Biology, 10.05.2020 17:57, droid5004

A geneticist discovers a new mutation in Drosophila melanogaster that causes the flies to shake and quiver. She calls this mutation quiver (qu) and determines that it is due to an autosomal recessive gene. She wants to determine if the gene encoding quiver is linked to the recessive gene for vestigial wings (vg). She crosses a fly homozygous for quiver and vestigial traits with a fly homozygous for the wild-type traits and then uses the resulting F1 females in a testcross. She obtains the following flies from this testcross.
vg+ sps+ 230
vg sps 224
vg sps+ 97
vg+ sps 99
total 650
1. Are the genes that cause vestigial wings and the spastic mutation linked?
2. Do a chi- square test of independence to determine if it is likely that the genes have assorted independently.
3. List the chi-square value, degree of freedom, P value, and whether you accept or reject the hypothesis of independent assortment.

Answers: 3

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Answers

A cross in which the purebred recessive is crossed with suspected heterozygous species to know the genotype of suspected species is called a test cross.

1. Yes the genes of vestigial wins and quiver mutations are linked as the percentage of recombination is less than 50% and are said to be linked.

2. The test for independent assortment can be done by segregating the two loci.

Test for vestigial wings:

$\begin{aligned}\text{Observed vg} &= 224 + 97 \\&= 321\end{aligned}$

$\begin{aligned}\text{Observed vg+ }&= 230 + 99\\&= 329\end{aligned}$

$\begin{aligned}\text{Expected vg or vg+ }&= 0.5 \times 650 \\&= 325 \end{aligned}$

Putting values we get:

$\begin{aligned}\text{Test} &= \dfrac {(321 - 325)2}{325} + \dfrac {(329 - 325) 2}{325 }\\\\&= \dfrac {16}{325} + \dfrac{16}{325}\\\\&= 0.098 \end{aligned}$

The degree of freedom = n – 1

Where,

The number of phenotypic classes (n) = 2

So, degree of freedom = 1

We know that value of P is between 0.7 and 0.8. So these results do not deviate from the predicted 1:1 segregation.Finally, we test for independent assortment where phenotypic ratios are = 1:1:1:1 or 162.5 of each.

The attached image below shows the values:

3. We have Phenotypic classes = 4

Degrees of freedom = 3

The chi-square value of 102.5 that is why we reject independent assortment. Instead, the genes are linked, and the RF is given by:

$\begin{aligned}\text{RF} &= \dfrac{(97 + 99)}{650} \times 100\% \\\\&= 30\%\\\\&= 30 \;\rm map \;units \end{aligned}$

To learn more about test cross and chi-square values follow the link:

link

answered: emma6892

from the parent crossing

vgvg spssps( FEMALE HOMOZYGOUS FOR SPASTIC AND VESTIGIAL TRAITS) x vg + vg + sps + sps{MALE WILD TYPE)

F1 -----------> vg+vg sps+sps

F1 gametes ------------> vg+ sps+, vg+ sps, vg sps+, vg sps

Now, if genes are assorting independently, we will observe each phenotype in 1:1:1:1 ratio in F2.

So, expected phenotypes of each type = Total number / 4 = 650 / 4 = 162.5

Now,X² = ∑(observed - expected)²/expected (230-162.5)2/162.5 + (224-162.5)2/162.5 + (97-162.5)2/162.5 + (99-162.5)2/162.5 = 28.04 + 23.28 + 26.40 + 24.81 (Up to 2 decimals) = 102.53

Now, degrees of freedom = n - 1 where, n = number of expected phenotypes

So, degrees of freedom = 4-1 (As 4 different phenotypes are expected) = 3

From the X² distribution table, we find that probability of X² = 102.53 with degrees of freedom 3 is much less than 0.05. As probability is less than 0.05, we can say that difference between observed & expected values are not due to chance alone & significant differences exist between them. So, genes are not assorted independently & they must be linked.

Explanation: