need help with these genetics problems. please help with what you can. thanks:

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

Question 2:
Assume that a cross is made beyween AaBb and aabb plants, and the offspring occur in the following numbers: 106 AaBb, 48 Aabb, 52 aaBb, 94 aabb. These results are consistent with which arrangement of genes?

jkn

for 2a - cis means the AB is paired on the same chromosome and the ab is on the other. you would expect 80% females AB/ab and 20% females ab/ab. For males, 80% AB/Y and 20% ab/Y.

2b use the same logic as 2a, expect now the chromosomes are Ab and aB to start.
for 2c - it is all AB/ab females and all ab/Y males.

how?

To solve these genetics problems, we first need to understand the concept of linkage and recombination.

Linkage refers to the phenomenon where genes that are located close to each other on the same chromosome tend to be inherited together more frequently. Recombination, on the other hand, refers to the exchange of genetic material between homologous chromosomes during meiosis, which can lead to the formation of new combinations of genes.

To determine the phenotypic frequencies in the offspring resulting from these crosses, we need to consider the distances between the loci and the type of cross being performed. In this case, we are given that loci A and B are sex-linked and 20 map units apart.

A) AaBb (cis) female X ab/Y male:
In this cross, the female is heterozygous for both loci, AaBb (cis). The male is recessive for both loci, ab/Y. Since these loci are sex-linked, the male only contributes a Y chromosome. Here, the phenotypes of the offspring will depend on whether crossing over occurs during meiosis.
Without crossing over: The female passes on one of each allele [AB] to half of the offspring and [ab] to the other half. The male contributes [Y] to all offspring. Therefore, the expected phenotypic frequencies will be 50% AaBb and 50% abb in both males and females.
With crossing over: Since loci A and B are 20 map units apart, the chance of recombination occurring between them is 20%. As a result, the expected phenotypic frequencies will be:
- 70% AaBb in males and females
- 10% Aabb in males and females
- 10% Aabb in males and females
- 10% aabb in males and females

B) AaBb (trans) X ab/Y male:
In this cross, the female is still heterozygous for both loci, AaBb (trans). The male is still recessive for both loci, ab/Y. Similarly, the phenotypes of the offspring will depend on the occurrence of crossing over.
Without crossing over: The expected phenotypic frequencies will be 50% AaBb and 50% abb in both males and females.
With crossing over: The expected phenotypic frequencies will be:
- 30% AaBb in males and females
- 20% Aabb in males and females
- 20% aaBb in males and females
- 30% aabb in males and females

C) aabb female X AB/Y male:
In this case, the female is homozygous recessive for both loci, aabb. The male is homozygous dominant for both loci, AB/Y. Since the female can only pass on the X chromosome, regardless of crossing over, all offspring will inherit [AB] allele from the male. Therefore, all offspring will have the AaBb phenotype.

For the second question, we are given the numbers of offspring resulting from a cross between AaBb and aabb plants. Using the Punnett square method, we can calculate the expected phenotypic ratios for each possible arrangement of genes.

The numbers of offspring obtained from the cross are:
- 106 AaBb
- 48 Aabb
- 52 aaBb
- 94 aabb

By examining these numbers, we can deduce that the arrangement of genes must be cis configuration, meaning that the A and B alleles are present on one chromosome, and the a and b alleles are present on the other chromosome. This configuration explains the higher frequency of AaBb and aabb phenotypes compared to Aabb and aaBb phenotypes.

Remember, understanding the basics of genetic concepts, such as linkage and recombination, and using Punnett squares can help you solve genetics problems like these.