Genetic linkage help

The R/r and S/s genes are linked and 10 map units apart. In the cross Rs/rS x rs/rs what percentage of the progeny will be Rs/rs?
incorrect

A. 5%
B. 10%
C. 25%
D. 40%
E. 45%

Could you explain this to me please with a cross so I know how to do it. Thank you :)

First you need to look at the parental chromosomes, to see which alleles are together on the same chromosome in this case Rs are on one chromosome and rS on the other one. These are known as parental types. Our second parent is always a double recessive so we can distinguish different genotypes. This is called a test cross.

When there is no recombination the parental gametes will either be Rs or rS and when combined with the rs recessive will give offspring Rs/rs or rS/rs. In this case where the map distance is 10 unites, our recombination frequency is 10% and so non recombinants are 90%, so we would expect 45% of each of those two genotypes.

When there is recombination we will end up with two new recombinant gametes RS and rs, When combined with our recessive alleles the genotypes will be RS/rs and rs/rs. These two groups together should be 10%, with 5% of each genotype expected.

In your question you were asked about Rs/rs, a parental type, so the expected frequency will be 45%.

10 map units = a total of 10% recombinants = 5% Rs/rs and 5% rs/rS.

Answer =A.