Can anyone help me with ny of these 4 genetics questions?

[stew]

1. The genes for mohogany eyes and ebony body are approximately 25 map units apart on chromosome III in Drosophila. Assume that a mahogany-eyed female was mated to an ebony-bodied male and the resulting F1 phenotypically wild type females were mated to mahogany, ebony males. Of 1000 offspring, what would be the expected phenotypes and in what numbers would they be expected?

2. Given that loci A and B in Drosophila are sex-linked and 20 map units apart, what phenotypic frequencies would you expect in male and female offspring resulting from the following crosses? (Assume A and B are dominant to a and b, respectively):
(a) AaBb(cis)female X ab/Y male
(b) AaBb(trans) X ab/Y male
(c) aabb female X AB/Y male

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.

 

[Moonbear]

Keep in mind our rules require you to show your own efforts first. Tell us how you got the answers you got.

 

[Blueshift5]

I am happy to assist you with your genetics questions. However, it is important to note that I am unable to provide direct answers to specific homework or exam questions. Instead, I can provide you with guidance and explanations to help you understand the concepts and solve the problems on your own.

1. To answer this question, we need to use the principles of Mendelian genetics and the concept of recombination. In this scenario, the genes for mahogany eyes and ebony body are located on the same chromosome, but are 25 map units apart. This means that during meiosis, there is a 25% chance that these genes will be separated and end up in different gametes. This is known as recombination.

In the first cross, a mahogany-eyed female is mated with an ebony-bodied male. This will result in F1 offspring that are all wild type because both parents are heterozygous for the two traits. When these F1 females are mated to mahogany, ebony males, the expected phenotypes and numbers are as follows:

- Mahogany-eyed, ebony-bodied: 25% (recombination frequency of 25%)
- Mahogany-eyed, wild type: 25% (recombination frequency of 25%)
- Wild type, ebony-bodied: 25% (recombination frequency of 25%)
- Wild type, wild type: 25% (no recombination)

Therefore, out of 1000 offspring, we would expect to see approximately 250 mahogany-eyed, ebony-bodied flies, 250 mahogany-eyed, wild type flies, 250 wild type, ebony-bodied flies, and 250 wild type, wild type flies.

2. In this question, we are dealing with sex-linked genes, which are located on the sex chromosomes (X and Y). In Drosophila, the sex chromosomes are denoted as X and Y, with X being larger and carrying more genes. In this scenario, loci A and B are located on the X chromosome and are 20 map units apart.

(a) AaBb(cis)female X ab/Y male
In this cross, the female is heterozygous for both traits (AaBb) and the male is recessive for both traits (ab). Since the genes are located on the same chromosome, there will be no recombination. Therefore, the expected phenotypic frequencies in the offspring will be:
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