Lim
Forum Elite
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A patient with sickle cell anemia presents to the clinic. This single base-pair mutation for sickle cell anemia destroys the MstII restriction enzyme recognition site represented by an asterisk in the image. The restriction enzyme-binding sites are shown as arrows on the map. DNA from this patient is treated with MStII and run on an electrophoresis gel. The DNA is then hybridized with a labeled probe that binds to the normal gene in the position shown on the map. In the southern blot shown in the image, which lane represents the patient?
A
B
C
D
E
F
click for the larger figure
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| docarchana
Forum Guru
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i dont know the ans but i think this info will be useful.
"restriction endonuclease digests abnormal allele. normal allele is shown as 2bands and abnormal allele is shown as 1band"
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| blueocean
Forum Senior
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A- is the only choice who has 2 allele which can not be digested by the endonuclease.
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| icemaiden
Forum Elite
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E---a larger and a smaller fragment
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| Ivil
Forum Senior
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A should be the answer. since the pt is present with the disease ( homozygous and not heterozygous) and MSTII cannot fragment the gene due to the mutation, both alleles will 1.35kb.
B will a carrier or sickle cell trait and C will be a Normal person. D,E and F is not possible as the probe is directed against 1.15 region. 0.2 Kb segment will not be detected.
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| icemaiden
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yeah right A
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| eytanraz
Forum Newbie
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A, since sickle cell anemia is an homozigous recessive disorder, no one of the two beta globin genes on the two 11th chromosome will be cleaved in the site marked by the asterisk.
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