Sickle cell anemia is caused by a single point mutation, in one of the exons that codes for the synthesis of β-globin. The picture below shows the DdeI restriction sites in the normal and sickle cell alleles of the β-globin gene. Notice that the mutation that causes sickle cell eliminates the second DdeI site. On the gel provide below draw what the restriction fragments would look like for an individual that is homozygous for the normal allele (AA), for an individual that is a heterozygous carrier (AS), and for an individual with sickle cell disease (SS). How many bands would appear on the gel of the heterozygote?

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yemmy yemmy · Answer 1 · 2020-04-24T06:37:19+02:00

Answer:

Note the fragments from DdeI restriction –

I. AA individuals have two copies of normal b-globin allele.

b-globin allele ----DdeI-----------> 176 bp + 201 bp + 600 bp

Since there are two copies of b-globin allele, the restriction of AA genome would produce 2 copies each of 176 bp, 201 bp and 600 bp fragments.

II. AS individuals have one normal and one mutated copy b-globin allele. So, following fragments would be obtained -

b-globin allele ----DdeI-----------> 176 bp + 201 bp + 600 bp

b-globin allele ----DdeI-----------> 376 bp + 600 bp

So, for such genotype, four fragments of MW as shown above would be produced.

III. SS individuals have two mutated copy b-globin allele. So, following fragments would be obtained -

b-globin allele ----DdeI-----------> 376 bp + 600 bp

So, for such genotype, two copies each of the fragments of MW as shown above would be produced.