Kimbrough_Lab Activity Inherited Traits PKU.docx

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Lab Activity BIOL 102 Lab Activity – Inherited Traits Data Analysis and Conclusions Learning Objectives: ● Differentiate between genotypes and phenotypes, homozygous and heterozygous traits, and dominant and recessive alleles. ● Construct and use Pedigrees to predict phenotypes and genotypes of individuals for select inherited traits. ● Using Pedigrees identify inheritance of single gene autosomal dominant traits and single gene autosomal recessive traits. Question: ● Which phenotype in PKU inheritance is the dominant allele and which is the recessive allele? Summary: ● In this lab you will study the inheritance of phenylketonuria (PKU). First, you will create a pedigree to visualize inheritance. From your pedigrees, you will predict which phenotype is the dominant allele and recessive allele. You will then use data from multiple families to confirm your results. Materials Needed for Lab ● Paper/pen ● Family Data Provided at end of Pre-Lab Inherited Traits Spring 2024-1 1
Lab Activity BIOL 102 Results and Analysis YOU NEED THE PRE-LAB PORTION OF THE ASSIGNMENT Step 1. Open your pedigree from the pre-lab. This is the pedigree for Family A and is needed for the data analysis. Background - Determining Dominant and Recessive Phenotypes Once we have collected our data from multiple family studies, how can we determine the dominant allele and the recessive allele? To determine the dominant and recessive allele, we need to compare our data from multiple families with known outcomes for single gene inheritance patterns using Punnett Squares. This is done by determining if one of our parental pairings in which both parents have the same phenotype, produces the expected outcomes for the homozygous recessive crossing. If both parents are homozygous recessive, then both parents have the same phenotype and genotype and all offspring produced have the same phenotype and genotype as the parents. The homozygous recessive pairing is the only parental crossing that produces these results as shown in Punnett square 6 in Figure 1 and Figure 2. Inherited Traits Spring 2024-1 2
Lab Activity BIOL 102 Figure 2. Offspring outcomes for homozygous recessive traits; Punnett square, genotypes, phenotypes and % of offspring with genotype and phenotype. In our class data, we do not know which phenotype is the recessive phenotype, therefore we will need to test both parental pairings in which the parents have the same phenotype. If one of these parental pairings outcomes follows the homozygous recessive outcomes, then we will conclude that this is the recessive trait. Only one of the parental pairings must follow this pattern; this phenotype would be the recessive phenotype and the other phenotype would be the dominant phenotype. Refer to example 1 below. Example 1 – Finding the Dominant and Recessive Allele This example is for PTC – a bitter compound that only certain people can taste. Even though our data is for a different trait, the same principles apply. Family studies were conducted to determine the inheritance patterns of phenylthiocarbamide (PTC). If a person can taste PTC, their phenotype is noted as Taster and if the person cannot taste PTC, their phenotype is noted as non-taster. The results of the family studies have been summarized based on the phenotypes of the parents and the offspring and are provided in Table 1. Table 1. Offspring and Parents of 103 Families Classified as Phenylthiocarbamide Tasters and Non-Tasters. Source The GENETICS Project. University of Washington, Department of Genome Sciences. Parents Offspring Taster Offspring Non-Taster Taster x Taster 20 5 Taster x Non-Taster 10 5 Non-Taster x Non-Taster 0 10 Step 1: Using the results from the family study located in Table 1, we will determine if the first parental pairing (Taster x Taster) where the parents have the same phenotype is the homozygous recessive Inherited Traits Spring 2024-1 3
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Lab Activity BIOL 102 pairing by looking at the percentage or number of offspring with each phenotype. We will compare the results of this parental pairing with known outcomes for single gene inheritance. a. To determine the total number of offspring, sum the number of offspring that are tasters and non-tasters for the parental pairing of Taster x Taster. This data is found in the first row of the table (highlighted in blue). Example: Number of Offspring Taster + Number of Offspring Non-Taster = total offspring 20 + 5 = 25 b. To determine the % of offspring with Taster phenotype; Example: (The number of offspring with taster phenotype / the total number of offspring for that parental pairing) * 100. % = (20/25) * 100 = 80% Step 2: Using the results from the family study located in Table 1, we will determine if the second parental pairing (Non-Taster x Non-Taster) where the parents have the same phenotype is the homozygous recessive pairing. Again, we will compare the outcomes of this parental crossing with the known outcomes for a homozygous recessive pairing. The data for this parental pairing is in the bottom row of the data table (highlighted in orange). Conclusion: From the data for the parental pairings of Non-Taster x Non-Taster, we can identify the Non-Taster phenotype as the recessive trait. When both parents are Non-Tasters, 100% of the offspring have the same phenotype as both parents. This parental pairing matches the expected outcomes for a homozygous recessive parental pairing; thus, Non-Taster is the recessive phenotype, and the genotype would be tt. The phenotype Taster is the dominant phenotype, and the possible genotypes would be TT and Tt. Inherited Traits Spring 2024-1 4
Lab Activity BIOL 102 Assignment – Exit Ticket Directions: 1. Complete Results and Analysis 2. Submit your completed assignment on Canvas Student Name, ID, and Date Name: Ben Kimbrough Student ID: 3014821 Date: Jan 26 2024 Activity 1 – Additional Pedigrees Below are three additional pedigrees for the inheritance of phenylketonuria. After reviewing the pedigrees, answer the question. 1. Review the pedigrees for Families B, C and D. Compare them to examples in Figures 1 and 2 in the pre-lab. Is the diagnosis of phenylketonuria the dominant or recessive phenotype? Explain which pedigree and parental pairing supports your conclusion. Inherited Traits Spring 2024-1 5
Lab Activity BIOL 102 Inherited Traits Spring 2024-1 6
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Lab Activity BIOL 102 Activity 2 –Determining Single Gene Inheritance Patterns The data from multiple family studies for phenylketonuria (PKU) have been combined and tabulated in the table below. Refer to Example 1 in the beginning of the lab document on finding the dominant and recessive alleles. Table 1. Number of offspring per parental pairings. Parent phenotypes are either phenylketonuria present (PKU present) or phenylketonuria absent (PKU absent). Summary of Data in Pedigrees Parent Pairing Number of Offspring that are PKU positive Number of Offspring that are PKU absent PKU present / PKU present 4 0 PKU present / PKU absent 3 8 PKU absent / PKU absent 6 15 Inherited Traits Spring 2024-1 7
Lab Activity BIOL 102 1. Complete the tables below by calculating the % of offspring with each phenotype. Family Study Offspring Characteristics when both Parents PKU Present: Phenotypes: Fraction of Offspring: Percent: PKU absent 0 out of 4 0% PKU Present 4 out of 4 100% Family Study Offspring Characteristics when both Parents PKU absent Phenotypes: Fraction of Offspring: Percent: PKU absent 15 out of 21 71.42% PKU present 6 out of 21 28.57% 2. Using supporting data from the lab, identify the dominant and recessive phenotype (allele). The dominant phenotype is PKU absent, and the recessive phenotype is PKU present Activity 3 – Confirming Results with Genetic Testing To identify the gene segment that is responsible for PKU, researchers analyzed the DNA sequence of the gene that codes for phenylalanine hydroxylase (PAH) enzyme in hundreds of individuals with PKU. The PAH gene is located at a specific location on the chromosome in two copies with one copy of the gene inherited from each parent. Each copy of the gene is called an allele, and these copies are not necessarily the same. Inherited Traits Spring 2024-1 8
Lab Activity BIOL 102 Table 1 shows the DNA sequences for one short region of DNA. As each person has two copies of the chromosome, there are two DNA sequences; the first is labeled Copy 1, Chromosome 12 and the second is labeled Copy 2 Chromosome 12. The researchers looked for differences in the DNA sequence – some individuals may have cytosine (C) nucleotide while others may have an adenine (A) nucleotide in specific locations in the genetic sequence. If the DNA sequences or alleles are the same, the person is homozygous for the trait. If the DNA sequences or alleles are different, the person is heterozygous. 1. Complete the phenotype column in Table 1 below. To determine phenotype, use the pedigrees from the Pre-Lab (Family A) and Activity 1 (Family B, C and D). For example, Family A IV-3 is individual 3 in the 4 th gen. in Family A. The phenotype will be either PKU present or PKU absent. 2. Complete the genotype column in Table 1 below. To determine the genotype, compare the Sequence of Copy 1 Chromosome 12 with Copy 2 Chromosome 12 for each individual in the table. If the sequences are the same, the genotype is homozygous. If the sequences are different, the genotype is heterozygous. Table 1. DNA sequences for Chromosome 12 for individuals in pedigrees A, B, C and D. Individual Phenotype (PKU absent or PKU present) Sequence Genotype (Homozygous or Heterozygous) Family A III-1 PKU Absent Copy 1, Chromosome 12: TTGTATACAA Copy 2, Chromosome 12: TTGTATACAA Homozygous Family C IV - 8 PKU Absent Copy 1, Chromosome 12: TTGTATACAA Copy 2, Chromosome 12: TTGTATACAA Homozygous Family B IV-3 PKU Absent Copy 1, Chromosome 12: TTGTATACAA Copy 2, Chromosome 12: TTGTATAAAA Heterozygous Family D III - 9 PKU Present Copy 1, Chromosome 12: TTGTATAAAA Copy 2, Chromosome 12: TTGTATAAAA Homozygous Inherited Traits Spring 2024-1 9
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Lab Activity BIOL 102 Family C III-3 PKU Absent Copy 1, Chromosome 12: TTGTATACAA Copy 2, Chromosome 12: TTGTATACAA Homozygous Family B II-3 PKU Present Copy 1, Chromosome 12: TTGTATAAAA Copy 2, Chromosome 12: TTGTATAAAA Homozygous **Sequence is a nucleotide sequence corresponding to nucleotides cDNA 12 1060 -1075 3. Is the variation associated with PKU found in people that have a homozygous or heterozygous genotype? The variation associated with PKU present is found on both chromosome copies in individuals with PKU present 4. Using the genetic analysis above, identify the recessive and dominant phenotypes (allele). Explain your reasoning. The recessive phenotype is PKU present, and the dominant is PKU absent. This is because when an individual is heterozygous, their phenotype is that they are PKU absent, indicating that PKU absent is the dominant allele. Only when a variation that is associated with PKU present occurs in both chromosomes, the recessive phenotype is shown (PKU present) 5. Does the genetic analysis agree with your conclusion using family studies (Activity 2)? Why or Why not? Yes, the genetic analysis agrees with my conclusion from activity 2, where the dominant phenotype/allele is PKU absent and the recessive phenotype is PKU present. In the second activity, when a PKU present was paired with another PKU present parents all of the children were PKU present, indicating that it is a recessive trait. This was further supported by the genetic analysis, as only when both chromosomes had a variation that was associated with PKU present, PKU present was the phenotype. When an individual had heterozygous genotype, it did not portray PKU present, indicating that it is a recessive phenotype. Inherited Traits Spring 2024-1 10

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