Final answer:
The probability that the first child of two a b / a b parents will be a b / a b is 16%, calculated by considering the 20% recombination frequency between the a and b loci and applying the principles of genetics.
Step-by-step explanation:
The question relates to genetic linkage and the calculation of recombination frequencies to predict the outcome of genetic crosses. If a and b loci are 20 map units apart, this indicates a 20% recombination frequency between these genes. In the given scenario, both parents are heterozygous (a b / a b), and we want to determine the probability that their offspring will also be heterozygous (a b / a b).
The probability of an offspring receiving an a allele from one parent is 0.5 (since each parent is heterozygous at this locus: Aa) and similarly, the probability of receiving a b allele from the same parent is also 0.5 (since each parent is heterozygous at this locus: Bb). However, due to the 20% recombination frequency, we need to take into account the recombinants, which occur at a rate of 20%.
The crossover event affects how alleles are inherited together. For alleles to be inherited as a b, they must either be inherited directly in their original configuration or result from a recombination event. Since there is a 20% recombination frequency, the chance of inheriting non-recombinant alleles is 80%, or 1 - 0.20. Therefore, the probability that the offspring will be a b / a b is (0.5 x 0.8) x (0.5 x 0.8), because each allele has a 50% chance of being passed on from each parent and the alleles are inherited in their original configuration 80% of the time.
Calculating this we get (0.5 x 0.8) x (0.5 x 0.8) = 0.4 x 0.4 = 0.16 or 16%. Thus, the probability that the child will inherit the a b / a b genotype is 16%.