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.
To answer question 3:
To calculate the expected number of oocytes with a chiasma between the two loci, we can use the concept of recombination frequency. In organisms, the recombination frequency between two loci is proportional to the distance between them on a chromosome.
The given information states that there are 12 map units (or centimorgans) between the two loci in question.
Recombination frequency (RF) can be calculated as RF = map distance / 100.
Therefore, the recombination frequency in this case would be RF = 12 / 100 = 0.12.
Now, if we assume independent assortment of chromosomes during meiosis, the recombination frequency gives us the probability of a chiasma forming between the two loci on each chromosome.
So for each primary oocyte, there is a 0.12 probability of observing a chiasma between the loci in question.
To calculate the expected number of oocytes with a chiasma, we can multiply the probability by the total number of oocytes examined:
Expected number = 0.12 * 200 = 24.
Therefore, the correct answer is 24, not 2.
Now let's move on to question 4:
To calculate the map distance between the light and straw loci in Drosophila, 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 22 light-straw (recombinant) offspring out of a total of 2000.
Map distance = (22 / 2000) * 100 = 1.1
Therefore, your answer of 1.1 mu for the map distance between light and straw loci is correct.
I hope this helps!