3. A couple is seeking paternity testing for a child. They went through two type
ID: 176543 • Letter: 3
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3. A couple is seeking paternity testing for a child. They went through two types of analyses: a fast one through testing a marker based or restriction enzymes and a slower one through testing seven STR markers. Results of test 1: A restriction digest on a 1000 bp DNA sequence showed the following genotypes: Mother: Heterozygote with one allele having a recognition site at 500 bp for the restriction enzyme used (found in 15% of the couple's population and another allele having no recognition site (found in 50% of the couple's population). Father: Heterozygote with one allele having no recognition site, and another allele having two recognition sites: at 250 and 750 bp (found in 7% of the couple's population) a) Illustrate how the genotype for each parent will appear on a stained agarose gel. Use the gel representation below: Lane 1 is the ladder, Lane2 isthe mother's sample, and Lane 3 is the father's sample. Assume complete digestion. bp, and one at 750 bp (not observed in the couple's population before) Draw the child's genotype under Lane4. Does the result support the child being the couple's biological child? Why? co Add lanes with bands that represent potential biological children of the couple. 1500 600 500 400 300 200 100Explanation / Answer
At lane 2 there is two bands will occur. First band will come in front of 1000bp because 50% have no recognition site for one of the both allele. Second band will be in front 500bp of marker, because the other allele has recognition sequence at 500 bp; so the DNA segment will divide into two equal parts [1000bpDNA/2 =500bp]
In lane 3, there is three bands will occur. First band will come in front of 1000bp because no recognition site for one of the both allele. Second band will be in front 250bp of marker, because the other allele has recognition sequence at 250 bp. That recognition will divide the DNA into two parts 250 bp and 750 bp. So the third band will occur at 750bp region in front of marker.
The results supports 89% [(85+93)/2 = 178/2] that the child is from the couple because one of the allele is 1000 bp and other allele is divided into 250 and 750bp. The second allele is having only 7% chances from a population so it must come from father. And chances will be cumulative 93 % from father and 85% from mother [the child has 1 allele of 1000bp which must come from mother]. It does not support strongly the biological child of the both.
The STR results strongly supports that child from father because the 108/120 region of child is having least frequency in a population. This will 99.993% chances that the child from both of them.
The STR method is more precise.
If he chances of being a child if over 99% then it count as biological child.
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