3. Assume that there are 12 map units between two loci in the mouse and that you are able to microscopically observe meiotic chromosomes in this organism. If you examined 200 primary oocytes, in how many would you expect to see a chiasma between the two loci mentioned above?

Question

3. Assume that there are 12 map units between two loci in the mouse and that you are able to microscopically observe meiotic chromosomes in this organism. If you examined 200 primary oocytes, in how many would you expect to see a chiasma between the two loci mentioned above?

I got 2 for an answer. Is that correct?

4. Phenotypically wild F1 female Drosophila, whose mothers had light eyes(Lt) and fathers had straw(stw) bristles produced the following offspring when crossed to homozygous light-straw males:
phenotype: number:
light-straw 22
wild 18
light 990
straw 970
total: 2000
Compute the map distance between light and straw loci.
I got 1.1 mu for an answer. Is that correct?
Thanks in advance for any help.

Answer 1

To answer question 3, we need to use the concept of recombination frequency. The recombination frequency is the percentage of offspring that show a recombination event (in this case, the presence of a chiasma between the two loci).

Given that there are 12 map units between the two loci, we can assume that each unit has an equal chance of undergoing a recombination event. Therefore, the recombination frequency is 12%.

To calculate the number of primary oocytes that would show a chiasma, we need to multiply the recombination frequency by the total number of primary oocytes examined.

In this case, 12% of 200 primary oocytes is:

0.12 * 200 = 24

So, you would expect to see a chiasma between the two loci in 24 primary oocytes, not 2.

Now, let's move on to question 4. To determine the map distance between the light and straw loci, we need to use the formula:

Map distance = (Number of recombinant offspring / Total number of offspring) * 100

From the given data, we can see that there are a total of 2000 offspring and the number of recombinant offspring (light-straw and straw-light) is 22 + 18 = 40.

Therefore, the map distance between the light and straw loci is:

(40 / 2000) * 100 = 2%

So, the correct answer would be 2.0 mu, not 1.1 mu.

It's important to double-check your calculations to ensure that you have the correct answers.