Mendelian Inheritance Patterns lab 6

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1 Mendelian Inheritance Patterns lab 6 Objectives Upon completion of this activity, you should be able to: Recognize Mendelian ratio s for monohybrid (3:1) and dihybrid crosses(9:3:3:1). Use probability and Punnett squares to solve genetics problems. Diagram family pedigree s and determine mode of inheritance and genotype for individuals in the pedigree. Since the rediscovery of Mendel s experiments in the early 1900 s, researchers and medical professionals have confirmed that in many cases, genes are inherited in predictable patterns. The Chromosome Theory of Heredity further supported Mendel s original ideas by demonstrating that genes reside on chromosomes and that the separation of homologous chromosomes during meiosis accounts for Mendel s law of segregation and law of independent assortment. Traits that follow simple patterns of inheritance are said to be Mendelian. Although many human traits show much more complex patterns of inheritance, predictions for Mendelian traits can be calculated using a device called a Punnett square. It is important to recognize that predictions from Punnett squares are statistical in nature and express probabilities not certainties. Punnett squares are constructed by writing the known maternal and paternal alleles for a given trait on the outside of the square. The combination of alleles from both parents are placed inside the square compartments. Possible homozygous and heterozygous genotypes can then be used to predict or calculate the chance that offspring will express the trait in question. In this activity you will be given the opportunity to practice solving genetics problems. You will also be introduced to several modifications to Mendelism and to pedigree analysis. You are encouraged to work with others to sharpen your problem-solving skills. Human Genetics Problems Procedure Work with a partner(s) to solve the following genetics problems. Number each problem and show all work. Once you are satisfied with your solution circle the correct answer. You may use marker boards to diagram and discuss the more difficult problems but your work and final answers should be transferred to your own paper. This lab is due one week from today and is worth 20 points. Probability 1. Suppose that a couple has five children all boys. a. What is the probability that their next child will be a girl? b. If they have a sixth child who is also a boy what is the probability that a family would consist of six boys; no girls? 2. How many possible gametes can be produced from the genotype NnMmKKDdFFGgSsQq?

2 Single Factor Crosses (one trait) 3. An Rh-positive man (RR) married an Rh-negative rr woman. Their first child was normal and their second child had hemolytic disease of the newborn or Rh incompatibility disease. a. What genetic explanation might be offered? b. What prediction might be made concerning future children by this couple? 4. A young female about to be married learns that she was born with hemolytic disease/rh incompatibility and is greatly worried that some of her children might be born with the same condition. Her fiance refuses to have his blood tested, saying it is all a lot of foolishness, that women have been having babies for centuries without all of this "newfangled Rh business. What could you tell her that might ease her anxiety? Double Factor Crosses (two traits) 5. Use the following symbols for the inheritance of thumb type and earlobes. T -normal thumb t -hitchhikers thumb E - free earlobes e - attached earlobes Two individuals heterozygous for both traits have children. a. List the genotypes of the parents. b. List the possible phenotypes of their offspring and the probability of each. 6. Tay-Sachs disease is an autosomal recessive inborn error of metabolism. Infants born with Tay-Sachs produce a defective enzyme needed for the metabolism of lipids and usually die within the first few years of life. Abnormally short fingers (brachyphalangy) is caused by a recessive allele (b) which is lethal in the homozygous condition (bb). Individuals with short fingers are Bb while BB individuals have normal fingers. a. Assume that a mating takes place between two people who are both brachyphalangic and carriers of Tay-Sachs. List the possible phenotypes of their offspring and the probability of each. b. What proportion of their live-born offspring could be expected to celebrate their tenth birthday? Sex-Linked (Genes found on the X chromosome; rarely on the Y) 7. In humans, the condition for normal blood clotting dominates the condition for prolonged clotting time known as hemophilia. The gene for Factor VIII associated with the most common form of hemophilia is on the X chromosome. A normal male marries a woman whose father had hemophilia. a. List the genotypes of the parents. b. What are the chances that they will have affected children? c. What is the probability that any of their sons will be affected with hemophilia? Multiple Alleles (traits having more than two forms of the same gene) 8. Abo blood typing in humans is a classic example of multiple alleles. The alleles (A or I A ) and (B or I B ) are codominant (both are expressed if present) while the allele (o or i) is recessive. A case was brought before a certain judge in which a woman of blood group O presented a baby of blood group O, which she claimed as her child, and brought suit against a man of group AB whom she claimed was the father of the child. What bearing might the blood type information have on the case?

3 9. A woman bears a child out of wedlock and sues a particular man for support of the child, claiming that he is the father. Blood typing shows that she is type A, her child is type O and the man is type B. The man says that this proves that he is not the father, but the judge says that it proves no such thing. Was the father or the judge correct? Include a Punnet square to support your answer. 10. A wealthy elderly couple die in the same accident. A woman shows up at the probate of the will, claiming the fortune they left behind. She contends that she is their only daughter who ran away from home when quite young and is now entitled to the fortune. Hospital records show that the deceased couple were of blood types AB and O. The woman has blood type O. Will this information help the judge in deciding the case? 11. In humans, kinky hair (H++), curly hair (H+), wavy hair (H), and straight hair (h) are dominant in that order. Two of the possible alleles are found at the same loci on chromosome 5. Dark hair dominates red hair. A curly, red-haired man, who had a straight, dark-haired mother, marries s straight, dark-haired female who had a father with curly, red-hair. What kind of children can they produce and what is the probability of each? Epistasis (gene interaction) 12. In humans, deafness is due to a homozygous condition of either or both recessive genes dd and/or ee. Both dominant alleles D (for normal ear bones) and E (for normal auditory nerves) are needed for hearing. A male with the genotype Ddee and a female with the genotype ddee have children. a. What is the hearing ability of the male, the female and their offspring? Sex Influenced (autosomal traits influenced by sex hormones) 13. Many people think that baldness follows only a maternal pattern of inheritance and is transmitted only from the mothers side of the family. Is this true? Why or why not...explain. More Than Two Genes Use the following information to help you solve the next four problems. S - normal hemoglobin F - six fingers C- cleft chin s - sickle cell anemia f - five fingers c - no cleft 14. Two parents heterozygous for sickle cell anemia and six fingers plan to have eight children. What is the probability that they will produce children that do not suffer from sickle-cell or have six fingers? 15. Two people heterozygous for six fingers and cleft chin mate. What is the probability that their first child will have five fingers and a chin unlike either parent? 16. Dick and Jane have always thought of themselves as healthy and normal. They have 9 children. Five of their children have sickle-cell anemia and three of their kids were born with polydactyly. What are the most likely genotypes for Dick and Jane? 17. Genotype of parents: SsFfCc x Ss FfCc What is the probability that the offspring of this mating could be considered normal and not have a cleft chin? think product rule!

4 Pedigree Analysis Since it is unethical to control mating in humans most traits are analyzed using a family pedigree. Pedigree charts are reconstructions of the inheritance of a single phenotype within several generations of a family. By analyzing how the phenotype is passed from generation to generation, it is possible to identify patterns of inheritance. Look for critical matings which exclude certain modes of inheritance to determine if your pedigree is conclusive. Inconclusive pedigrees do not allow you to draw an accurate conclusion based solely on the information within the family. Once the mode of inheritance has been determined, genotypes can be determined and predictions of phenotype and genotype for individuals of future generations can be made. Genetic counselors can use this information to help prospective parents make difficult reproductive decisions if genetic diseases such as Tay-Sachs appear in the family pedigree. Here are some principles for analyzing pedigrees. 1. The emphasis is on possibilities not probabilities. The number of individuals in a family is too small to expect typical Mendelian ratios. Instead, use the logic of Punnett squares to determine possible genotypes. 2. The first question in the analysis of any pedigree is "Is the condition of interest dominant or recessive? Several Rules of Thumb will help you make this decision. Mode of Inheritance Rule of Thumb Examples (McKusick #) Autosomal Dominant (AD) Autosomal Recessive (AR) X-linked Dominant (XD) X-linked Recessive (XR) Y-linked trait cannot hide; unaffected parents cannot have affected progeny; both sexes affected often skips generations; unaffected parents can have affected progeny; both sexes affected more females affected than males; affected males pass to all daughters more males affected than females; carrier mothers pass to sons no females affected; affected fathers pass to all sons Free earlobes (128900) Achondroplasia (100800) Cystic fibrosis (219700) PKU (261600) Vitamin D resistant rickets (307800) Colorblindness (303800) SRY gene Sex-limited traits found in males only or females only Abundant chest hair Sex-influenced male dominant - all sons of affected mothers will be affected female dominant - all daughters of affected fathers will be affected Male pattern baldness (109200)

5 Problem Solving Using Pedigrees See page 78 for how to use pedigree symbols. Answer applied questions 2, 3, 5, 7, 8, 14, & 15 found at the end of Chapter 4 (p. 84). Include the answers and pedigrees in your lab report. Probability and Genetic Counseling Most genetic counseling situations involve the preparation of a pedigree for the family seeking advice. The counselor will try to determine a phenotype and genotype for each individual in the pedigree. The counselor will then apply genetic principles to determine the chance that an affected child will be produced in future matings. To demonstrate the application of these concepts, consider the pedigree below for a family with cystic fibrosis. Assume that individuals who have married into the family do not carry the recessive allele for the trait, unless there is evidence to the contrary. In the general population, the chance that two random individuals would mate and produce a child with cystic fibrosis can be calculated as follows: 1/25 carrier rate x 1/25 carrier rate x 1/4 = 1/2500 chance of Aa chance of Aa probability of affected child In this example, you will be calculating the increased risk for a child with cystic fibrosis if consanguinity occurs. Determine the risk for a child with cystic fibrosis if the following cousins and second cousins should mate. First number the generations from I-III, then number the individuals in the pedigree from left to right starting over each generation. Then, use the Product Rule to calculate the risk for a CF child in the following matings III-1 x III-7 III-2 x III-11 III-6 x III-15 III-16 x III-17

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