Phenotypes: Numbers Observed: spineless 321 wild 38 claret, spineless 130 claret
ID: 94024 • Letter: P
Question
Phenotypes: Numbers Observed:
spineless 321
wild 38
claret, spineless 130
claret 18
claret, hairless 309
hairless, claret, spineless 32
hairless 140
hairless, spineless 12
2. How many sets/pairs of reciprocal phenotypes are shown?_________________
3. How many mutant traits are being studied?____________________________
4. This chart represents F2 phenotypes obtained from a ____________ (choose one: single, double, triple) crossover.
5. What is the most likely allelic arrangement of the phenotypically wildtype female parent that was crossed with a male homozygous recessive for all traits?
6. What is the genetic distance between the genes claret and hairless? Show all of your work
Explanation / Answer
Answer 2- There are three sets of phenotyopes. These are wing bristles, body type and eyes.
Answer 3- There are three mutants here, namely spineless means wings are without bristles, claret means darker shade of eyes and hairless means body has no bristles.
Answer 4- This is a result of double crossover. We can see this from the minimum result we got, i.e. claret 18 and hairless spineless 12. These are the results of double cross overs. In double crossovers, we get 8 phenotypes at the end, and this is what we have in the question.
Answer 5-
Parental gentoypes-
Spineless 321 = s + +
Claret, hairless 309 = + c h
DCO-
Claret 18 = + c +
hairless, spineless 12 = s + h
By comparing the two above genotypes we can find the allelic arrangement, which was as follows:
s ---- h ---- c
Spineless ----hairless ---- claret.
It means hairless allele is in between.
Answer 6-
Distance between c and h = SCO + DCO / Total
= 130 + 140 + 12 + 18 / 1000 = 300/1000 = 0.3 or 30 cM.