Consider the following data from a hypothetical test-cross for linkage of an ima

Question

Consider the following data from a hypothetical test-cross for linkage of an imaginary laboratory insect. Assume Morgan's typical annotation, where a superscript + (+) = wild type phenotype, and lower-case letters indicate that the mutation is recessive. Genes: b = black body mutation vg = vestigial wings mutation P0 cross: (male parent) b vg+/b vg+ × (female parent) b+ vg/b+ vg F1 offspring: b vg+/b+vg F1 testcross: b vg+/b+ vg × b vg/b vg (homozygous recessive individual) F1 testcross sample results: offspring types number: b vg+/b vg 243 b+ vg+/b vg 233 b+ vg/b vg 257 b vg/b vg 267 What is the map distance between these two genes?

Explanation / Answer

Answer:

F1 testcross: b vg+/b+ vg X b vg/b vg (homozygous recessive individual)

b vg+/b+ vg genotype produces recombinant gametes as b vg & b+ vg+ and non-recombinanat gametes as b vg+ & b+ vg.

Recombination frequency RF = (no. of recombinants / Total progeny) 100

RF = (267+233 / 1000)100

= (500/1000)100

= 50%

RF (%) = Map distance (m.u.)

So the map distance between these two genes = 50 map units

b vg b vg b vg / b vg---267 (recombinant) b+ vg+ b+ vg+ / b vg--233 (recombinant) b vg+ b vg+ / b vg -243 (non-recombinant) b+ vg b+ vg/ b vg---257(non-recombinant)

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