Question

genetics

In rabbits, there are many genes that control coat appearance. The EN gene controls whether the rabbits are solid or spotted. The C gene controls whether rabbits tend to be brown or grey. The alleles of these genes follow a simple dominant/recessive pattern. The following test-cross/mapping-cross was set up: spotted and brown rabbits x solid and grey rabbits. The resulting offspring and numbers of each type are listed here... Offspring of test cross: 38 solid, grey 28 spotted, brown 70 solid, brown 64 spotted, grey If you assume that the gene for pattern and the gene for color are linked, what is the genetic distance between these genes? (answer in M... just list the number)

what is the distance between these genes? also set up a chi square test to examine data. show chi square value and p value. please show work.

Answer 1

The data does not seems to match for the results expected, there is something wrong in either the offspring frequencies or the parents genotype, but let us apply what has to be done:

1.- Linkage distance

Neither the genotype of the parents nor which trait is recessive and dominant in each loci is given, and that is important information, but given the phenotypes of the parents then it means spotted and brown are dominant traits while solid and grey are recessive, and the first parent must be heterozygous. Let us write down their genotype:

Spotted brown: En C / en c

Solid grey: en c / en c

Now we have to label the different offsprings phenotypes as either Parental or Recombinant. The parentals will be either EN C or en c (that is either spotted brown or solid grey, but not the mixture between them). Let us label them then:

38 solid, grey P 28 spotted, brown P 70 solid, brown R 64 spotted, grey R

Now, we have to divide the number of recombinants by the total number, and then multiply it by 100:

((70+64)/200)(100) = 67

The map units are 67. That actually indicates the loci are not linked

2.- Chi square

To solve this one we are need both the observed data (which we already have) and the expected data. Let us calculate the expected data by making a punnet square:

	EN C	EN c	en C	en c
en c	EN en Cc	EN en cc	en en Cc	en en cc

That means we have 1/4 for each phenotype, and considering the total number of offspring is 200, then the expected for each phenotype is 50 (200 x 1/4 = 50).

The chi square formula is:

$X^{2}= \sum \frac{(O-E)^{2}}{E}$

Phenotype	O	E	(O-E)2/E
Solid grey	38	50	2.88
Spotted brown	28	50	9.68
Solid brown	70	50	8
Spotted grey	64	50	3.92
Sum			24.48

Our chi square value is 24.48, we have 3 degrees of freedom (number of alleles minus 1), and by looking at a chi square table our significance value at 0.05 is 7.815.

Our value is larger, that means the model we tested (simple dominant/recessive pattern) does not fit reality. This means the traits are not inherited by a simple dominant/recessive pattern