BIO 202 Midterm Exam Winter 2007
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1 BIO 202 Midterm Exam Winter 2007 Mario Chevrette Lectures : Question 1 (1 point) Which of the following statements is incorrect. a) In contrast to prokaryotic DNA, eukaryotic DNA contains many repetitive elements. b) Mouse satellite DNAs are located near the centromere of the mouse chromosomes. c) Cesium chloride gradients can be used to isolate telomeric repeated sequences. d) Insertion of Alu sequences into genes (transposons) is the cause of human diseases. e) A complete digestion of a cosmid DNA that has 2 restriction sites for a given restriction endonuclease will generate 2 DNA fragments. Answer is c) Questions 2 and 3 are derived from the following figure. Question 2 (2 points) The following figure represents an ethidium bromide stained gel containing molecular weight markers ( M ) and where fragments of DNA from 5 samples (1 to 5) have been digested with restriction enzymes, and then separated according to their size. 10 kb M kb 7 kb 6 kb 4 kb 3 kb 1 kb Which one of the following statements will always be true. a) This gel could represent the digestion of a plasmid with 5 different restriction enzymes. b) The 1 kb fragment seen in samples 2 and 4 are identical and will have the same DNA sequence.
2 c) If one was to use the 4 kb fragment of sample 3 as a probe in a Southern blot obtained from this gel, there will be a hybridization signal in samples 3 and 5 only. d) If one was to use the 7 kb fragment of sample 2 as a probe in a Southern blot obtained from this gel, there will be a hybridization signal in all samples. e) Assuming that the length of the undigested DNA of sample 1 is 14 kb, a 2 kb fragment will never be generated from this DNA using a restriction enzyme that recognizes a 8 base pairs restriction site. Answer is a) Question 3 (2 points) Please read carefully the following statements, all derived from the same gel as above. 1) Sample 2 can be obtained by digesting sample 1 with another restriction enzyme. 2) Sample 3 can be obtained by digesting sample 1 with another restriction enzyme. 3) Sample 4 can be obtained by digesting sample 5 with another restriction enzyme. 4) Sample 1 can be obtained by digesting sample 5 with another restriction enzyme. 5) None of these 5 samples could have been obtained by using two different restriction enzymes. Based on the previous statements, which one of the following analyses is RIGHT: a) Only statement 5) is true, all others are false. b) Statements 1), 2) and 3) are true, while statements 4) and 5) are false. c) All statements are false. d) Statements 1) and 4) are true, while statements 2), 3) and 5) are false. e) Only statement 5) is false, all others are true. Answer is b) Question 4 (2 points) Your mother, a molecular biology expert, has brought home a photocopy of something she is working on (illustrated here)
3 Please read carefully the following statements, all derived from the photocopy above. 1) Such results could have been generated by VNTR analysis. 2) Such results could have been generated by Southern blot analysis. 3) Such results could have been generated by PCR amplification using one set of primers. 4) Such results could have been generated by DNA sequencing. 5) Such results could have been generated by microsatellite analysis at a single locus. 6) Such results could have been generated from an ethidium bromide stained DNA gel. Based on the previous statements, which one of the following analyses is RIGHT: a) Only statement 6) is true, all others are false. b) Statements 1), 2) and 3) are true, while statements 4), 5), and 6) are false. c) All statements are true. d) Statements 1), 2) and 6) are true, while statements 3), 4) and 5) are false. e) Statements 3) and 5) are false, while statements 1), 2), 4) and 6) are true. Answer is d) Question 5 (1 point) Which of the following statements regarding VNTRs is incorrect. a) They can vary in number in the same individual. b) They can be used in DNA fingerprinting. c) PCR amplification of the VNTR sequence using primers complementary to the VNTR sequence has replaced the need of using Southern blot to detect the VNTRs. d) In DNA fingerprinting, the restriction enzyme will digest the DNA on each side of a VNTR sequence. e) Since VNTRs varies from one individual to another, they can be used to identify a specific individual. Answer is c) Question 6 (1 point) Which of the following statements is incorrect. a) Slippage during replication is the main cause of microsatellite polymorphism. b) Although they exist, long DNA deletions are rarely seen in normal human DNA.
4 c) A rapid analysis of RFLPs among individuals within a population can be performed by combining PCR amplification, restriction enzyme digestion and gel electrophoresis. d) About 1,00000,000 different microsatellite sequences are found in the human genome and they result from unequal crossingovers. e) PCR is the method of choice to detect microsatellite variation between individuals. Answer is d) Question 7 (2 points) DNA obtained from different members of the same family (mother (M), father (F) and four children (1, 2, 3, and 4)) was tested for a recessive disease using ASOs. Two different disease alleles, identified by two different ASOs (ASO 1 and ASO 2), lead either by themselves or in combination, to the disease. The results of the analysis are illustrated on the following figure where hybridization with the ASOs is indicated by a black circle, while no hybridization is represented by a white circle. M F M F M F Hybridized to labeled normal ASO Hybridized to labeled disease ASO 1 Hybridized to labeled disease ASO 2 From these results, you can conclude: a) Only children 2 and 3 have the disease. b) All members of this family have the disease. c) Only the father and children 1 and 4 carry a defective gene implicated in this disease. d) Although not having the disease now, every members of this family will eventually get the disease. e) Only the father, children 1 and 4 have the disease. Answer is e) Question 8 (1 point) Which of the following statements is incorrect. a) The presence of an origin of replication (ori) in a plasmid will allow its replication as an extrachromosomal entity in bacteria. b) In a plasmid containing an ampicillin resistance gene, cloning a DNA fragment into the sequence of the β-galactosidase gene and plating the bacteria on media
5 containing ampicillin and X-Gal will allow selection for the bacteria that have been transformed with plasmids that contain a foreign piece of DNA. c) A cosmid vector is a plasmid containing the cos sites of the bacteriophage λ and can, upon ligation with DNA of appropriate size, be packaged in virus particles. d) Genomic libraries can be obtained by generating genomic DNA fragments without the use of restriction endonucleases. e) The number of clones to be screened to identify a particular gene in a genomic library depends only on the size of the genomic DNA that can be inserted in the cloning vector. Answer is e) Question 9 (2 points) You have just cloned a cdna from a new gene, performed a Northern blot analysis and obtained the illustrated results in five different tissues: L: liver P: prostate O: ovary B: brain K: kidney L P O B K - 2 kb -1.5 kb Your gene actin Please read carefully the putative following conclusions from these results: 1) Your gene could be alternatively spliced in kidney. 2) Your gene is not expressed in liver. 3) Your gene is expressed in all tissues tested. 4) Your gene is highly expressed in prostate but expressed at very low level in brain. 5) The protein encoded by your gene could be phosphorylated in prostate. 6) Your gene exhibits a RFLP in the kidney. Based on the previous statements, which one of the following analyses is RIGHT: a) Only statement 1) is true, all others are false.
6 b) Statements 1), 2) and 4) are true, while statements 3), 5) and 6) are false. c) All statements are true. d) Statements 1) and 4) are true, while statements 2), 3), 5) and 6) are false. e) Only statement 5) is false, all others are true. Answer is a) Question 10 (2 points) A cdna has been cloned in the unique EcoRI site of the multiple cloning sites of a puc38 plasmid (length of 2,7 kb). This multiple cloning site contains unique restriction sites for EcoRI, SalI, MstII, HindIII, and ApaI. Note that puc38 itself has no restriction site for PstI. After transformation of bacteria, plasmid DNA (containing the cdna insert) has been digested with few restriction enzymes, and the size of the fragments resulting from an ethidium bromide stained agarose gel is illustrated below. Please note that the resolution of the gel will not allow you to differentiate fragments of the same length. EcoRI PstI HindIII HindIII & PstI HindIII & PstI & EcoRI 6.7 kb 6.2 kb 5.2 kb 2.7 kb 2.0 kb 2.7 kb 2.5 kb 1.0 kb 1.0 kb 0.5 kb 0.5 kb 0.5 kb Based on these results, digesting the puc38 plasmid that contains the cdna insert with PstI and EcoRI will generate the following fragments: a) One single fragment of 6.7 kb b) Two fragments of 2.5 and 1.5 kb c) Three fragments of 2.7, 2,5, and 1.5 kb d) Three fragments of 6.7, 2.7 and 2.0 kb e) Three fragments of 5.2, 1.0 and 0.5 kb We will accept all answers since there was a mistake in the fragments generated by the triple digestions (HindIII, EcoRI and PstI). Instead of the written fragments (2,7; 2,5; 1,0 and 0,5) which obviously represent a partial EcoRI digest,
7 the gel should have shown fragments of 2,7; 2,0; 1,0 and 0,5 (Note that there are two 0,5 fragments). Note that this result does not affect the answer of the next question Mario Chevrette Question 11. (1 point) For subcloning purposes, and from the gel above, your supervisor asked you to isolate all fragments derived only from the cdna that are flanked (ended) by both HindIII and PstI sites (thus the wanted fragment(s) will have a PstI site at one end and a HindIII site at the other end). If the fragments are present in more than one sample, you need only to isolate it once. Thus to obtain such fragment(s), you will cut out of the gel: a) The three fragments of 5.2, 1.0 and 0.5 kb b) The two fragments of 2.5 and 1 kb. c) The two fragments of 1.0 and 0.5 kb d) Only the fragment of 1 kb. e) The two fragments of 6.7 and 1.0 kb Answer is d) Question 12 (1 point) The karyotype of a male germ cell (sperm) from a new wild cat found in the deep forest of Lac-St-Jean has just been revealed. Analysis of this karyotype shows that it contains 18 chromosomes plus a chromosome X and a chromosome Y. Based on this data, the minimum number of contigs that a female from this animal species will have upon full sequence and analysis of its genome will be: a) 18 b) 38 c) 20 d) 10 e) 19 Answer is e)
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