

Suppose we have a situation where we are dealing with 4 genes. The capital letter is representing a dominant allele and a lower-case letter represents a recessive allele. If a cross is performed with parents BbHhSsGg x BBHhSsgg, what is the probability of getting an F1 individual with the BbhhSSGg genotype?

*Enter your response as either a fraction in the lowest common denominator to earn credit OR a decimal rounded to the nearest one hundredth when necessary. Example: 1/4 or 0.25 would be correct. If an answer given was 2/8 the answer does not follow the directions and stays incorrect.

Answer 1

The probability of getting an offspring with the BbhhSSGg genotype from the given cross is 1/64 or 0.015625. Each gene is considered independently, and the individual probabilities are multiplied to find the joint probability.

Step-by-step explanation:

The question involves calculating the probability of obtaining an offspring with the BbhhSSGg genotype from a cross between parents with BbHhSsGg and BBHhSsgg genotypes. To find this probability, we need to look at each gene independently because of the law of independent assortment.

For the B gene, the chance to get Bb is 1/2 since one parent is Bb and the other is BB. Only the Bb parent can pass down the 'b' allele.

For the H gene, getting hh is 1/4 because there is a 1/2 chance of getting 'h' from each Hh parent (1/2 * 1/2 = 1/4).

For the S gene, getting SS when crossing Ss with Ss is 1/4, again by the product of independent probabilities for each parent to provide the dominant 'S' allele (1/2 * 1/2).

For the G gene, Gg is achieved with a probability of 1/2 because one parent is Gg and the other is gg. Only the Gg parent can contribute a 'G' allele.

Lastly, we find the joint probability by multiplying the individual probabilities together: (1/2) * (1/4) * (1/4) * (1/2) = 1/64. Therefore, the probability of getting an offspring with the BbhhSSGg genotype is 1/64 or 0.015625 when converted to a decimal.