In sweet pea, flower color is controlled by two genes which are epistatic to each other- c and p are recessive alleles that encode a defective enzyme, and C and P are the respective dominant alleles. A true-breeding white variety CCpp is crossed to a true-breeding white variety ccPP. Diagram a test cross of the resulting F1, and Label the obtained phenotypic classes.
The defective enzyme part is what is confusing me...I did the 1x4 test cross but don't understand my results.
Given that the recessive alleles c and p are epistatic to each other. Thus, note that whenever we have cc__ or __pp, the phenotype of flower will be white.
Color phenotype will be observed only when at least a single copy of both the dominant alleles are present in an individual (C_P_).
Parental cross: CCpp × ccPP
Gametes: Cp, and cP.
Resulting F1: CcPp.
The test cross is done with a recessive individual, ccpp.
The required cross:
CcPp × ccpp
Gametes: CP, Cp, cP and cp from CcPp, and cp from ccpp parent.
CP | Cp | cP | cp | |
cp | CcPp (colored) | Ccpp (white) | ccPp (white) | ccpp (white) |
Thus, the phenotypic ratio obtained will be 1 (colored) : 3 (white), as shown in the punnett square above.
Get Answers For Free
Most questions answered within 1 hours.