All B N N AIIN AN 13. W 14. W 4 All B All B All B 2 B, N 15. W ty a. Is it reces

Question

All B N N AIIN AN 13. W 14. W 4 All B All B All B 2 B, N 15. W ty a. Is it recessive or dominant? b. Is it autosomal or sex-linked? c. What are the genotypes of all parents and progeny? Now which all flies have brown eyes are available. Similarly list of Inspe yours 31. The normal eye color of Drosophila is red, but strains in wings are normally long, but there are strains with short does wings. A female from a pure line with brown eyes and short wings is crossed with a male from a normal pure line. The Fi consists of normal females and short-winged 32. I males. An F2 is then produced by intercrossing the Fi. Both sexes of F2 flies show phenotypes as follows: 3 red eyes, long wings 3 red eyes, short wings brown eyes, long wings 1 brown eyes, short wings Deduce the inheritance of these phenotypes; use clearly defined genetic symbols of your own invention. State the genotypes of all three generations and the genotypic proportions of the Fi and F2.

Explanation / Answer

31.

1.Normal is used to mean wild type or red eye color or long wings .

2.Both line and strain are used to denote pure breeding fly stocks, and the words are interchangeable.

4.Three different characters are there:eye color, wing length and sex.

5.For eye color there are two phenotypes :red and brown, for wing length there are two phenotypes:long and short and for sex there are two phenotypes:male and female.

6.The F1 females designated normal have red eyes and long wings.

7.The F1 males that area called short winged have red eyes and short wings.

8.The F2 ratio is:

3/8 red eyes, long wings

3/8 red eyes, short wings

1/8 brown eyes, long wings

1/8 brown eyes, short wings

9.Because there is no expected 9:3:3:1 ratio, one of the factors that distorts the expected dihybrid ratio must be present. Such factors can be sex linkage, epistasis, genes on the same chromosome, environmental effects, reduced penetrance or a lack of a complete dominance in one or both genes.

10.With sex linkage traits are inherited in a sex specific way.With autosomal inheritance males and females have the same probabilities of inheriting the trait.

11.The F2 does not indicate sex-specific inheritance.

12.The F1 data does not show sex-specific inheritance-all males are short winged, like their mothers while all females are normal winged like their fathers.

13.The F1 suggests that long is dominant to short and red is dominant to brown. The F2 data shows a 3 red:1 brown ratio indicating the dominance of red but 1:1 long :short ratio indicative of a test cross. Without the F1 data it is not possible to determine which of the wing character is dominant.

14.If mandalian notation is used then the red and long alleles need to be designated with upper case letters for example R and L, while the brown(r) and short(l) needs to be designated with the lower case letters. If drosophila notation is used then the brown allele may be designated with the lower case b and wild type (red) with a b+.The short wind length gene with an s and wild type(long) allele with an s+.(Genes are often named with their mutant phenotype.

15.ToTo deduce the inheritance of these phenotypes means to provide all genotypes for all animal in the three generations discussed and account for the ratios observed.

Solution for the next problem:

Write crosses and results to clear the details:

P brown short female,red long male

F1 red long females

Red short males

These results tell you that red eyed is dominant to brown eyed and since both females and males are red eyed ,this gene is autosomal. Since males differ from females in their genotype with regard to the wing length, this trait is sex-linked. Knowing that drosophila females are XX and males are XY the long winged females tells us that long is dominant to short and that the gene is X linked.

Let B=red, b=brown, S=long and s=short.The cross can be written as follows:

P b/b,s/s'   B/B,S/Y

F1 1/2B/b S/s females

1/2 B/b. s/Y males

F2. 1/16B/B S/s red long females

1/16 B/ B s/s red short female

1/8 B/b s/s. Red long female

1/8 B/b s/s. Red short female

1/16 b/b S/ s . Brown long female

1/16 b/b s/s. Brown, short, female

1/16 B/B S/Y. Red, long, male

1/16 B/B s/Y. Red, short, male

1/8 B/b S/Y. Red, long, male

1/8 B/b . s/Y. Red, short, male

1/16 b/b .S/Y. Brown, long, male

1/16 b/b s/Y. Brown, short ,male

The final phenotype ratio is:

3/8 red, long

3/8 red, short

1/8 brown, long

1/8 brown, short

With equal number of males and females in all classes.

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