Assume a strain of flies carrying the dominant eye mutation ‘Lobe’ (L) was cross

Question

Assume a strain of flies carrying the dominant eye mutation ‘Lobe’ (L) was crossed with a strain homozygous for recessive mutations ‘smooth abdomen, straw body’ (ab, stw). All three genes are carried on the second chromosome. The F1 L females were then crossed with homozygous ab, stw males, and the following phenotypes were observed: smooth abdomen, straw body 820 ____________ Lobe 780 ____________ smooth abdomen, Lobe 42 ____________ straw body 58 ____________ smooth abdomen 148 ____________ Lobe, straw body 152 ____________ 6. List the FULL phenotypes of the parents: 7. List the FULL phenotypes of the F2s in the blanks above. 8. What is the gene order and map units between these three loci? SHOW ALL OF YOUR WORK

Explanation / Answer

Here we can consider the wild type phenotypes with "+" for abdomen, eyes and body.
The complete phenotypic representations are as follows:-
smooth abdomen, straw body 820 = ab stw +
Lobe 780 = + + L
smooth abdomen, Lobe 42 = ab + L
straw body 58 = + stw +
smooth abdomen 148 = ab + +
Lobe, straw body 152 = + stw L

Missing phenotypes-

These are result of single crossover between ab/L.
straw body, smooth abdomen, lobe= stw ab L
wild type= + + +

Here we can see that the maximum frequencies are of ab stw + and + + L, which are the PARENTAL PHENOTYPES. And the least are + stw + and ab + L, which are the DOUBLE CROSS OVERS.

All these genotypes are arranged in same order i.e. abdomen--body--eyes.
According to the arrangement of the maximum and the least frequencies, we can conclude that the order of gene is-
EYES--ABDOMEN--BODY.

Distance between stw/ab = 148+152+42+58 / = 2000 = 20 cM.

Distance between L and ab can't be determined because the phenotypic frequencies are not mentioned in the question.

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