The following partial DNA Sequences encode distinct 3 or V region genes of a kap

Question

The following partial DNA Sequences encode distinct 3 or V region genes of a kapp light chain Use these sequences to illustrate the following Identify the heptamer sequences in the V1, V2.31 and 32 region genes shown above and show where your hairpin is a) Choose 1 V and 1 J gene from above to use in B and C b) Re-write the sequence in (a) (use dsDNA) with P-nucleotide additions of 2 or 4 nacleotides as hairpin loop c) N-Region Diversity- rewrite the sequence in (b) with random nucleotide additions (Use TTC t follow sample G) Recombinatorial Diversity-Use the other combinations of V and J genes provided above (that you didn't use in B and C), show the other combinations you could get with the same additions as above.

Explanation / Answer

Nucleic acid sequence

it is a succession of letters that indicate the order of nucleotides within a DNA or RNA molecule.

DNA SEQUENCING

It is the process of determining the precise order of nucleotides within a DNA molecule. It is placed in the following order- Adenine, Guanine, Cyotisne and Thymine.

An heptamer is a compound whose molecules are formed by seven monomers.

Here, V1 = CCACATGGCACAGTGX12ACAAAAACC

Heptamer sequence is CACAGTG. Hairpin sequence is at the position of X12 in between G and A.

HERE, J1 = GGTTTTTGTX23CACTGTGGTTA

Heptamer sequence is CACTGTG. Hairpin sequence is at X23 between T and C.

HERE, V2 = GGTACATTCACAGTGX12ACAAAAACC

Heptamer sequence is CACAGTG. Hairpin sequence is at X12 between G and A.

HERE, J2 = GGTTTTTGTX23CACTGTGCCAA

Heptamer sequence is CACAGTG. Hairpin sequence is at X23 between T and C.

B)

Let's take V1 Gene, After addition of 4 nucleotides in hairpin sequece it would be represented in dsDNA form as follows:

CCACATGGCACAGTGX16ACAAAAACC

GGTGTACCGTGTCACX16TGTTTTTGG

Let's take J1 Gene, so after addition of 2 nucleotides at hairpin loop it would be represented in dsDNA form as follows:

GGTTTTTGTX25CACTGTGGTTA

CCAAAAACAX25GTGACACCAAT

C)

Let's add up CTC in the above dsDNA

GGTTTCTCTTGTX25CACTGTGGTTA

CCAAAGAGAACAX25GTGACAGGAAT

D)

now you know the basics, you know how we will add or cut the nucleotides and hence you can complete the sequence.

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