Fluff is a mutation in Drosophila Melanogaster with fine sho

Fluff is a mutation in Drosophila Melanogaster with fine, short bristles. Carmine eyes is a mutation that has a different color than the normal brick red, and ocellarless is a mutaton with ocellars missing. A drosophilia male with ocellarless is crossed to a female with carmine eyes and fluff bristles. F1 males all have carmine eyes. F1 Females are all ocellarless. F1 females with Ocellarless are crossed to males that are homozygous or hemizygous for the recessive allele at each locus segregrating in the F1 female and produce the following progeny:

59 Ocellarless, Fluff, Carmine

52 Ocellarless, Carmine

167 Ocellarless

163 Carmine

155 Ocellarless, Fluff

156 Fluff, Carmine

55 Fluff

57 Normal

P1: Which mutant alleles, if any, are dominant to the respective normal allele?

P2: Which mutant gene alleles, if any, are x-linked?

P3: How many loci are segregating (heterozygous) in the F1 female?

P4: What is the chi-square value and the degree of freedom for the data above?

P5: If any genes are linked, what is the gene order?

P6: What is the total length of the gene map in Question 5?

Please help, this is due soon and I can figure it out for the life of me.

Solution

Let the symbol for Carmine eyes be “c”; fluff be ‘f’ and ocellarless be ‘o’.

A drosophila male with ocellarless (++/oo++) is crossed to a female with carmine eyes and fluff bristles (cc/++ff).

All F1 males have carmine eyes (cc). This means that the carmine mutation for eye color is X-linked, and is present on the X-chromosome of the mother; which is inherited by the F1 males.

F1 Females are all ocellarless. Now, if fluff was also X-linked, then all F1 males would have been fluff. This is not the case. So, the genes for fluff and ocellarless are autosomal.

This means that ‘o’ mutation is autosomal and dominant over ‘f’. An individual with ‘of’ genotype will be ocellarless, and that with “o+” genotype will also be ocellarless. This is interpretation of F1 results. Please note that F2 results are showing that ocellarless and fluff are codominant. There are 59 phenotypes with ocellarless, carmine and fluff. This further implies that ocellarless and fluff, both are dominant over there normal alleles (Answer P1). I think some misprint is there. Genotype ‘of’ will is also resulting in ocellarless.

Results: Males- c (carmine eyes. Note: Information about ocellarless/fluff in F1 males is not given. Seems they are all fluff).

Females: oo (ocellarless. Note: Information about eyes is not given. This means that the eyes are wild type. This information is required for further explanation).

F1 females with Ocellarless(c+/O_) are crossed to males (c/oo) that are homozygous or hemizygous for the recessive allele at each locus segregating in the F1 female and produce the given progeny.

P1: Carmine mutation is dominant to normal allele. This is because normal allele phenotypes are 57, and that of carmine are 163. Similarly, fluff is also dominant to normal allele.

P2: Carmine is X-linked. That is why F1 males are all carmine. They inherited the X-linked character from their mothers, not fathers.

P3: It is given that all F1 females are ocellarless. They have taken this character from their fathers, and their mothers were fluff. So, F1 females are heterozygous at ocellarless locus. But ocellarless and fluff is codominant, that is why, in F2 result, 59 are ocellarless, fluff and carmine.

So, F1 females are all ocellarless. Information about carmine locus is not given. This means that either they are wild type. So, they cannot be heterozygous at this locus also.

So, F1 females are not heterozygous. They are homozygous.

P4: For further calculations, complete genotype information of F1 females and males is required.

** Hope it helps.