GENE20001 Lecture Notes - Lecture 4: Wild Type, Zygosity
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In this diagram above we see that the disease causing allele is linked to SSR-A5
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Expect incorporation of molecular markers in linkage analyses
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The mutant phenotype black wing, red eye and wing-less exist
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True breeding black wing female mated with male showing red eye and wingless
phenotypes. Half of F1 progeny show wing-less phenotype. Wing-less females progeny
mated with male showing red eyes and black wings. 198 progeny from this cross were
scored phenotypically:
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Wildtype
12
Wing-less
48
Red eye
38
Black wing, red eye
11
Red eye, wing-less
50
Black wing
39
The recombination frequency between black wing and red eye is
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The recombination frequency between black wing and wing-less is
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The recombination frequency between wing-less and red eye is
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Because winglessness appears in the F1it is probably dominant. WL=wingless,
wl+=winged.
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Since only half f F1is wingless the father must be a heterozygote (WL/wl+)
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Wild type b+ r+ wl+
12
Wingless ? r+ WL
48
Red eye b+ r wl+
38
Black wing, red eye b r wl+
11
Red eye, wingless ? R WL
50
Black wing b r+ wl+
39
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Wingless could be b+ r+ WL or b r+ WL
Question:
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Genetics Page 4
Document Summary
In this diagram above we see that the disease causing allele is linked to ssr-a5. Expect incorporation of molecular markers in linkage analyses. The mutant phenotype black wing, red eye and wing-less exist. True breeding black wing female mated with male showing red eye and wingless phenotypes. Wing-less females progeny mated with male showing red eyes and black wings. 198 progeny from this cross were scored phenotypically: The recombination frequency between black wing and red eye is. The recombination frequency between black wing and wing-less is. The recombination frequency between wing-less and red eye is. Because winglessness appears in the f1 it is probably dominant. Since only half f f1 is wingless the father must be a heterozygote (wl/wl+) Wingless could be b+ r+ wl or b r+ wl. Red eye wingless could be b+ r wl or b r wl. To work out the amount of each just take the number of the reciprocal (exact opposite genotype)
