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A 23-year-old female patient who is a carrier of sickle cell trait presents to the clinic. In this patient’s case, the single base-pair mutation for sickle cell anemia destroys the MstII restriction enzyme recognition site represented by an asterisk (*) in the image below. The restriction enzyme-binding sites are shown as arrows on the map. DNA is taken from this carrier patient and is treated with MstII and run on an electrophoresis gel. The DNA is then hybridized with a labeled probe that binds to the gene in the position shown on the map.
In the Southern blot shown to the right in the image above, which lane represents the patient?
A. A
B. B
C. C
D. D
E. E
F. F
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B
The probe only binds to one segment, so I would expect to see only one band. The two MstII sites surrounding the probe site are still intact, so I would expect the probe-bound segment to say 1.15kb. Therefore I think the answer choice is C.
Just realized my reading error, B!
B
The correct answer is B. Lane B represents the Southern blot of a heterozygous carrier of sickle cell anemia (also called sickle cell trait). Restriction recognition sites are located on DNA molecules and are recognized by restriction enzymes for DNA cleavage. In this question, MstII cleavage of the ?A-globin gene will result in two bands measuring 1.15 kb and 0.2 kb, respectively. However, only the band for the 1.15-kb fragment will show up on the gel, because the probe only hybridizes to that segment. The ?S-globin gene will produce a 1.35-kb fragment that will appear on the gel. Because this patient is heterozygous for sickle cell anemia, DNA analysis of her carrier status will reveal a 1.15-kb and a 1.35-kb band, consistent with lane B.
A is not correct. The band in lane A is from a sickle cell anemia patient with two copies of the ?S-globin gene. This gene results in a 1.35-kb band because the single base-pair mutation responsible for sickle cell anemia eliminates an MstII restriction site. Without the middle MstII restriction site cut, only one band is seen in this patient’s lane, since both copies of the gene in this patient are defective. A diagnosis of sickle cell anemia is confirmed by hemoglobin electrophoresis and the sickling tests. PCR is preferred for prenatal diagnosis.
C is not correct. The band in lane C is from an unaffected patient with two copies of the ?-A-globin gene. The gene results in a 1.15-kb fragment of DNA cut by the MstII restriction enzyme.
D is not correct. The bands in lane D could not result from any patient. The labeled DNA probe does not bind to the 0.2-kb DNA fragment and therefore would not be visualized on the Southern blot. Thus the 0.2-kb DNA fragment is “lost” because it is not detected by the probe.
E is not correct. The bands in lane E could not result from any patient. The labeled DNA probe does not bind to the 0.2-kb DNA fragment and therefore would not be visualized on the Southern blot.
F is not correct. The bands in lane F could not result from any patient. The labeled DNA probe does not bind to the 0.2-kb DNA fragment and therefore would not be visualized on the Southern blot. A heterozygote will have two bands, indicating one normal allele with an intact MstII site (two fragments), and a mutant allele with a missing MstII site (one fragment).
Kindly clarify my doubt..
What is the minimum length of DNA fragment that can bind a labelled DNA probe??
Thank you for the explanation
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