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?
If you put biology, or genetics in the subject title, you might be more successful getting someone to look and answer.
To determine the phenotypic frequencies for the given crosses and the arrangement of genes in the second question, we need to understand the concepts of sex-linkage and genetic mapping.
Sex-linkage refers to the inheritance of genes located on the sex chromosomes (in Drosophila, the X and Y chromosomes). In this case, loci A and B are sex-linked, meaning they are located on the sex chromosomes.
Genetic mapping measures the distance between genes on a chromosome through the phenomenon of recombination. One map unit (also known as a centimorgan) corresponds to a 1% recombination frequency between two loci. The higher the recombination frequency, the further apart the genes are on the chromosome.
Now, let's determine the phenotypic frequencies in the given crosses:
A) AaBb (cis) female X ab/Y male:
In this cross, the female parent has genes AaBb, and the male parent has genes ab/Y. Since the genes are sex-linked, the male offspring will inherit their single X chromosome from the mother and the Y chromosome from the father. The female offspring will inherit one X chromosome from each parent.
The possible combinations of alleles in the offspring are:
Female offspring: AaBb (50%), AabB (50%)
Male offspring: ab (100%)
B) AaBb (trans) X ab/Y male:
In this cross, the female parent has genes AaBb, and the male parent has genes ab/Y. Again, the male offspring will inherit their single X chromosome from the mother and the Y chromosome from the father. The female offspring will inherit one X chromosome from each parent.
The possible combinations of alleles in the offspring are:
Female offspring: AaBb (25%), aaBb (25%), AabB (25%), aabB (25%)
Male offspring: ab (100%)
C) aabb female X AB/Y male:
In this cross, the female parent has genes aabb, and the male parent has genes AB/Y. Once more, the male offspring will inherit their single X chromosome from the mother and the Y chromosome from the father. The female offspring will inherit one X chromosome from each parent.
The possible combinations of alleles in the offspring are:
Female offspring: AaBb (50%), Aabb (50%)
Male offspring: AB (100%)
Next, let's determine the arrangement of genes based on the given numbers of offspring in the second question. We have:
106 AaBb
48 Aabb
52 aaBb
94 aabb
By summing up the frequencies of the different gene combinations, we find that the total number of offspring is 300.
To determine the arrangement of genes, we compare the observed frequencies to the expected frequencies based on different gene arrangements. The four possible arrangements are:
1) AB/ab = 106 (AaBb) + 0 (Aabb) + 0 (aaBb) + 94 (aabb) = 200
2) Ab/ab = 0 (AaBb) + 48 (Aabb) + 52 (aaBb) + 0 (aabb) = 100
3) AB/aB = 52 (AaBb) + 0 (Aabb) + 48 (aaBb) + 0 (aabb) = 100
4) aB/ab = 0 (AaBb) + 0 (Aabb) + 0 (aaBb) + 0 (aabb) = 0
Since the observed frequencies match only the first arrangement (AB/ab), we can conclude that the genes are in a cis configuration.
In summary:
A) Phenotypic frequencies: Female offspring - AaBb (50%), AabB (50%); Male offspring - ab (100%)
B) Phenotypic frequencies: Female offspring - AaBb (25%), aaBb (25%), AabB (25%), aabB (25%); Male offspring - ab (100%)
C) Phenotypic frequencies: Female offspring - AaBb (50%), Aabb (50%); Male offspring - AB (100%)
Answer to Question 2: The observed frequencies of offspring are consistent with the genes being in a cis arrangement (AB/ab).