3. Wild-type mice are black. Crosses of pure-breeding parents produced the follo

Question

3. Wild-type mice are black. Crosses of pure-breeding parents produced the following F1 generations, and intercrossing the F1 from each cross gave the ratios shown for the F2 generation. The phneotypes are: Bl (black), Y (yellow), Br (brown), and W (white). Cross Parents F1 F2 1 Bl x Br Bl 3 Bl: 1 Br 2 Bl x Y Bl 3 Bl: 1 Y 3 Br x Y Br 3 Br: 1 Y 4 W x Br Bl 9 Bl: 3 Br: 4 W 5 W x Y Bl 9 Bl: 3 Y: 4 W 6 Bl x W Bl 9 Bl: 3 Y: 4 W

a. From these results, determine the inheritance of these colors. b. What are the genotypes of the parents, F1, and F2 offspring of each cross?

Explanation / Answer

Crosses 1-3 show that there is normal monohybrid dominance between the three alleles involved, with Black dominant to Brown, and Yellow recessive to both. Calling the alleles B for black and R for brown, with lowercase for non-functional recessive alleles and - for any allele. A genotype of B--- gives black phenotype, a genotype of bbR- gives brown and a genotype of bbrr gives yellow. For crosses 4-6, a third allele with an epistatic role is introduced, W- is normal and ww is no colouration (i.e. white). So for all crosses, the parental white genotype is BBrrww Cross 4: x bbrrWW (so we can ignore the r allele). This gives an F1 of BbWw, which is black. The F2 is then 9 B-W-, 3x bbW-, and 4--ww. Cross 5: x bbRRWW. This gives an F1 of BbRrWw, which is black. The F2 is then 9 B---W-, 3x bbR-W-, and 4--ww. For cross 6 the white genotype must be different, bbrrww (otherwise the progeny ratios don't make sense): x BBrrWW (so we can ignore the r allele). This gives an F1 of BbWw, which is black. The F2 is then 9 B-W-, 3x bbW-, and 4--ww.

Related Questions

Navigate

• Browse (All)
• Browse 3
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.