The type of wounds that are healing and the dressing that are being examined.
Answer:
10%
Explanation:
The distance between the given two loci is 20 mu. This means that the recombination frequency between these two loci is 20% and the heterozygote will form 20% recombinant gametes and rest 80% parental type gametes. Therefore, the heterozygous woman AB/ab will form a total of 20% recombinant gametes. The proportion of each recombinant gamete will be 20/2= 10%.
Gametes formed by the woman= 10% Ab, 10% aB, 40% AB, and 40% ab
Gametes formed by the man= All ab (since the man is homozygous for both loci).
Therefore, proportion of Ab/ab genotype= 10% Ab (female gamete) x ab (male gamete)= 10% Ab/ab
