Blackett Family DNA Activity 2

Genetics of STR Inheritance

Since there are no phenotypes associated with the CODIS STR loci, understanding the genetics of STR inheritance is simplified compared to other genetic problems. We need only consider the genotypes of the parents and their offspring. The alleles of different STR loci are inherited like any other Mendelian genetic markers. Diploid parents each pass on one of their two alleles to their offspring according.

Here is brief review of the genetic concepts and terms important for understanding STR allele inheritance. For an in depth tutorial, visit our Monohybrid Cross problem set.

• Allele. The different forms of a gene. Different STR repeat lengths represent different alleles at a genetic locus, i.e. 8 and 9 are different alleles of the THO1 locus.
• Locus. The position on a specific chromosome where the different alleles of a genetic marker are located. The plural is loci.
• Monohybrid Cross. Genetic cross involving parents differing in only one trait. Inheritance of each of the 13 STR loci can be treated as a separate Monohybrid Cross.
• Genotype. The genetic composition of the alleles at a locus. Since we are diploid, we each have two alleles at each locus.
• Homozygous. Both alleles at a locus are the same, i.e. Fred has a genotype of 29, 29 at the D21S11 locus.
• Heterozygous. Alleles at a locus are not the same, i.e. Normal has a genotype of 29, 31 at the D21S11 locus.
• Multiple Allelic Series. Many different alleles at a locus, i.e. the known alleles at the vWA locus are 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, and 21.
• Punnett Square. A diagram used to determine all possible genotypes that can occur in a genetic cross. All of the diagrams on this page are Punnett Squares.

Here are some examples of the how STR data can be interpreted in a family DNA study. The numbers outside the Punnett Squares are the parental alles that can be present in the egg or sperm of the parents. The numbers inside the squares are the genotypes possible for the resulting children.

Case 1

If the genotypes of both parents are known, we use a Punnett Square to predict the possible phenotypes of their offspring. Each child inherits one allele of a given locus from each parent. Panel (a) - At the D21S11 locus, the children of Bob Blackett and wife Anne can have four different genotypes. Son David is 28, 31. Daughter Katie is 29, 30. Panel (b) - Bob Blackett inherited the 31 allele from his mother, Norma. Therefore the 29 allele is paternal. If Bob's paternal was not 29, what would be your conclusion?

Case 2

In the genotypes of a mother and several children are known, it is often possible to unambiguously predict the genotype of the father. In this case, Karen is the mother with a genotype of 9, 9.3 at the THO1 locus. From the Punnett Square we can determine that the paternal alleles of Tiffany, Melissa, and Amanda are 8, 9.3, and 9.3, respectively. Therefore, their father Steve must have a genotype of 8, 9.3. If the three daughters had three different paternal alleles, what would be your conclusion?

Case 3

Sometimes only one allele of the father can be predicted when the genotypes of a mother and several children are know. In this example, the genotype of Karen, the mother, is 16, 17 at the D18S51 locus. The genotypes of the daughters are either 16, 18 or 17, 18. In each case, Melissa, Tiffany, and Amanda inherited the 18 allele from their father, Steve. We cannot determine if the genotype of Steve is homozygous, 18, 18 or 18, ? where the ? means any other allele.

Case 4

Is it possible to determine parental genotypes when only the genotypes of their children are known? Consider the case of Bob Blackett's 4 first coursins, Marilyn, Buddy, Dick and Janet. Bob did not have DNA samples for their parents, Bud and Louise, who are both deceased. In a real forensic case, Bud and Louise might represent "missing persons". In panel (a) we can arrange the 3 known genotypes of the 4 children. In panel (b) we predict the only two paternal genotypes for the parents that can account for the children. Note that we cannot determine which genotype goes with which parent.

Case 5

A variation on Case 4 is when there are only two genotypes known for the children, and both parental genotypes must be predicted. Panel (a) - Marilyn and Janet are 15, 16 at the locus D3S1358. Buddy and Dick are 18, 18. Panel (b) - The only parental genotypes that can give this result are 15, 18 and 16, 18. Once again, we cannot predict which parent as which genotype.

Case 6

Sometimes the parental genotypes cannot be predicted unambigously from the genotypes of their children. Marilyn is 16, 17 at the locus vWA. Buddy, Dick, and Janet are 16, 18. What are the parental genotypes? Panel (a) - One interpretations is that the parents are 16, 18 and 16, 17. Panel (b) - Another possibility is that one parent is 17, 18 and the other is 16, ?, where ? is any allele.