Based on the following data from a three point testcross:
1. Determine the gene order and phase for the genes E, F and G.2
2. Determine the distance between genes E, F and G.
3. Draw a linkage map with the correct gene order, allele combinations on each chromatid and distance between the three genes.
For this exercise, phenotypes with capital letters are dominant and phenotypes in lower case letters are recessive. For example, âphenotype Aâ means the dominant phenotype for the gene and âphenotype aâ means the recessive phenotype for that gene.
Show all your calculations. Remember that the three genes can be in any order. Remember that the dominant alleles do not necessarily need to all be on the same chromosome in the triple heterozygote parent.
Phenotypes
N
Efg
250
eFG
231
EFg
32
efG
38
EFG
11
efg
9
EfG
53
eFg
48
Total
672
Based on the following data from a three point testcross:
1. Determine the gene order and phase for the genes E, F and G.2
2. Determine the distance between genes E, F and G.
3. Draw a linkage map with the correct gene order, allele combinations on each chromatid and distance between the three genes.
For this exercise, phenotypes with capital letters are dominant and phenotypes in lower case letters are recessive. For example, âphenotype Aâ means the dominant phenotype for the gene and âphenotype aâ means the recessive phenotype for that gene.
Show all your calculations. Remember that the three genes can be in any order. Remember that the dominant alleles do not necessarily need to all be on the same chromosome in the triple heterozygote parent.
Phenotypes | N |
Efg | 250 |
eFG | 231 |
EFg | 32 |
efG | 38 |
EFG | 11 |
efg | 9 |
EfG | 53 |
eFg | 48 |
Total | 672 |
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Homozygous recessive loss of function mutations in any one of 4 different genes in C. elegans worms (genes A, B, C, or D) results in female worms that don't form proper egg-laying structures, called vulvas, as shown in Lines 2-5 of the data table below. In vulvaless females, the fertilized eggs hatch and baby worms develop inside the mother, killing her in the process.
genotype | phenotype | |
1 | AA BB CC DD (wild type) | normal vulva |
2 | aa BB CC DD | No vulva |
3 | AA bb CC DD | no vulva |
4 | AA BB cc DD | no vulva |
5 | AA BB CC dd | no vulva |
You would like to figure out the order in which these genes act during the creation of the vulva (you cannot assume it will be A--->B--->C--->D as the gene names are arbitrary). Your friend tells you to perform epistasis analysis by making double mutants between the different homozygous recessive mutants and analyzing the phenotype of the double mutant. For example, she asks you to examine the phenotype of aabbCCDD (double mutant of genes A and B) and compare this to the single mutants to figure out whether gene A acts earlier than gene B, or vice versa. You know this won't work. Why not? Pick the ONE BEST choice:
a. The phenotype of the single mutant is already pretty severe. The double mutant will most likely be dead, therefore epistasis analysis won't be possible.
b. Each of the single mutants (aaBBCCDD) and (AAbbCCDD) has the same mutant phenotype i.e., no vulva. Epistasis analysis between any 2 genes is only possible when the mutant phenotypes for each gene is different.
c. Epistasis analysis involves making triple mutants (such as aabbccDD) in order to learn something about how the genes are ordered.
d. Epistasis analysis can never be carried out with null loss of function mutations. The mutations being analyzed all have to be dominant gain of function alleles.
To address your concern, you first generate overactive alleles of either gene A (denoted as A*) or gene B (denoted as B*). As shown in lines 6-7 of the table below, C. elegans that have one of these overactive alleles produce multiple vulvas.
Genotype | Phenotype | |
6 | A*A BB CC DD | multiple vulvas |
7 | AA B*B CC DD | multiple vulvas |
To find out the order in which these genes act, you combine the overactive alleles with different loss of function alleles and observe the phenotype in double mutants (see Lines 8-11).
Genotype | Phenotype | |
8 | A*A bb CC DD | multiple vulvas |
9 | A*A BB cc DD | multiple vulvas |
10 | A*A BB CC dd | no vulva |
11 | AA B*B cc DD | multiple vulvas |
Based on the data in the table, what is the order in which these 4 genes normally act in wild-type C. elegans in order to produce a wild-type/normal vulva?
a. A----> B----->C----->D ---> Vulva
b. D----> B----->C----->A----> Vulva
c. B----> C----->A----->D----> Vulva
d. C----> B----->A----->D----> Vulva
e. C----> B----->D----->A----> Vulva
f. don't have enough data to make any conclusions
Based on the vulva formation phenotype of the double mutants, which of the following statement(s) accurately describes the genetic interactions between the A* allele and alleles of other genes affecting vulva formation? Pick ALL that apply:
a. The A* allele is epistatic to homozygous recessive loss of function mutations in gene B
b. A homozygous recessive loss of function mutation in gene B is epistatic to the A* allele
c. Homozygous recessive loss of function mutation in gene D is epistatic to the A* allele
d. The A* allele is epistatic to homozygous recessive loss of function mutation in gene D
e. The A* allele enhances the homozygous recessive loss of function mutation in gene D