8. A homozygous strain of Drosophila that has dumpy (dp) wings is crossed to a h

Question

8. A homozygous strain of Drosophila that has dumpy (dp) wings is crossed to a homozygous strain that expresses cinnabar (cn) eyes and shaven bristles (sv). The F1 are all phenotypically wild-type. F, females are then test crossed to a strain that is homozygous for all three of these recessive mutants. The results of this cross follow: Phenotype # observed | Phenotype # observed | Phenotype #observed | Phenotype # observed +cn+208 dp cn46 +cn sv 202 dp cn sv 54 ++SV dp sv 200 dp ++ 197 35 Total = 1000 What is the map distance between dumpy and cinnabar? a. 7.4 b. 14.8 c. 50 d. 19.3 e. Genes are not linked 9. What is the map distance between cinnabar and shaven? b. 14.8 a. 7.4 C. 50 d. 19.3 e. Genes are not linked

Explanation / Answer

Hi,
The genotype of parents are:
dpdp CNCN SVSV   x    DPDP cncn svsv
    (dpdp ++ ++)        (++ cncn svsv)

F1: DPdp CNcn SVsv

Test cross:    DPdp CNcn SVsv x dpdp cncn svsv

So, the parental type = dp + + (197 in the given table )
another parental type = + cn sv (202 in the given table)

Now, if the cross over takes place between dp and + , then we get,
dp cn sv (35) and
+ + + (58)

If cross over occurs between cn and sv =
dp + sv (200) and
+ cn + (208)

If cross over occur between both the places ( Double cross over DCO) =
dp cn + (46) and
+ + sv (54)

Recombination frequency between dp and cn
= cross over products of dp and cn + DCO / total
= 35+58+46+54 / 1000
= 0.193
Gene map = RF x 100
= 0.193*100
= 19.3 mU

similarly :
Recombination frequency between dp and cn
= cross over products of cn and sv + DCO / total
= 200+208+46+54 / 1000
= 0.50
Genetic map = 50 mU.

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