In the nematode Caenorhabditis elegans there are two mutatio

In the nematode, Caenorhabditis elegans, there are two mutations, dumpy (dmp) and uncoordinated (unc). These animals have two sexes, male and hermaphrodite. Hermaphrodites will reproduce by self-fertilization. A heterozygous hermaphrodite dmp* dmp unc* unc is allowed to self-fertilize and her offspring are examined. The data are shown in Table 1. Do these two genes appear to assort independently? Calculate a x^2 value to answer this question. Justify your answer. Assort independently (circle correct answer) Yes No Justification______ _____ _____ _____

Solution

Assume that, dmp+ = A; dmp = a; unc = B; unc = b.

The self-cross of heterozygotes will have the progeny in the ratio of 9:3:3:1.

Total number of offspring is = 1600

The expected frequencies are,

Aa Bb -à 1600 *9/16 = 900

Aa bb -à 1600 *3/16 = 300

aa Bb -à 1600 *3/16 = 300

Aa bb -à 1600 *1/16 = 100

CHI - SQUARE (X²):

X² = (O - E)² / E

Where O = Observed frequency

E = Expected frequency

Phenotype

O

E

(O-E)

(O-E)^2

(O-E)^2/E

AaBb

864

900

-36

1296

1.44

Aabb

339

300

39

1521

5.07

aa Bb

311

300

1.5

2.25

0.155172

aa bb

86

100

-14

196

1.96

6.665172

The chi-square value is = 6.665

Degrees of freedom is = n-1 = 3

The calculated P value is 0.083378, which is not significant at p< 0.05. So, we accept the null hypothesis.

The difference between the expected an observed phenotypic ratios is not significant and the genes are assorting independently.

Phenotype	O	E	(O-E)	(O-E)^2	(O-E)^2/E
AaBb	864	900	-36	1296	1.44
Aabb	339	300	39	1521	5.07
aa Bb	311	300	1.5	2.25	0.155172
aa bb	86	100	-14	196	1.96
					6.665172