Mendelian Genetics, Probability, Pedigrees, and Chi-Square Statistics

MENDELIAN GENETICS, PROBABILITY, PEDIGREES, AND CHI-SQUARE STATISTICS INTRODUCTION Hemoglobin is a protein found in red blood cells (RBCs) that transports oxygen throughout the body. The hemoglobin protein consists of four polypeptide chains: two alpha chains and two beta chains. Sickle cell disease (also called sickle cell anemia) is caused by a genetic mutation in the DNA sequence that codes for the beta chain of the hemoglobin protein. The mutation causes an amino acid substitution, replacing glutamic acid with valine. Due to this change in amino acid sequence, the hemoglobin tends to precipitate (or clump together) within the RBC after releasing its oxygen. This clumping causes the RBC to assume an abnormal “sickled” shape. Individuals who are homozygous for the normal hemoglobin allele (HBA) receive a normal hemoglobin allele from each parent and are designated AA. People who are homozygous for normal hemoglobin do not have any sickled RBCs. Individuals who receive one normal hemoglobin allele from one parent and one mutant hemoglobin, or sickle cell allele (HBS), from the other parent are heterozygous and are said to have sickle cell trait. Their genotype is AS. Heterozygous individuals produce both normal and mutant hemoglobin proteins. These individuals do not have sickle cell disease, and most of their RBCs are normal. However, due to having one copy of the sickle cell allele, these individuals do manifest some sickling of their RBCs in low-oxygen environments. People with sickle cell disease are homozygous for the sickle cell allele (SS genotype); they have received one copy of the mutant hemoglobin allele from each parent. The resulting abnormal, sickle-shaped RBCs in these people block blood flow in blood vessels, causing pain, serious infections, and organ damage. MATERIALS critical values table (see page 12) PROCEDURE 1. Watch the short film The Making of the Fittest: Natural Selection in Humans. While watching, pay close attention to the genetics of sickle cell trait and the connection to malaria infection. 2. Answer the following questions regarding genetics, probability, pedigrees, and the chi-square statistical analysis test. The Making king of the Fittes ttest: Na tura tur a l Selec Selectio ti on i n Huma uma ns QUESTIONS MENDELIAN GENETICS AND PROBABILITY 1. If two people with sickle cell trait have children, what is the chance that a child will have normal RBCs in both highand low-oxygen environments? What is the chance that a child will have sickle cell disease? Write the possible genotypes in the Punnett square. Normal RBCs in high- and 2. AA AS low-oxygen environments AS SS Sickle cell disease 1 of 4 ____________ 1 out of 4 ____________ a. 2 of 4 What is the chance that a child will carry the HbS gene but not have sickle cell disease? ___________________ b. What are the chances that these parents will have three children who are homozygous for normal RBCs? (Show 25 % chance 1 of 4 your work.) ___________________________ c. What are the chances that these parents will have three children who have both normal and mutant 1/2 x 1/2 x 1/2 = 1/8 12.5% hemoglobin beta chains? (Show your work.) ___________________________ d. What are the chances that all three of their children will show the disease phenotype? (Show your work.) 1/4 x 1/4 x 1/4= 1/64 1.56% ___________________________ e. What are the chances that these parents will have two children with sickle cell trait and one with sickle cell 1/2 x 1/2 x 1/2= 1/16 6.25% disease? (Show your work.) ___________________________ f. In the cross above, if you know that the child does not have sickle cell disease, what is the chance that the child has 2/3 66.67% sickle cell trait? ___________________ An individual who has sickle cell trait has children with an individual who does not have the HbS allele. a. AS and AA What are the genotypes of the parents? ____________________ b. In the Punnett square, show all the possible genotypes of their children. State the genotype and phenotype ratios of the offspring. Genetic: AA 50% ,AS 50% AA AS AA Phenotype: 50% normal ,50% sicklecell trait AS c. 0% What are the chances that any one of this couple’s children will have sickle cell disease? __________________ d. If this couple lives in the lowlands of East Africa, what are the chances that one of their children would be 1/2 50% resistant to malaria if exposed to the malaria parasite? _________________ The Making kinng of the Fittes ttest: Na tura tur al Selec Selectio ti on i n Huma uma ns 3. A woman with sickle cell disease has children with a man who has sickle cell trait. Answer the following questions. a. SS and AS What are the genotypes of the parents? ____________________ b. What is the genetic makeup of the gametes the mother can produce? ___________ c. d. S A or S What is the genetic makeup of the gametes the father can produce? ___________ In the Punnett square, show all the possible genotypes of their children. Then summarize the genotype and phenotype ratios of the possible offspring. Genotype: AS, 50% SS,50% AS AS SS SS Phenotype: sickle cell disease 50% , sickle cell trait 50% 50% 1/2 _______________________________________ 2 of the 4 children or 50% would surrive if they had sickle cell to protect against malaria. • IA is dominant over iO, • IB is dominant over iO, and • IAIB are codominant. Two parents who are heterozygous for type A blood and have sickle cell trait have children. Answer the following questions. Ia io AS a. What is the genotype of the parents? ____________________ b. What are the genetic makeups of all the possible gametes they can produce? ____________________ c. Complete the dihybrid Punnett square to determine the frequency of the different phenotypes in the offspring. (Note: Consider blood type and normal versus mutant hemoglobin in the various phenotypes.) Ia Ia AA Ia Ia AS Ia Ia AS Ia Ia SS Ia io AA Ia io AS Ia io AS Ia io SS Ia io AA Ia io AS io io AA io io AS Ia io AS Ia io SS io io AS io io SS IaA, IaS,ioA, ioS A type, no sickle cell, 3/16 18.75% A type , sickle cell trait, 3/8 37.5% A type, sickle cell disease , 3/16 18.75% O type, no sickle cell , 1/16 6.25% O type sickle cell trait, 1/8 The Making king of the Fittes ttest: Na tura tur a l Selec Selectio ti on i n Huma uma ns 5. Now try a different way of solving a dihybrid cross. Because of Mendel’s (second) law of independent assortment, you can work with the blood type gene and the hemoglobin gene separately. Set up two monohybrid crosses with the following parents: the mother is heterozygous for type B blood and has sickle cell trait, while the father has type AB blood and also has sickle cell trait. IA IB IA IO AA AS IB IB IB IO AS SS a. What are the chances that a child of this couple will have type B blood and sickle cell trait? (Show your work.) 1/2 x 1/2= 1/4 25% _________________________________________ b. What are the chances that a child will have type AB blood and will not have sickle cell disease? (Show your work.) 1/4 x 3/4= 3/16 18.75% _________________________________________ 1/2 x 1/4= 1/8 12.5% 1/2 x 3/4= 3/8 37.5% The Making king of the Fittes ttest:: Na tura tur a l Selec Selectio ti on i n Huma uma ns PEDIGREES 6. The following pedigree traces sickle cell disease through three generations of a family. Use the pedigree to answer the following questions. Female Male AS SS AS because individual 3's father had sickle cell and he passed it on to the male in generation 3. Individual 3 had to be a carrier. 50% 50% 0% Sickle cell disease __________ Sickle cell trait __________ Completely normal hemoglobin __________ In Africa the four males who are heterozygous have 2 advantages, they dont have sickle cell disease and they are more resistant to malaria. The Making king of the Fittes ttest: Na tura tur a l Selec Selectio ti on i n Huma uma ns 7. The following pedigree traces sickle cell disease through four generations of a family living in New York City. Use the pedigree to answer the following questions. SS AA ,AS I second generation the children are herterozygous for sickle cell, so they all have one normal gene and inheritated the S from the mother. Because they live in New York there is no advantage or disadvantage but if one of these individuals had a child with someone who also carried the sickle cell gene they would have a higher percentage of having a child that exhibit sickle cell disease. AS AS Because 2 out of the 4 children have sickle cell disease so both parents would have to be carriers. AS, AA 25% Having sickle cell disease __________ Having sickle cell trait 50% __________ 25% Being homozygous for normal RBCs __________ 50% Being resistant to malaria and not having sickle cell disease __________ The Making king of the Fittes ttest:: Na tura tur a l Selec Selectio ti on i n Huma uma ns 8. The following pedigree traces sickle cell disease through four generations of a family living in the highlands of eastern Africa. Use the pedigree to answer the following questions. a. What are the genotypes of the following individuals? (If more than one genotype pertains, include all possibilities.) AA ,AS AS AS Individual 1 __________ Individual 10 __________ Individual 2 SS __________ Individual 13 __________ Individual 7 AS __________ AS Individual 17 AA, __________ 1/4 or 25% ______________ 1/2 x 1/2 x 1/2= 1/8 or 12.5% Since they live in a relatively dry Highlands of East Africa there is low chance of malaria so the heterozygous genotype has no genetic advantage They would have a 0% chance of having a child who has normal hemoglobin. Because the father has sickle cell disease and he can only pass on a sickle cell allele so each child will have the disease or be a carrier The Making king of the Fittes ttest: Na tura tur a l Selec Selectio ti on i n Huma uma ns 9. Imagine that you are a genetic counselor, and a couple planning to start a family comes to you for information. Jerome was married before, and he and his first wife have a daughter with sickle cell disease. The brother of his current wife, Michaela, died of complications from sickle cell disease, but neither of her parents has the disease. a. Draw a pedigree representing this family. Be sure to clearly label Jerome and Michaela. J. b. M What is the probability that Jerome and Michaela will have a baby with sickle cell disease? Note that neither Jerome is heterozygous for sickle cell so for Michaela and Jerome to have a child with sickle cell anemia they must both be heterozygous. Based on the pedigree Michaela has more of a chance of being heterozygous for sickle cell. 2/3 x 1/4= 2/12= 1/6 or 16.67% 10. Natasha and Demarcus are planning on having children. Each has a sister with sickle cell disease. Neither Natasha nor Demarcus nor any of their parents have the disease, and none of them has been tested to see if they have sickle cell trait. a. Draw a pedigree representing this family. Be sure to clearly label Natasha and Demarcus. D. b. N Based on this incomplete information, calculate the probability that if this couple has a child, the child will have Each parent has a 2/3 chance of being heterozygous in there is a 1/4 chance of having a child with the disease therefore all values must be multiplied together. 2/3 x 2/3 x 1/4= 4/36= 1/9 or 11.11% The Ma kii ng of the Fittes ttest: Na tura ma nss tur a l Selec Selectio tion i n H Hum CHI-SQUARE STATISTICS 11. Multiple couples living in a small village in the eastern African lowlands, all of whom are heterozygous for the HbS allele, have 500 children among them. Of these children, 139 are homozygous for HbA, 279 are heterozygous for HbS, and 82 suffer from sickle cell disease. Are these data statistically significant? Explain using a chi-square statistical analysis test. Chi-Square Data Table Observed (o) Expected (e) (o − e ) (o − e ) 2 /e Normal AA Sickl cell trait AS 139 125 14 1.57 279 250 29 3.36 Sickle disease SS 82 125 -43 14.79 Phenotype/Genotype 19.724 2 0.01 P is less than 0.05 which means the null hypothesis is rejected, and there is a difference between the observed and expected data. Because the family lives in Africa in a hotspot for malaria the children who are heterozygous for sickle cell allele have an advantage over the homozygous and homozygous for sickle cell anemia genotypes The Ma kii ngg of the Fittes ttest: Na tura tur a l Selec Selectii on i n Huuma ns 12. Suppose there are 50 couples with the same blood type and hemoglobin genotypes. They live on a small, isolated Pacific island on which very few mosquitoes have been identified. All the individuals are heterozygous for type A blood and have sickle cell trait. The 50 couples had 224 children over the years. The children were all tested for blood type and for the presence of the sickle cell allele. Here are the results. Testing Results Blood Test Results Number of Children Type A, normal RBCs 48 Type O, normal RBCs 18 Type A, sickle cell trait 92 Type O, sickle cell trait 33 Type A, sickle cell disease 27 Type O, sickle cell disease 6 Are these data significant? Explain using a chi-square statistical analysis test. (Use the table below if you need assistance.) Chi-Square Data Table Phenotype Observed (o) Expected (e) (o − e ) (o − e ) 2 /e Type A, normal 48 42 6 0.86 Type O, Normal 18 18 4 1.14 A ,sickle trait O ,sickle trait 92 84 8 0.76 33 28 5 0.89 Type A , disease Type O, disease 27 42 6 14 -15 -8 5.36 4.57 13.58 5 < 0.05 The null hypothesis is rejected therefore the difference between observed and expected data is not solely due to chance Since healthcare is scarce there's a selection against the children with sickle cell disease this also shows why the no hypothesis was rejected because something else is acting on the population and it is the selection against sickle cell disease The Making king of the Fittes ttest: Na tura tur a l Selec Selectio ti on i n Huma uma ns f. Due to the increase in global travel and the prevalence of invasive species, the Anopheles mosquito carrying the malaria parasite was inadvertently introduced to this isolated Pacific island. A researcher, 100 years from the present day, decides to complete a follow-up study and monitors another 50 couples who are all heterozygous for type A blood and have sickle cell trait. These couples had 136 children. Based on the introduction of the Anopheles mosquito carrying the malaria parasite, predict scientifically logical observed numbers of children for each genotype possibility and complete a chi-square statistical analysis test. Chi-Square Data Table Phenotype Predicted Observed (o) Expected (e) (o − e ) (o − e ) 2 /e Type A, normal Type O, normal 19 5 25.5 8.5 -6.5 -3.5 1.66 1.44 Type A,sickle trait Type O, s-trait 70 25 51 17 21 8 8.65 3.76 Type A, disease 13 4 25.5 -12.5 6.13 8.5 -4.5 2.38 Type O, disease i. What is your predicted chi-square value (χ2)? 5 ii. Calculate df. __________ 24.02 __________ < 0.01 iii. Using the critical values table, determine the predicted P value. __________ reject iv. From your predicted numbers, do you accept or reject the null hypothesis? __________ v. Based on what you know about hemoglobin, sickle cell disease, blood type, and malaria, what selection pressures are acting on this population of children? Explain your answer. These people are facing selection pressures caused by limited healthcare as it relates to sickle cell disease in the introduction of malaria, there are selection pressures against sickle cell disease genotype and the normal genotype is selected against due to higher susceptibility to malaria infection. The Making king of the Fittes ttest: Na tura tur a l Selec Selectio ti on i n Huma uma ns CRITICAL VALUES TABLE P 0.995 0.975 0.9 0.5 0.1 0.05 0.025 0.01 6.635 df 1 0.000 0.000 0.016 0.455 2.706 3.841 5.024 2 0.010 0.051 0.211 1.386 4.605 5.991 7.378 9.210 3 0.072 0.216 0.584 2.366 6.251 7.815 9.348 11.345 4 0.207 0.484 1.064 3.357 7.779 9.488 11.143 13.277 5 0.412 0.831 1.610 4.351 9.236 11.070 12.832 15.086 6 0.676 1.237 2.204 5.348 10.645 12.592 14.449 16.812 7 0.989 1.690 2.833 6.346 12.017 14.067 16.013 18.475