You are studying the expression of genes linked to a segment of the mouse sarcom
ID: 77834 • Letter: Y
Question
You are studying the expression of genes linked to a segment of the mouse sarcoma virus, which is regulated by glucoticoid hormones. You made a series of constructs with the viral segment in both orientations, upstream and downstream of a reporter gene (chloramphenicol acetyl transferase, CAT) You then transfected the constructs into two cell lines obtained from different tissues and measured CAT activity in the presence and absence of glucorticoid (dexamethasone). The orientation and location of the viral segment made relatively little difference in the expression of the reporter gene (CAT), which is the expected result if the viral DNA segment carries an enhancer of transcription. Results from the CAT gene alone and a gene construct with the viral DNA segment in one orientation are shown in Figure 1 You are puzzled by the results with cell line 1 because the viral DNA segment increased CAT expression by 20-fold in the absence of dexamethasone.Explanation / Answer
I] Yes, both the cell lines have glucocorticoid receptors. We, can say so because in the presence of dexamethason, a glucocorticoid agonist, the expression of the receptor gene (CAT) goes up.
ii] The 2 cell lines differ from each other in the basal expression of the enhancer protein (which will bind to the enhancer site). Probably, in cell line 1 the enhancer protein is already present in some amount which facilitates the expression of CAT even in the absence of dexamethason.
iii] The 2 cell lines would respond in a similar fashion i.e. they will not differ in the CAT activity, as far as the viral gene segment is not an enhancer region.
6B.
i] 1st lane.The cells would could contain 2 wild type copies of the gene before transformation. Enzymatic digestion with X would result in 3 segments : the largest segment (B,C), A which is considerably smaller and then the smallest D fragment. Southern blotting would show up onlu the BC fragment since the probe binds specifically to that region. Hence 1 band will show up as in lane 1.
ii] 2nd lane. Upon expected mutagenesis, the wild type longer BC fragment would get replaced by the shorter mutant one. So now, upon digestion by X the recombined chromatid would form A and D fragments similar to wild type but a rather shorter fragment for BC while the non-recombined chromatid would give a wild type BC segment. The probe now binds both of them hence you get a band corresponding to the one in lane 1 plus another band at a lower posiyion corresponding to the mutant BC segment.
iii] Only lane 1 would be visible. this is because 2 cross overs would result in the undoing of mutagenesis i.e. the normal copy of the BC fragment will be replaced. hence only one band would be visible.
I] exected sizes of DNA fragments = 7000bp
Digestion with PstI would cut at site 2, 4, 6 and result into 3 fragments = 1st >5000bp, middle 7000 bp (3000+4000) and last 7000bp (2000+5000). Out of these 3 only the last 2 fagments have sites homologous to the probe
II] exected sizes of DNA fragments = 14000bp
Digestion with PstI would cut at site 2, 6 and result into 2 fragments = 1st >5000bp and the other of 14,000bp (3000+4000+2000+5000). The latter has sites homologous to the probe.
III] exected sizes of DNA fragments = 7000bp
Digestion with PstI would cut at site 2, 4, 6 and result into 3 fragments = 1st >5000bp, middle 7000 bp (3000+4000) and last 7000bp (2000+5000). Out of these 3 only the last 2 fagments have sites homologous to the probe
Presence or absence of site 5 which is an EcoRI cut site doesnot matter when digesting with PstI.
IV]
(a) A ->homozygous wild type
B -> The mutation would be recessive one and individual B must be a heterozygote for the mutation
(b) A-> homozygous wild type
C -> either the individual is homozyous for a recessive mutation or the mutation is dominant in which case being heterozygous or homozygous wouldnt count.
(c) B -> The mutation would be recessive one and individual B must be a heterozygote for the mutation
C -> either the individual is homozyous for a recessive mutation or the mutation is dominant in which case being heterozygous or homozygous wouldnt count.
V] RFLP = restriction fragment length polymorphism
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