Drosophila melanogaster has one pair of sex chromosomes (XX or XY) and three pairs of autosomes, referred to as chromosomes II, III, and IV. A genetics student discovered a male fly with very short (sh) legs. Using this male, the student was able to establish a pure breeding stock of this mutant and found that it was recessive. She then incorporated the mutant into a stock containing the recessive gene black (b, body color located on chromosome II) and the recessive gene pink (p, eye color located on chromosome III). A female from the homozygous black, pink, short stock was then mated to a wild-type male. The F₁ males of this cross were all wild type and were then backcrossed to the homozygous b, p, sh females. The F₂ results appeared as shown in the following table. No other phenotypes were observed. Wild Pink* Black, Black, Pink, Short* Short Females 63 58 55 69 Males 59 65 51 60 *Other trait or traits are wild type. Based on these results, the student was able to assign short to a linkage group (a chromosome). Which one was it? Include your step-by-step reasoning.

Video transcript

Hello everyone here. We have a question telling us that genetic linkage is the possibility of genes that are located close together on a chromosome being inherited together during diagnosis of sexual reproduction. And our question is linkage likely occurs if the recombination frequencies are blank, so we're looking for linkage to likely occur. So recombination frequencies can determine whether genes are linked or not. Linkage is likely to occur when recombination frequencies are less than 50%,, And that makes sense because it is less than half, but it is likely not to occur if the recombination frequencies are closer to or equal 50%. So our answer here is C less than 50%. Thank you for watching. Bye.