4. Microsatellites are repeating sequences of 2-6 base pairs of DNA. KAH is a hy

Question

4. Microsatellites are repeating sequences of 2-6 base pairs of DNA. KAH is a hypothetical human microsatellite locus with a repeating sequence of CAGA. The locus is shown in the figure with 20 bp of flanking DNA sequences that can serve as primers for amplification of the locus by PCR.

5’ GTTACTTAGGTATTGCCGAT   KAH     ATCTTGTAACCTAT ACTGTG 3’

3’ CAATGAATCCATAACGGCTA LOCUS TAGAACATTGGATATGACAC 5’

a. You will use PCR to genotype individuals for the KAH locus. The primers should be 20 nucleotides long. Determine the sequences for the two primers required to amplify the KAH locus. Note that you must label the 5’ and 3’ ends of both primers. Also, primer sequences, by designation, are always written in the 5’ to 3’ direction. 1 pt

b. Assume two KAH alleles, one with 8 and another with 5 copies of the repeating unit. Using the primers you have designed, what will be the sizes of the amplified PCR products for each allele? 1 pt

c. There are actually four known alleles of the KAH locus with 13, 10, 8, and 5 copies of the repeating unit. How many possible genotypes are there for these alleles, and what are they? 1 pt

d. One parent is heterozygous for the 13 and 8 alleles of the KAH locus and the other parent is heterozygous for the 8 and 5 alleles. The two parents live with three children. When you determine the genotype of the two children, the two genotypes are (8, 5) and (13, 10). What can you conclude about the parentage of the two children? 1 pt

Explanation / Answer

a.

Forward primer: 5’ GTTACTTAGGTATTGCCGAT 3'

Reverse primer: 5' CACAGTATAGGTTACAAGAT 3'

b.

one with 8 copies of the repeating unit: 72 b.p

one with 5 copies of the repeating unit: 60 b.p

c.

number of possible combinations with 5: 5-5, 5-6, 5-7, 5-8, 5-9, 5-10, 5-11, 5-12, 5-13 (total = 9)

number of possible combinations with 8: 8-8, 8-9, 8-10, 8-11, 8-12, 8-13 (total = 6)

number of possible combinations with 10: 10-10, 10-11, 10-12, 10-13 (total = 4)

number of possible combinations with 13: 13-13 (total = 1)

d.

number of possible allelic combinations in 13, 8 = 8-8, 8-9, 8-10, 8-11, 8-12, 8-13

number of possible allelic combinations in 5, 8 = 5-5, 5-6, 5-7, 5-8

The one withg genotype (8, 5) is child of both parents (13, 8 and 8, 5); however the other one with genotype (13, 10) is child of only one parent (13, 8)

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