HMB265H1 Lecture Notes - Restriction Fragment Length Polymorphism, Silent Mutation, Southern Blot
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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
1. Characters that show a continuous range of variation, such as height and eye color, usually are controlled:
a. | by a single gene with two alleles that are codominant. |
b. | by many genes with an additive effect. |
c. | by epistatic interactions between two genes. |
d. | mainly by the environment, with only a small genetic component. |
2. In humans, red-green colorblindness is inherited as a sex-linked recessive trait. In order for a woman to be red-green colorblind, which of the following statements must be true.
a. | Her mother must be red-green colorblind. |
b. | All of her brothers must be red-green colorblind. |
c. | Her father must be red-green colorblind. |
d. | All of the above statements must be true if a woman is red-green colorblind. |
3. The x-ray crystallography data collected by Rosalind Franklin suggested to Watson and Crick that the:
a. | structure of DNA is a double helix. |
b. | two strands of the DNA molecule are joined by hydrogen bonds between the bases. |
c. | four bases within DNA pair in a specific way. |
d. | two strands of the DNA molecule are joined by covalent bonds between the bases. |
4. In the genetic code, _________ one amino acid.
a. | one nucleotide specifies |
b. | two nucleotides specify |
c. | three nucleotides specify |
d. | four nucleotides specify |
5. During Meiosis I, a homologous pair of chromosomes may not separate, resulting in daughter cells that have extra chromosomes or are missing chromosomes. This can lead to genetic disorders, including Down Syndrome. This phenomenon is called:
a. | independent assortment. |
b. | nondisjunction. |
c. | segregation. |
d. | crossing over. |
6. You are a human geneticist studying the incidence of retinitis pigmentosa in the residents of Tristan de Cunha, a group of small islands in the middle of the southern Atlantic Ocean. The allele for retinitis pigmentosa, which causes a form of blindness, is inherited as an autosomal recessive. You have determined that the frequency of this allele (r) in the population is 0.4 (40%). Using the principles of the Hardy-Weinberg rule, you would estimate the frequency of individuals who are heterozygous for this allele (Rr) in the population to be:
a. | 0.16 (16%) |
b. | 0.24 (24%) |
c. | 0.36 (36%) |
d. | 0.48 (48%) |
7. Natural selection acts at the level of the:
a. | phenotype. |
b. | gene. |
c. | population. |
d. | nucleotide. |
8. You are working with pea plants, trying to recreate the experiments that Mendel performed. You are doing a dihybrid cross with a plant that is heterozygous for both seed shape and seed color, with the genotype RrYy. Which allelic combinations would you expect to find in the gametes produced by this plant?
a. | This plant would produce only RY and ry gametes. |
b. | This plant would produce only RrYy gametes. |
c. | This plant would produce RY, Ry, rY, and ry gametes. |
d. | You cannot determine which gametes this plant can produce without knowing the genotypes of its parents. |
9. Biochemist Erwin Chargaff found that in DNA there is a special relationship between the four bases that we now call Chargaff's rule. His observation was that, in an organism's genome the:
a. | percentage of A nucleotides = the percentage of T nucleotides, and the percentage of C nucleotides = the percentage of G nucleotides. |
b. | four bases all occur in an equal frequency (25%) within each organism. |
c. | percentage of A nucleotides = the percentage of G nucleotides, and the percentage of C nucleotides = the percentage of T nucleotides. |
d. | genetic material is composed of proteins, not DNA. |
10. During DNA replication:
a. | each strand of the double helix acts as a template for the synthesis of a new strand. |
b. | the enzyme DNA polymerase adds nucleotides to the strand being synthesized. |
c. | the bases A,C,G and T are required. |
d. | All of the above are true of DNA replication. |
11. During translation, amino acids are joined by peptide bonds to make polypeptides. The formation of these peptide bonds is catalyzed by:
a. | DNA. |
b. | mRNA. |
c. | tRNA. |
d. | rRNA. |
12. If an allele (R) at a gene with two alleles shows complete dominance, individuals with the genotypes ______ will have the same phenotype.
a. | RR and rr. |
b. | RR and Rr |
c. | Rr and rr |
d. | Each of the three possible genotypes will have a different phenotype. |