BIOL 461 Lecture Notes - Lecture 18: Transfer Rna, Chromosome, Tetrahymena
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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
Please show detailed work. Thanks in advance.
16. Syntenic genes can assort independently when
A) they are very close together on a chromosome.
B) they are located on different chromosomes.
C) crossing over occurs rarely between the genes.
D) they are far apart on a chromosome and crossing over occurs frequently between the genes.
E) they are far apart on a chromosome and crossing over occurs very rarely between the genes.
17. The alleles of linked genes tend to
A) segregate together more often than expected by random assortment
B) assort independently.
C) be mutated more often than unlinked genes.
D) experience a higher rate of crossing over.
E) assort independently and show a higher rate of crossing over.
18. If you know that the frequency of recombination between genes X and Y is 34% and between X and Z is 25%, can you predict the order of the three genes?
A) Yes; the order is X-Z-Y.
B) Yes; the order is X-Y-Z.
C) Yes; the order is Z-X-Y.
D) No; based on this data alone, the order could be Z-Y-X or X-Y-Z.
E) No; based on this data alone, the order could be X-Z-Y or Z-X-Y.
Question 19 - 20. You have performed the following dihybrid cross in Drosophila using the black body color (b) and vestigial wing (vg) mutations. The b+ (grey body) and vg+ (normal wing) are dominant wild type alleles. These genes are autosomal.
Female â b+ vg+/b vg à male â b vg/b vg
Progeny:
Phenotype # of Progeny
Grey body normal wing 965
Black body vestigial wing 944
Grey body vestigial wing 208
Black body normal wing 195
19. Assuming linkage between black and vestigial, the estimated recombination frequency would be:
0.17
0.09
0.82
1.00
0.50
20. What key test could you use to determine whether the observed offspring frequencies deviate from those expected by chance alone?
A) Pascal's triangle
B) The product rule
C) The Chi-square (Ï2) test
D) The law of random assortment
E) The sum rule
21. In a genome wide association study (GWAS) designed to map the gene(s) that control height you divide subjects into a group of 1000 who are all more than seven feet tall and a control group of 1000 people of average height. You find the following associations between two genetic markers and the height trait:
| Tall group | Control group | ||
| Marker 1 | Allele A | 20% | 50% |
| Allele T | 80% | 50% | |
| Marker 2 | Allele G | 15% | 15% |
| Allele C | 85% | 85% |
What is your best guess for which marker is more closely linked to a gene that influences height?
A) Marker 1
B) Marker 2
22. Two pure breeding parents produce red and white flowers. They are crossed and the F1 produces pink flowers. When the F1 are selfed to produce the F2, nine distinct classes of pigmentation are present among F2 individuals. What is your best guess of the minimum number of genes that underlie flower pigmentation in this species?
A. 2
B. 3
C. 4
D. 5
E. 6
23. In a quantitative genetic experiment you identify two genes that confer bands of color on the back of a fly. At each gene, a dominant allele causes one band of color. If flies that are heterozygous at both loci are crossed, what ratio of offspring do you expect in each phenotype (i.e., number of color bands) class? (answer options are given from lowest to highest band number)
A) 1:1:1:1:1
B) 1:2:2:2:1
C) 1:4:6:4:1
D) 4:4:4:4:4
