Biology Forums - Study Force

Using this data could you help me figure out if these genes are linked and if they are what’s the map distance between them? 
This is a dihybrid cross working with fruit flies.  Parental generation is a sepia eyed male x apterous female.  The F1 are all wild type.  The F2 are as follows:
637 +;+   
175 +;se
63 ap;+
17 ap;se
Overall F2 ratio observed: 37 wild type: 10 sepia: 3 apterous: 1 apterous sepia.  Lewis suggested that this might be a case of epistasis. 

When in dihybrid crosses, the epistatic interactions occur between two genes, less than four phenotypes appear in F2. Such bigenic (two genes) epistatic interactions may be of following types:

1. Dominant Epistasis (12: 3 : 1)

- In your case, you get 37 : 10 : 3 : 1 or... 12:3:1

When out of two genes, the dominant allele (e. g., A) of one gene masked the activity of alleles of another gene (e. g., B) and expressed itself phenotypically, then A gene locus is said to be epistatic to the B gene locus. Because, the dominant allele A can express itself only in the presence of either B or b allele, therefore, such type of epistatis is termed as dominant epistatis. The alleles of hypostatic locus or gene B will be able to express themselves phenotypically only when gene locus A may contain two recessive alleles (aa). Thus, the genotype AA BB or Aa Bb and AA bb or Aa bb produce the same phenotype whereas the genotype aa BB or aa Bb and aa bb produce two additional phenotypes. The dominant epistasis modify the classical ratio of 9 : 3 : 3 : 1 into 12 : 3 : 1 ratio.

http://www.microbiologyprocedure.com/genetics/genetic-interaction/kinds-of-epistatic-interaction.htm (http://www.microbiologyprocedure.com/genetics/genetic-interaction/kinds-of-epistatic-interaction.htm)

A is the number of  flies with both dominant phenootypes

B is the umber of flies with one homozygous recessive phenotype

C is the number of flies with the other homozygous recessive phenotype

D is the number of flies with both homzoygous recessive phenotypes.

For coupling cross: Use the formula Z = B*C / A*D

For repulsion cross Z = A*D/ B*C

Map units = [recominants / total] * 100 = x map units

So using the formula above, I get 26 map units bio_man...

One centimorgan or "map unit" corresponds to about 1 million base pairs in humans on average, so does that mean they are ~26 map units apart?

Gene1                           Gene2
| ---------------------------------- |
               26 Map Unit

A recombinant frequency (RF) of 1 % is equivalent to 1 m.u.

I remember using this formula for this question (below) which I did during my undergrad where we did use this formula. It was for 2 traits (dihybrid) which are LINKED (similar to ur problem)...

Shaggy hair in dogs is dominant to its recessive allele, short hair, and black coat color is dominant to its recessive allele, tan. When heterozygous black, shaggy-haired dogs are crossed to short-haired tan dogs, progeny with the following phenotypes are obtained: 46% shaggy and black, 44% short and tan, 5% shaggy and tan and 5% are short and black. What is the distance between the hair length and hair color loci?

answer was 10 map units (or centimorgans)