Question

In Drosophila, a gene for bristle shape, thread, is linked to a gene for eye color, sepia. A wild type phenotype (normal bristles, red eyes) F1 female fruit fly that is heterozygous for the thread bristles (th+/th) trait and the sepia eye color (se+/se) trait is mated to a th se/th se tester male. 1000 F2 offspring are of 4 types in the following numbers: 408 normal bristles & red eyes, 402 thread bristles & sepia eyes, 98 normal bristles & sepia eyes, and 92 thread bristles and red eyes.

For each of the following statements, fill-in the blank with True or False (type in the entire word to receive credit).
A) Considering the data from the cross, the mutant alleles were in coupling (cis) in the F1 female:
a. True
b. False

Answer 1

The statement that the mutant alleles were in coupling (cis) in the F1 female is false based on the frequencies of the phenotypes observed in the F2 offspring.

Step-by-step explanation:

The statement provided states that the F1 female fruit fly is heterozygous for the thread bristles (th+/th) trait and the sepia eye color (se+/se) trait. This means that the alleles for both traits are located on the same chromosome. To determine if the alleles are in coupling (cis) or repulsion (trans), we can compare the phenotypes of the F2 offspring. In this case, the F2 offspring consist of four types: normal bristles & red eyes, thread bristles & sepia eyes, normal bristles & sepia eyes, and thread bristles and red eyes. If the mutant alleles were in coupling (cis) in the F1 female, we would expect to see a higher frequency of individuals with the mutant phenotypes together (thread bristles & sepia eyes and normal bristles & red eyes). However, the data shows that the frequencies of individuals with the mutant phenotypes together (thread bristles & sepia eyes and normal bristles & red eyes) are lower than the frequencies of individuals with the wild-type phenotypes together (normal bristles & red eyes and thread bristles & sepia eyes). Therefore, the statement is False.