I. Interpreting Your Gel

1. Observe the photograph of the stained gel containing your sample and those from other students. Orient the photograph with the sample wells at the top. Interpret the band(s) in each lane of the gel:

a. Scan across the photograph to get an impression of what you see in each lane. You should notice that virtually all student lanes contain one to three prominent bands.

b. Now locate the lane containing the pBR322/BstN I markers. Working from the well, locate the bands corresponding to each restriction fragment: 1,857bp, 1,058 bp, 929 bp, 383 bp, and 121 bp (may be faint).

c. One band visible. Compare its migration to that of the 929-bp and 383-bp bands in the pBR322/BstN I lane. If the PCR product migrates slightly ahead of the 929-bp band (lane 4), then the person is homozygous for the PV92 Alu insertion (+/+). If the PCR product migrates well behind the 383-bp band (lane 2), then the person is homozygous for the absence of the PV92 Alu insertion (-/-).

d. Two bands visible. Compare migration of each PCR product to that of the 929-bp and 383-bp bands in the pBR322/BstNI lane. Confirm that one PCR product corresponds to a size of about 715-bp and that the other PCR product corresponds to a size of about 415-bp (lanes 1 and 3). The person is heterozygous for the PV92 Alu insertion (+/-).

e. It is common to see an additional band lower on the gel. This diffuse (fuzzy) band is "primer dimer," an artifact of the PCR reaction that results from the primers overlapping one another and amplifying themselves. Primer dimer is approximately 50 bp, and should be in a position ahead of the 121 bp marker. (The presence of primer dimer confirms, in the absence of other bands, confirms that the reaction contained all components necessary for amplification, but that there was insufficient template to amplify the PV92 locus.)

f. Additional faint bands at other positions occur when the primers bind to chromosomal loci other than PV92 and give rise to "nonspecific" amplification products.

2. Determine the genotype distribution for the class by counting the number of students of each genotype (+/+, +/-, and -/-).

3. What can you say about any person in the class who has at least one + allele?

4. An allele frequency is a ratio comparing the number of copies of a particular allele to the total number of alleles present. Imagine a class of 100 students that list their genotype distribution as follows: +/+ 20 +/- 50 -/- 30 Since humans are diploid, the total number of alleles in the class is 2 x 100 = 200. The allele frequency for PV92+ is: 2 x 20 (homozygotes) + 50 (heterozygotes) / 200 = 90 / 200 = 0.45 Likewise, the allele frequency for PV92- is: 2 x 30 (homozygous) + 50 (heterozygotes) / 200 = 110 / 200 = 0.55 Using the genotype distribution from your class, calculate the frequencies of the + and - alleles of PV92.

5. If a population is genetically stable, then the allele frequencies will remain constant from one generation to the next. Such a population is said to be in Hardy-Weinberg equilibrium. Once the allele frequencies have been determined, the distribution of genotypes are described by the equation: p + 2pq + q = 1 where p and q represent the allele frequencies; p and q are the homozygote frequencies; and 2pq is the heterozygote frequency. Use the allele frequencies calculated for your class in Step 3 to determine the expected genotype frequencies. How do they compare with the actual genotype frequencies? How can you account for differences?

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