Question

In: Biology

In Drosophila melanogaster, black body (b) is recessive to gray body (b+), purple eyes (pr) are...

In Drosophila melanogaster, black body (b) is recessive to gray body (b+), purple eyes (pr) are recessive to red eyes (pr+), and vestigial wings (vg) are recessive to normal wings (vg+). These three genes are linked with pr located between b and vg at a genetic distance of 6 cM from b and 13 cM from vg. Previous experiments have shown that the interference associated with these three genes is 0.5. A fly with a black body, purple eyes and vestigial wings is crossed with a fly with a gray body, red eyes and normal wings. All of their F1 progeny is wild-type. The F1 females are mated with male flies that have black bodies, purple eyes and vestigial wings. Ten thousand progeny are produced from this testcross. How many of the testcross progeny are expected to have purple eyes but be wild-type for body color and wing type?

A. 19.5 B. 39 C. 78 D. 280.5 E. 561

Expert Solution

Interference = 1- Coefficient of coincidence

So Coefficient of coincidence = 1-interference

=1-0.5

0.5

Coefficient of coincidence = 0.5

Coefficient of coincidence is given by

The observed frequency of double recombinant/ expected frequency of double recombinant

Map distance between color and body type gene is 10cM

Map distance between body type and disposition is 14cM

expected frequency of double recombinant is given by

Map distance between color and body type /100 * map distance body type and disposition/100

6/100 * 13/100 = 0.0140

Observed frequency of double recombinant we need to find out

Coefficient of coincidence = the observed frequency of double recombinant/ expected frequency of double recombinant

0.5 = X /0.0078

X = 0.0039

The observed frequency of double recombinant = sum of the frequency of both the double recombination

Since nothing is given about the progeny numbers so we will take the double recombinant frequency as half of the total double recombinants that we got.

So the frequency will be 0.0039/2 = 0.00195

black body, purple eyes and vestigial wings and gray body, red eyes and normal wings are the parent type.

purple eyes black body normal wing is the phenotype.

Gene arrangement in wild-type is

b pr vg

b+ pr+ vg+

b pr+ vg

Since the purple gene is the middle gene so it can only be translocated to another chromosome if has undergone double crossover.

So the frequency of a purple eyes black body normal wing is the phenotype. will be 0.00195

Given that the total number of progenies are 10000

so purple eyes black body normal wing will be 10000*0.00195

19.5

So the answer is option A.

gladiator answered 1 year ago

Black body (b) and purple eye (pr) are recessive autosomal mutations in Drosophila. Bridges crossed b/b...

Black body (b) and purple eye (pr) are recessive autosomal mutations in Drosophila. Bridges crossed b/b females and pr/pr males. F2 cross produced 684 wild-type, 371 black-bodied, and 300 purple-eyes flies. Do these result indicate that the b and pr genes are closely linked? Explain. Remember that there is no crossing over in male Drosophila

In Drosophila black body (B) is dominant to gray (b) and long wings (L) is dominant...

In Drosophila black body (B) is dominant to gray (b) and long wings (L) is dominant to short wings (l). The F1 flies were produced by crossing a homozygous black, homozygous long winged fly with a gray short winged fly. (4) a) Give the genotypes of the Parents and the F1. b) If an F1 is testcrossed, what is the expected phenotypic ratio? c) If the F1 are allowed to interbreed what are the expected phenotypes and probably proportions in...

In Drosophila , a recessive autosomal allele, ebony (e) produces a black body colour when homozygous....

In Drosophila , a recessive autosomal allele, ebony (e) produces a black body colour when homozygous. Another recessive autosomal allele, black (b) also produces a black body colour when homozygous. Wild type flies are grey. Flies with genotypes e/e b+/-, e+/- b/b, and e/e b/b show identical colour phenotype. If e/e b+/b+ ebony flies are crossed with e+/e+ b/b black flies than; c) What phenotypic ratios would be observed from the crosses F1 x true-breeding ebony, and F1 x true-breeding black? d) What would be...

32) Normal (wild-type) Drosophila eyes are red. In a cross in Drosophila involving the X-linked recessive...

32) Normal (wild-type) Drosophila eyes are red. In a cross in Drosophila involving the X-linked recessive eye mutation white (resulting in a white eye) and the autosomal recessive eye mutation sepia (resulting in a dark eye), predict the sex and color of the eyes of the offspring from crossing true-breeding parents for parts A) and B) below. White is epistatic to the expression of sepia (i.e., if an eye is white, it cannot also be sepia; i.e., white masks the...

In D. melanogaster, a homozygous female with mutations for vestigial wings (vg), black body (b) and...

In D. melanogaster, a homozygous female with mutations for vestigial wings (vg), black body (b) and purple eye color (pr) was mated to a homozygous wild-type male. All the F1 progeny, both male and female, had normal (wild-type) phenotypes for the three characteristics, indicating that the mutations are autosomal recessive. In a testcross of the F1 females with males that have vestigial wings, black body and purple eyes, the progeny were of eight different phenotypes reflecting eight different genotypes. The...

2. In panthers, black coat (g) is recessive to golden coat (G), pink eyes (b) is...

2. In panthers, black coat (g) is recessive to golden coat (G), pink eyes (b) is recessive to blue eyes (B), and short tail (l) is recessive to long tail (L). A female heterozygous for all three genes is test crossed to a homozygous recessive male and the following offspring were obtained: BGL 70 BgL 78 bGL 358 BGl 4 bgL 8 Bgl 344 bGl 70 bgl 68 a. How many genes are linked? b. Which groups are the parentals?...

4) In fruit fly (Drosophila melanogaster), grayish eye color and wingless bodies are caused by recessive...

4) In fruit fly (Drosophila melanogaster), grayish eye color and wingless bodies are caused by recessive genes located on different chromosomes. Red eyes (gr+) and normal wings (wl+) are dominant to gray eyes (gr) and wingless (wl), respectively. Suppose that a true-breeding male with gray eyes and normal wings is crossed with a true-breeding female with red eyes and wingless. a) What is the genotype and phenotype of the F1? b) If these F1 individuals were allowed to intermate, what...

In Drosophila the recessive alleles for white eyes is X-linked (normal eye color is red) and...

In Drosophila the recessive alleles for white eyes is X-linked (normal eye color is red) and the recessive allele for vestigial wings is autosomal. A female heterozygous for both of these traits is crossed with a white-eyed vestigial male. What proportion of the FEMALE progeny will be white-eyed with vestigial wings.? (a) 100 % (b) 75 % (c) 50 % (d) 25 % (e) 0 %

The allele b gives Drosophila flies a black body, and b+ gives brown, the wild-type phenotype....

The allele b gives Drosophila flies a black body, and b+ gives brown, the wild-type phenotype. The allele wx of a separate gene gives waxy wings, and wx+ gives nonwaxy, the wild-type phenotype. The allele cn of a third gene gives cinnabar eyes, and cn+ gives red, the wild-type phenotype. A female heterozygous for these three genes is testcrossed, and 2,424 progeny (you could also use n=2,423) are classified as follows: cn+ · wx+ · b 101 cn · wx+...

In Drosophila, the X-linked recessive mutation vermillion (v) causes bright red eyes, in contrast to the...

In Drosophila, the X-linked recessive mutation vermillion (v) causes bright red eyes, in contrast to the brick-red eyes caused by the wild-type allele. A separate autosomal recessive mutation, brown (bw), causes flies to have brown eyes. Flies with both mutations have no eye pigmentation and are white-eyed. Determine the F1 and F2 phenotypic ratios for each sex from the following crosses: (a) vermillion females X brown males (b) wild-type females X white-eyed males