In addition to C. elegans and Drosophila, Zebrafish are also another model organism - and have...

In addition to C. elegans and Drosophila, Zebrafish are also another model organism - and have a spine. We're interested in two phenotypes, a curved spine and a gene that causes skin tumors to form on the fish's skin. We'd like to know if we could use the curved spine as an indicator if a fish may develop tumors (if the genes are linked). We'll assume that each of these traits is controlled by a single gene where:

a curved spine is the result of a dominant allele S (and a normal spine is the result of recessive alleles s),
and tumor growth is the result of a recessive allele t (and no tumor growths are a dominant T allele).

A testcross is performed with a fish that is heterozygous for both genes and the resulting progeny are given below.

TtSs x ttss

curved spine & tumors 18
curved spine & no tumors 23
normal spine & tumors 26
normal spine & no tumors 15

1. Assume for a moment that these genes are linked. If so, what allele combinations are parental in the TtSs genotype listed in the cross?

Are these two genes following Mendelian inheritance patterns? Use Chi-Squared analysis to test them. (X² = the sum of (o-e)²/e)

2. X² value :

3. Degrees of Freedom :

4. p-value :

5. Determination : NO they appear to be unlinked, or YES they may be linked

Are these two genes linked? Use Chi-Squared analysis to test them. (X² = the sum of (o-e)²/e)

6. X² value :

7. Degrees of Freedom :

8. p-value :

9. Determination : NO they appear to be unlinked, or YES they may be linked

Solutions

Expert Solution

Hint: Always non-recombinant genotypes are large numbered than the recombinant genotypes.

Hence, the parental (non-recombinant) combinations would be ST/st

If the genes are assorted independently, all the progeny of testcross would be in equal number according to Mendel’s law of independent assortment.

Phenotypes	Observed(O)	Expected (E)	O-E	(O-E)2	(O-E)2/E
curved spine & tumor	18	20.50	-2.50	6.25	0.305
curved spine & no tumor	23	20.50	2.50	6.25	0.305
normal spine & tumor	26	20.50	5.50	30.25	1.476
normal spine & no tumors	15	20.50	-5.50	30.25	1.476
Total	82	82			3.561

Chi-square value = 3.561

Degreees of freedom = no. of outcomes – 1

DF = 4-1=3

p-Values = 7.814

The chi-square value of 3.561 is less than the p-value of 7.814. Hence, the null hypothesis is accepted and the two genes are not appeared to be linked. It means they are unlinked.

gladiator answered 1 year ago

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