Question

In a cross in Drosophila, a female heterozygous for the autosomally linked genes a, b, c, d, and e (abcde/ +++++) was testcrossed with a male homozygous for all recessive alleles. Even though the distance between each of the loci was at least 3 map units, only four phenotypes were recovered, yielding the following data:
Table showing four Drosophila phenotypes with corresponding fly counts totaling 1000: 440, 460, 48, and 52 flies.

Table showing four Drosophila phenotypes with corresponding fly counts totaling 1000: 440, 460, 48, and 52 flies.

Why are many expected crossover phenotypes missing? Can any of these loci be mapped from the data given here? If so, determine map distances.

Accepted Answer

Step 1: Identify the parental and recombinant phenotypes from the data. The two most frequent phenotypes (+++++ with 440 flies and abcde with 460 flies) represent the parental types, while the less frequent phenotypes (++ ++e with 48 flies and abcd+ with 52 flies) represent single crossover recombinants. The absence of other recombinant phenotypes suggests double or multiple crossovers are not observed. Step 2: Understand why many expected crossover phenotypes are missing. Since the genes are autosomally linked and the distance between each locus is at least 3 map units, double or multiple crossovers are expected but not observed here. This is likely due to crossover interference, where one crossover event reduces the probability of another nearby crossover, leading to fewer double crossover phenotypes than expected. Step 3: Determine which loci can be mapped from the data. Because only single crossover classes involving the last gene 'e' are observed, it suggests that only the distance between gene 'd' and gene 'e' can be reliably estimated from this data. Other distances cannot be mapped accurately without additional recombinant classes. Step 4: Calculate the recombination frequency between loci 'd' and 'e'. Use the formula for recombination frequency (RF):

$$	ext{RF} = \frac{	ext{Number of recombinant offspring}}{	ext{Total number of offspring}} 	imes 100$$

Here, the recombinant offspring are those with phenotypes ++++e and abcd+, totaling 48 + 52 flies. Step 5: Convert the recombination frequency to map units (centiMorgans, cM). Since 1% recombination frequency corresponds to 1 map unit, the distance between loci 'd' and 'e' is numerically equal to the recombination frequency calculated in Step 4.

비슷한 문제에 대한 검증된 영상 답변:

주요 개념

Genetic Linkage and Recombination

Testcross and Phenotypic Ratios

Interference and Double Crossovers