Unformatted text preview: BIOL14 Quiz Name______________________ Practice Quiz A 1 point free: just to write their name at the top! 1. You are trying to clone a particular gene from a novel microbe. You are using a plasmid vector 4361 bp in length, whose restriction map is shown below. Amp and tet are both antibiotic resistance genes. a. (3 points) Which of the restriction sites in this plasmid could be used as a cloning site to construct a useful recombinant vector? Assume that you will need both antibiotic resistance genes in your recombinant vector. Choose one or more of the options below. i. PstI v. EcoRV ii. NdeI vi. BamHI iii. EcoRI vii. SalI iv. HindIII Other choices are incorrect: HindIII is not found in this vector at all, and the other restriction enzymes would cut in the middle of the antibiotic resistance genes. Cloning the insert into the middle of either antibiotic resistance gene would disrupt the antibiotic resistance gene, so that it could not be expressed. Either one, but not both at once, could be used. Full credit for either NdeI, EcoRI, or both. b. (2 points) You digest your plasmid with an enzyme that cuts it at one site. i. What are the size(s) of the product(s)? Cutting a circular DNA at one site will yield one linear DNA product of the same size as the circle, in this case: 4361 bp. 1 point ii. Are the product(s) linear or circular? It is linear. 1 point BIOL14 Quiz Name______________________ c. (3 points) Cutting the plasmid at the cloning site produced 5’ overhangs of sequence: ..TTAA-‐5’ How will you prepare your insert DNA? Choose one or more of the options below: i. Cut it with the same restriction enzyme you use to cut the plasmid ii. Cut it with a restriction enzyme that produces 5’ overhangs of sequence: ..TTAA-‐5’ iii. Cut it with a restriction enzyme that produces 3’ overhangs of sequence …TTAA-‐3’ Note that a 5’ overhang and a 3’ overhang are not complementary to each other: nnnnTTAA-‐5’ would anneal to this: 3’nnnn not this: 3’AATTnnnn nnnn-‐3’ 5’-‐AATTnnn 5’nnnn Full credit for either I, ii, or both. d. (2 points) The vector and insert are joined together by what enzyme? Name that enzyme. DNA ligase e. (3 points) You induce bacterial cells to take up the vector (by transformation), and grow the resulting cells on growth medium including ampicillin. To your dismay, no colonies of cells grow in the presence of the drug. Assuming that many of the cells did pick up the recombinant vector, what part(s) of the vector could have had mutations which would cause this problem? Choose one or more of the options below. i. Cloning site ii. Tet gene iii. Amp gene iv. Origin v. Promoter controlling Amp expression A mutation in the Amp gene might prevent expression of the ampicillin resistance gene. A mutation in the origin might prevent the replication of the plasmid, since a DNA needs an origin of replication to be replicated. 3 points to choose III, IV and V 2 points to choose two of the correct choices 1 point to choose 1 of the correct choices -‐2 for each incorrect choice (but only down to zero), so for example choice of II, III and IV would be 0 BIOL14 Quiz Name______________________ f. After correcting the problem in part (e), you get several colonies. You select two of them and test their plasmid DNA by restriction digest. The empty vector map is reproduced here for your convenience. After digestion of the plasmid by PstI and BamHI, you get the following results (gel below). One colony has the original vector (with no insert DNA); the other has the recombinant vector. i. (3 points) Which of these colonies has the empty vector (without the insert)? Choose one or more of the options. i. Colony 1 ii. Colony 2 The empty vector should be cut into two fragments by this double-‐digest, so it must be lane 1 and not lane 2. You can also confirm that the sizes fit the information given in the map above. ii. (3 points) Which of the following is true of the insert DNA which was cloned into the recombinant vector? Choose one of the options: i. The insert includes a PstI site ii. The insert includes a BamHI site iii. Either (i) or (ii) is true iv. Both (i) and (ii) are true v. Neither (i) nor (ii) is true For this double-‐digest to produce three fragments instead of 2 or more than 3, there must be one additional site that was cut, either by PstI or by BamHI (but we cannot tell which from this data). BIOL14 Quiz Name______________________ Practice Quiz B 3 points free: just to write their name at the top! You want to find out what phenotype results from loss of function of the XPV gene in mice. g. (5 points) Describe the genotype of a transgenic mouse that would allow you to answer this question? 3 points: It would have to be a knock-‐out mouse, in which the mouse’s normal XPV gene is deleted or disrupted so that it cannot produce a functional protein product. 1 point: To study a complete loss of function, we would need mice homozygous for the knockout allele of XPV. h. (3 points) Which approach could be used to produce mice of this genotype? i. Direct injection of transgenic DNA into a pronucleus of a fertilized mouse egg, followed by breeding of the mouse that develops from that egg. ii. Transfection (which just means introduction of DNA) of transgenic DNA into mouse embryonic stem cells, selective growth of those cells, addition of surviving cells to a mouse blastocyst, and finally breeding of the mouse that develops from that blastocyst. i. (4 points) Explain why you chose as you did in response to the preceding question. You must have the selective growth of ESCs to select the very rare homologous recombination that must occur in order to insert the transgenic DNA at the site of the mouse’s normal XPV gene. Even if they got (b) wrong, student can get credit here by demonstrating that they understand that they need to select site-‐specific integration of the transgene into the mouse genome. j. (5 points) In the process of creating the transgenic mice you want to study, several chimeric mice are born. Why might you not focus your study on the phenotypes of these chimeric mice? Many of their cells are not transgenic (in other words, do not have the genotype we are interested in). Only some random parts of these mice are lacking XPV function. Therefore, some phenotypic effects of the knockout would not be apparent in these mice. Also, it’s possible that phenotypic effects of knockout allele are recessive (in this actual case, most but not all are recessive), so they’d need to breed the mice to get homozygotes. Student should demonstrate: BIOL14 Quiz Name______________________ 1. (2 points) They know what chimeric means: having cells of two genetically distinct populations (in this case, because the non-‐transgenic blastocyst had transgenic ESCs added to it, though they might not explain that). So some tissues would not have the knockout genotype. 2. (1 points) They understand that expression of the normal XPV gene in some parts of the mouse would prevent expression of the knockout phenotype 3. (2 point) Some phenotypes may be observed only the homozygote Note that even if one didn’t get points 1 and 2, they can still come up with point 3. ...
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- Fall '13
- Genetics, SalI, Restriction enzyme, embryonic stem cells