First, the recombination equation is:
cM = recombinants/(recombinants AND parentals) * 100
Let's assign the following genotypes for clarity's sake:
RR = black bristles
Rr = black bristles
rr = red bristles
SS = pebbly eyes
Ss = pebbly eyes
ss = shiny eyes
F1 cross: rrSS x RRss = RrSs (all progeny)
F2 cross: RrSs (F1) x rrss = ?
You would have the following gametes:
Parent 1, RrSs = RS, Rs, rS, rs
Parent 2, rrss = rs, rs, rs, rs
Draw your punnet square and you get the following genotypes of the offspring.
So your GENOTYPES of your F2 cross will be RrSs, Rrss, rrSs, rrss
Let's translate those to phenotypes.
RrSs: black pebbly (parental)
Rrss: black shiny (recombinant)
rrSs: red pebbly (recombinant)
rrss: red shiny (parental)
Now let's look at recombination frequency.
15cM = (recombinants/(recombinants+parentals)*100
recombinants/(parentals+recombinants) = 15/100
This tells you for every 15 recombinants, you have 85 parentals. If you scale this up to 1000, that's 150 recombinants for every 850 parentals.
Your 150 recombinants comes from 75 black/shiny and 75 red/pebbly.
Your 850 parentals ceoms from 425 black/pebbly and 425 red/shiny.
Anyways that was my attempt which appears to be directly opposite to what your answer key says. Did you copy it down correctly?