BSC 114 Lecture Notes - Lecture 23: Exon Shuffling, Intron, Tata Box
Document Summary
Get access
Related Documents
Related Questions
Question 1
1 pts
Cystic fibrosis is caused by nonsense and missense mutations in the CFTR gene, which encodes for a chloride channel. You are studying cystic fibrosis patients to determine what mutation they possess in the CFTR gene. The difference between the mutant and wild type CFTR genes can be uncovered by examining the CFTR:
DNA | |
mRNA | |
protein | |
tRNA |
Question 2
1 pts
You decide to identify the CFTR mutation by analyzing the genomic DNA of your patients compared to healthy individuals. You specifically are looking to see whether a specific 3' gene truncation has occurred in the patients. You will determine this using hybridization techniques with samples from healthy and CF patients. Which of the following will allow you to accomplish this?
Using an RNA probe complementary to the region not removed by the truncation. | |
Using an RNA probe complementary to the region removed by the truncation. | |
Using an DNA probe complementary to the region not removed by the truncation. | |
Using an DNA probe complementary to the region removed by the truncation. |
Question 3
1 pts
You would like to ensure that this experiment (to determine whether patients have a specific CFTR gene truncation using hybridization) is properly controlled. Which of the following samples must you test?
The genomic DNA of a healthy individual who does not have cystic fibrosis. | |
The genome of a CFTR patient known to have the specific truncation you are trying to identify. | |
The genome of a CFTR patient with a missense mutation but full length gene. | |
The genome of a healthy individual married to a CFTR patient with the specific truncation you are trying to identify. | |
The genome of a patient with muscular dystrophy, which can be due to a trucation in the dystrophin gene. |
Question 4
1 pts
To conduct the hybridization experiment, you are trying to decide between using a DNA or RNA probe. Which would be ideal to use and why?
As both are composed of nucleic acids, using either would result in identical results. | |
An RNA probe because RNA has uracil bases. | |
An RNA probe because it could also be used in a translation experiment. | |
A DNA probe because it is more stable than RNA. | |
A DNA probe because RNA cannot bind to DNA. |
Question 5
1 pts
Which of the following will lower the Tm of a given DNA strand?
Increasing the percentage of GC base pairs. | |
Raising the pH of the solution from neutral to basic. | |
Decreasing the buffer concentration from 50mM NaCl to 5mM NaCl. | |
None of the above. |
Question 6
1 pts
One step of the Hershey/Chase experiment involved blending the virus/cell mixture before centrifugation and probing the pellet for radioactivity. Why was the blending step necessary?
To collect the bacteria at the bottom of the tube. | |
To break open the bacteria to release the genome. | |
To separate the bacteria from the bacteriophages. | |
To be able to detect the radioactivity. |
Question 7
1 pts
Imagine Hershey/Chase had used an RNA virus (genome composed of RNA) instead of a DNA virus in their experiment. Would radioactivity still have been found in the pellet?
No, because only DNA can be labeled with radioactivity. | |
No, because the RNA genome would not enter the bacteria upon infection. | |
No, because while DNA and RNA nucleotides are similar, they are not identical. | |
Yes, because DNA and RNA nucleotides are similar. | |
Yes, because genome in any form (DNA, RNA, protein) would be labeled similarly. |
Question 8
1 pts
Griffith and Avery's transformation experiments allowed us to identify that DNA is our genetic information. Which of the following scenarios would result in bacterial cells that are capable of killing mice upon injection?
Heat killed non-virulent bacteria is added to a live virulent bacteria strain. | |
Heat killed virulent bacteria is added to a heat killed non-virulent bacteria strain. | |
A heat killed virulent bacteria that is treated with a nuclease, is then added to a non-virulent bacteria strain. | |
Heat killed mouse cells are added to a non-virulent bacteria. |
Question 9
1 pts
The human genome consists mostly of non-coding DNA. Which of the following are benefits of this?
Random DNA mutations generally won't affect RNA and protein function. | |
It is faster to duplicate the genome when these are present. | |
The existence of introns can lead to multiple variations of proteins encoded by a single gene. | |
It is unlikely transposons would exist in the genome if there was too much protein coding DNA. |
Question 10
1 pts
Andrew Murray's sister, Andrea, is adding to her brother's work on chromosomes. She is using cells that are unable to synthesize adenine (âade) and histidine (âhis). The plasmid she is currently working with consists of an origin of replication and the Ade gene.
Following her transformation of the plasmid into her yeast, what media will the cells be plated on to select for cells that have picked up the plasmid?
Media containing histidine | |
Media containing adenine | |
Media lacking adenine | |
Media lacking histidine |
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. |