Question

In: Biology

In carrying out crosses with the nematode Caenorhabditis elegans, a geneticist hypothesizes that the allele for...

In carrying out crosses with the nematode Caenorhabditis elegans, a geneticist hypothesizes that the allele for dumpy body shape (dpy) is recessive to the allele for normal body (dpy+), and the allele for uncoordinated (unc) is recessive to the coordinated body movement allele (unc+). Moreover, she hypothesizes that these two genes are unlinked (i.e., they assort independently). In a mating between a normal bodied, coordinated worm and a dumpy, uncoordinated worm, all of the F1 worms were normal bodied and coordinated in their movements. Selfing these F1's (it should be noted here that C. elegans are capable of self- fertilization) yielded an F2 generation with the following phenotypes:

o normal body, coordinated 95

o normal body, uncoordinated 21

o dumpy body, coordinated 19

o dumpy body, uncoordinated 25

TOTAL 160

a. What is the expected ratio and number of each class of F2 worms for this cross?

b. Determine the Chi-square (X2) value for the observed data under the proposed hypothesis.

c. What is the probability value for this Chi-square value?

d. Do these data support the hypothesis that these loci are unlinked? Explain.

Solutions

Expert Solution

Answer:

Based on the given information:

dominant recessive
normal body dpy+ dpy dumpy body
coordinated body movement unc+ unc uncoordinated body movement

F1: dpy+ unc+ x dpy unc

Progeny are all: dpy+ dpy unc+ unc

F2 Self cross:  dpy+ dpy unc+ unc x dpy+ dpy unc+ unc

Punnett Square for F2 cross:

dpy+ unc+ dpy+ unc dpy unc+ dpy unc  
dpy+ unc+ dpy+ dpy+ unc+ unc+ dpy+ dpy+ unc+ unc dpy+ dpy unc+ unc+ dpy+ dpy unc+ unc
dpy+ unc dpy+ dpy+ unc+ unc dpy+ dpy+ unc unc dpy+ dpy unc+ unc dpy+ dpy unc unc
dpy unc+ dpy+ dpy unc+ unc+ dpy+ dpy unc+ unc dpy dpy unc+ unc+ dpy dpy unc+ unc
dpy unc   dpy+ dpy unc+ unc dpy+ dpy unc unc dpy dpy unc+ unc dpy dpy unc unc

Phenotype:

normal coordinated normal coordinated normal coordinated normal coordinated
normal coordinated normal, uncoordinated normal coordinated normal, uncoordinated
normal coordinated normal coordinated dumpy coordinated dumpy coordinated
normal coordinated normal, uncoordinated dumpy coordinated dumpy uncoordinated
Phenotype Frequency
normal body, coordinated movement 9
normal body, uncoordinated movement 3
dumpy body, coordinated movement 3
dumpy body, uncoordinated movement 1

Phenotypic ratio 9:3:3:1

Chi Square Test:

Phenotype Observed Frequency (O) Expected Frequency ('E) (O-E) (O-E)^2/E
normal body, coordinated movement 95 90 5 0.278
normal body, uncoordinated movement 21 30 -9 2.700
dumpy body, coordinated movement 19 30 -11 4.033
dumpy body, uncoordinated movement 25 10 15 22.500
Total 160 160 Chi Square (sum) 29.511

Degree of freedom = number of categories - 1 = 4 - 1 = 3

P-Value corresponding to Chi square value of 29.511 with 3 degree of freedom is P-Value is < 0.00001.

The result is significant at a significance level of 0.05. Therefore Null Hypothesis is rejected. Based on the evidence, there is significant difference in observed frequency from expected frequency.

gladiator answered 2 years ago
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