Final answer:
The expected frequency of the AaBBCC genotype in the progeny of a cross between two AaBbCc individuals is 1/32, calculated by multiplying the probabilities of each gene's contribution using the product rule (1/2 for Aa, 1/4 for BB, and 1/4 for CC). Therefore , the correct answer options is a)
Step-by-step explanation:
The question asks about the expected frequency of the genotype AaBBCC in the progeny of a cross between individuals with genotype AaBbCc and AaBbCc. To find this, we use the principles of Mendelian genetics, specifically the product rule of probability.
For gene A, the probability of an offspring being Aa is 50% (1/2) because it results from either an A from one parent and a from the other, or vice versa. For gene B, the probability of an offspring being BB (homozygous dominant) is 25% (1/4) as it can only come from a B from both parents. For gene C, the probability of an offspring being CC is again 25% (1/4), for the same reason as with gene B.
To find the probability of the combination of AaBBCC, we multiply the probabilities for each gene together, using the product rule. 1/2 (for Aa) multiplied by 1/4 (for BB) multiplied by 1/4 (for CC) equals 1/32. Therefore, the expected frequency of AaBBCC progeny in the cross between two AaBbCc individuals is 1/32.