. The R/r and S/s genes are linked and 20 map units apart. P generation is RR(s)(s) X (r)(r)SS ---> F1 R(r)S(s). The F1 is used for the test cross R(r)S(s) X (r)(s)/(r)(s) what fraction of the progeny will be R(r)S(s)?
5%, 10%, 20%, 40%, or 45%?
2. Among the offspring which is a parental type in this test cross?
R(r)(s)(s) & (r)(r)S(s),
RrS(s) & rr(s)(s),
rrS(s) and R(r)S(s),
Rr(s)(s) & rr(s)(s)
3. Use the chi square test to determine if the O and P genes linked. The P1 cross was OOpp x ooPP. The dihybrid is crossed to oopp (OoPp X oopp). The numbers show the outcome of each phenotypic class. All of the progeny are shown. The phenotype and counts of the offspring are OP 152, O(p) 348, (o)P 353, (o)(p) 147. How many degrees of freedom should you use in your chi square calculation?
4, 3, 2, or 1