1. For each of these genetic traits which is the dominant allele and which is the recessive allele:

Transcription

1 Biology 114 Name: Lab Section Prelab questions for Lab 8 1. For each of these genetic traits which is the dominant allele and which is the recessive allele: being a tongue roller having attached earlobes having a Widow s peak interlocking fingers left over right hand having dimples having mid digital hair having freckles being a non tongue roller having unattached earlobes not having a Widow s peak interlocking fingers right over left hand not having dimples not having mid digital hair not having freckles 2. Huntington s Disease in humans is a lethal disease caused by inheriting a dominant allele. If one of your parents carried a single allele for Huntington s what is the probability that you would develop the disease? 3. If Huntington s Disease was the result of inheriting a recessive allele instead of a dominant allele and one of your parents carried a single allele for the disease, what would be the probability that you would develop the disease? 4. What is a restriction enzyme? 5. The process of gel electrophoresis can be used to separate DNA fragments into bands on the gel. How is the DNA in each band different from the other bands? Lab DOC 1

2 Biology 114 Lab 8 Genetics Objectives 1. Gather data on heritable human traits and use the Hardy Weinberg Principle to estimate allele frequencies in the human population. (Part I) 2. Use information provided by molecular biology techniques to determine which alleles an individual possess. (Part II) 3. Use agarose gel electrophoresis to separate fragments from various samples of DNA and use this information to establish possible paternity. Safety Precautions 1. In Part III use gloves and goggles when handling the agarose gel or loading samples. 2. Keep UV light box cover closed when the UV light is on. Materials Table for collecting class data on 7 phenotypes Colored sheets containing DNA sequences (a and b) representing Sam s chromosome 4 Scissors Scotch tape Micropipetters, 10µl Electrophoresis chamber and power source agarose gel tray and comb 10X TAE buffer (dilute to 1X for running buffer) DNA samples (Bste II, EcoR I, Pst I, Hind III) Introduction (Part I) Before coming to lab today you were asked to gather data on seven specific human heritable traits. Today you will pool these data with your classmates data and estimate the frequencies of the various alleles present in the human population. When we speak of the frequency of an allele we are not referring to how many individuals show the dominant phenotype and the recessive phenotype. Rather, we are referring to the relative abundance in the population of each of the kinds of genes diploid individuals possess for a trait. Consider the dominant tongue rolling gene (T) and its recessive non tongue rolling allele (t) in a very small hypothetical population of fifty people. In this population there are: 2 individuals with the genotype tt who are non tongue rollers, Lab DOC 2

3 32 with the genotype TT who are tongue rollers, 16 with the genotype Tt who are also tongue rollers. The frequencies of these alleles are the ratios of the total number of each type of allele (T or t) to the total number of genes for the trait. Since there are 50 people and each person is diploid, having two genes for each trait, that totals to 100 genes. In this population the number of T genes is 80 ( ). The frequency of T is 80/100 (80% or 0.8). The number of t genes is 20 ( ) and the frequency of t is 20/100 (20% or 0.2). If we are considering a trait with a single dominant allele and a single recessive allele, then the sum of the frequencies of each of these alleles should total 100% or 1 depending on how we express the frequencies. In discussing the Hardy Weinberg Principle typically the letter p is used to represent the dominant allele and q is used to represent the recessive allele. So we can write this equation: p + q = 1 and for the hypothetical population = 1. According to the Hardy Weinberg Principle if certain criteria are met in a population, then the frequency of the alleles for a given trait will remain the same over time. The Hardy Weinberg Principle applies to a biological population which is defined as a group of individuals of the same species who live in the same region and can reproduce with each other. The criteria that must be met are 1) the population must be large and randomly mating 2) none of the alleles can have an adaptive advantage over the other alleles and 3) there cannot be any immigration or emigration to and from the population. Often students of genetics not familiar with this principle expect recessive genes in a population to decrease in frequency over time. The mathematics from which the Hardy Weinberg Principle is derived demonstrates that this expectation is incorrect. We will not study this mathematical problem now but use the equations involved in demonstrating the Hardy Weinberg Principle for another purpose. We will use these equations to determine what proportion of a population is homozygous dominant and heterozygous. In the hypothetical population described above you were told how many individuals were TT and Tt. Both of these different genotypes have the same phenotype and they cannot be distinguished by observation. If an individual is a tongue roller, you cannot tell by how they do it whether they are TT or Tt. The data you have gathered on tongue rolling tells us how many tongue rollers and non tongue rollers there are. You will use it to calculate how many of the tongue rollers are TT and how many are Tt. You will not be able to tell who is TT or Tt however. In order to determine how many individuals in a population are homozygous dominant and how many are heterozygous, we use the frequencies of the alleles. These frequencies can be determined from the number of individuals who are homozygous recessive. In another hypothetical population there are 9 homozygous recessives in a group of 100. This is a frequency of.09 for the homozygous recessive phenotype. The Lab DOC 3

4 frequency of the recessive allele is the square root of this number.09 = 0.3. Why? If the frequency of the recessive allele is any number q then the probability of possessing two of those alleles or being homozygous recessive would be q X q or q 2. This comes from the law of probability which states that the probability of 2 events occurring together (inheriting a recessive allele from both parents) is the product of the probability of each event. The frequency of the homozygous recessive phenotype in the population (0.09) then is q 2 and the frequency of the recessive allele (0.3) is the square root. Since p + q = 1 we can also calculate p to be 0.7. Knowing these frequencies allows us to determine the frequencies of the homozygous dominant and heterozygous individuals. The homozygous dominant would be p x p or p 2 which in this population would be If there are 100 people and we now know 9 are homozygous recessive and 49 are homozygous dominant that means the remaining 42 people must be heterozygous. Notice that 2pq or 2 X 0.7 X 0.3 = The p X q represents a heterozygous individual. Can you figure out why the formula for determining the number of heterozygous individuals is 2pq? Procedure (Part I) 1. Enter your data on the seven traits in the table provided in class and calculate the class totals for each trait. Enter these numbers in Table 1 of the assignment sheet. This must be completed during the lab period but the remainder of Part I can be completed outside of the scheduled lab time. 2. Calculate the frequency of the recessive phenotype in the population of individuals polled by the class. Enter these numbers in Table Using the frequency of the recessive phenotype (=q 2 ) calculate the frequency of the recessive allele (q) and the frequency of the dominant allele (p). Enter these numbers in Table From the values of p and q determine the frequency of homozygous dominant and heterozygous individuals in the population. Enter these numbers in Table Using these frequencies determine the actual number of individuals who were polled that are homozygous dominant and heterozygous. Enter these numbers in Table 1. You can check your work by adding these numbers and the number individuals with the recessive phenotype. They should total to the number of individuals polled. Introduction (Part II) In the previous section we worked with a mathematical process that allowed us to determine how many people possessed each of the different genotypes but we could not tell who had a homozygous dominant genotype and a heterozygous genotype. In the next section we will use some more recently developed molecular techniques that allow us to analyze the DNA of individuals and determine if they carry specific alleles. The trait we will be working with is Huntington s Disease. Lab DOC 4

5 Huntington s disease is caused by an autosomal dominant allele. The disease is characterized by involuntary muscular movement and mental changes. It is a progressive illness; walking, swallowing, speaking, reasoning become increasingly impaired. The effect on families can be devastating because no cure has been developed and it cannot be treated effectively. Because the onset is typically late in life, individuals bear offspring and pass this gene on before they realize they carry it. Consequently this disease is easily transmitted to later generations. A diagnostic test has been developed using DNA analysis that allows individuals carrying an allele for Huntington s to be recognized. It involves using enzymes to cut the DNA into fragments called restriction fragments and then determining the length of these fragments using a process called gel electrophoresis. In gel electrophoresis the DNA fragments are pulled through a gel matrix by an electric current. DNA is negatively charged and the fragments then move toward the positive pole created by the current. The gel matrix impedes the larger fragments more than the smaller fragments so different sized fragments travel through the gel at different rates and separate into bands according to their size. The allele for Huntington s disease has a repetition of a CAG base sequence that is longer than the non-huntington s disease allele. This causes the Huntington s disease allele fragment to migrate more slowly than the non-huntington s disease allele in the gel matrix and it can be recognized. Individuals who have a history of this disease in their family have an agonizing decision to make whether to be tested for this incurable disease or not. Testing bears consequences because insurance companies and HMO s have in the past disqualified or reduced benefits to individuals known to carry this disease. However, those individuals who are tested do add to the overall knowledge of the disease and hope for a cure or effective treatment becomes more concrete. Whether to be tested or not is an individual decision. The following case study is true, however, the names have been changed to protect Sam and his family. Sam received a registered letter from Columbia University requesting that he submit a blood sample for their Huntington s disease study. His mother was healthy and his father died when he was a young child, so he thought the letter was meant for someone else, not himself. However, when the second registered letter from Columbia University arrived, he began to wonder. So he called his mother. His mother told him that it was not a mistake, Sam s paternal grandfather had died of complications of the disease at age 58. When Sam s father, Dave, died his mother had moved back to the Midwest and severed nearly all contact with Dave s family. Sam knew that his father had died in a car accident. Road conditions were slippery and Sam s father had lost control of his car and slammed into a bridge embankment. Sam s mother, Cheryl, told Sam that she always wondered if Dave purposefully did not try to compensate when the car began to skid. He had been depressed. Dave knew his own father had Huntington s disease and he thought that he was displaying some of the early symptoms. Sam s mother thought that Dave had, in effect, committed suicide. Sam later learned that one of his uncles, Mike, had Huntington s disease. Each of his cousins, like himself, had a chance of inheriting the disease. Lab DOC 5

6 Sam felt tremendous conflict and irrational guilt. He and his wife, Cindy, had two children with a third on the way. He wondered if he had the allele for Huntington s disease. He worried that he had passed it on to his children. And he wondered whether he really wanted to know if increasing physical disability and dementia were his fate. Put yourself in Sam s shoes and think about what you would have decided: would you have had the DNA analysis performed? In this exercise, you will model the DNA testing that would diagnose whether Sam inherited the allele for Huntington s disease. Procedure (Part II) 1. Work in groups of two. Obtain the colored sheets containing information about Sam s DNA. These are print outs of the base sequence in a short segment of human chromosome 4 that includes part of the Huntington gene. There are 2 different print outs (a and b) each representing one of the chromosomes of his diploid chromosome 4 pair. Take a copy of each. 2. Number the columns on each print out and assemble each into a long chains of DNA bases by taping the columns together in sequential order. 3. The restriction enzyme used to cut this DNA segment recognizes the sequence GGTGT ACACC and cuts between the last T and the first A. Mark the cuts with a pencil. Count the resulting number of base pairs in each fragment and record the number for each fragment in Table 2 of the assignment sheet. 4. Figure 1 is a diagram of a gel showing the migration pattern of the DNA of members of Sam s family who have been tested for Huntington s. Using the information you have recorded in Table 2, diagram Sam s DNA. Use a thin line to represent unique fragments and a thick line to represent fragments that both chromosomes have in common. 5. Compare Sam's DNA with the band patterns of other members of his family and his family's pedigree (Figure 2) and determine if Sam inherited the disease. Lab DOC 6

7 Figure 1 Lab DOC 7

8 Figure 2 Lab DOC 8

9 Introduction (Part III ) In the previous section you used information about alleles of genes obtained through molecular techniques such as DNA sequencing, producing restriction fragments and gel electrophoresis. You did not perform the actual techniques. In this section you will have an opportunity to perform one of the techniques, gel electrophoresis, and use the information obtained to do some genetic analysis. We have prepared samples of DNA isolated from cells. We also have prepared gels made of a substance called agarose that are used to separate DNA fragments cut with enzymes. You will load these gels with the DNA samples and expose the gels to an electric current that will cause the fragments to move through the gel and separate into bands according to the fragment lengths. The samples provided will represent DNA taken from a child, its mother and several people who are the possible fathers of the child. You will use the results of the gel electrophoresis to determine which of the men are most likely to be the father. Procedure (Part III) Your instructor will explain how the electrophoresis process works and show you how to use the apparatus. The gel has been prepared ahead of time as well as the buffer solution. 1. Obtain the five samples of DNA (child, mother and 3 possible fathers) that are contained in microcentrifuge tubes. In addition to DNA these samples contain a purple dye that will allow us to monitor how fast the electrophoresis is progressing and a stain for DNA that is visible under UV light. This stain is toxic and you must wear gloves!!!!! 2. Prepare the gel for loading with the DNA. Fill the chamber with the running buffer. Both ends of the chamber should be filled and the surface of the gel should be covered with buffer. Remove any large air bubbles under the tray by sliding the tray back and forth. 3. Pipette 10µl of each sample into a different well on the gel. Keep track of the location (lane number) of each sample. 4. Run the electrophoresis apparatus at 100V for 60 minutes. Periodically check the progress of the dye in the samples. We do not want the dye to run off the bottom of the gel. 5. After 60 minutes turn off the power and unplug the electrodes from the power supply. 6. Wearing gloves and goggles, remove the tray from the chamber. Allow the gel on the tray to cool for several minutes. Gently slide the gel from the tray onto the surface of the UV light box. Close the lid and observe your gel. Be sure not to lift the lid of the light box when the light is on or to look directly at the UV light. It can damage your eyes. Lab DOC 9

10 7. Make a diagram of the bands that have formed in each lane in the space provided on your assignment sheet and use this information to determine which man is most likely to be the father of this child. Lab DOC 10

11 Assignment Table 1 Total # polled # dominant phenotype # recessive phenotype Frequency recessive phenotype Frequency recessive allele Frequency dominant allele Frequency of homozygous dominant Frequency heterozygous # homozygous dominant # heterozygous Tongue rolling Ear lobes Widow s peak Finger locking Dimples Digital hair Freckles Lab DOC 11

12 1. Are dominant alleles always more common (have a higher frequency in a population) than recessive alleles? Support your answer with information from the data collected. 2. In order to have the heterozygotes be the largest group in a population what criteria is necessary in the gene frequencies of the alleles? 3. If the frequency of the dominant allele in a population is some number p and the frequency of the recessive allele is some number q, then why is the probability of being a heterozygote in the population 2pq rather than just p X q? Table 2 Fragment lengths (number of bases) Chromosome (a) Chromosome (b) 4. Did Sam inherit the Huntington allele? Lab DOC 12

13 Drawing of DNA restriction fragment bands after electrophoresis. Mother Child Candidate 1 Candidate 2 Candidate 3 5. Which of the men (indicate lane number from your drawing) is most likely the father of the child? Lab DOC 13