The above data are from a landmark student looking into the relationship between sickle cell anemia and malaria. These data are from a population where malaria is prevalent. A is the allele for normal red blood cells; S is the sickle cell allele (you should know what this means). Is this population in H-W equilibrium? Make sure to use chi-square. Why or why not (if not, what assumption(s) violated). Genotype AA AS SS Total Observed 9365 (75.6%) 2993 (24.2%) 29 (0.2%) 12,387

The above data are from a landmark student looking into the relationship between sickle cell anemia and malaria. These data are from a population where malaria is prevalent. A is the allele for normal red blood cells; S is the sickle cell allele (you should know what this means). Is this population in H-W equilibrium? Make sure to use chi-square. Why or why not (if not, what assumption(s) violated). Genotype AA AS SS Total Observed 9365 (75.6%) 2993 (24.2%) 29 (0.2%) 12,387

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter19: Population Genetics And Human Evolution

Section: Chapter Questions

Problem 9QP: Using the HardyWeinberg Law in Human Genetics Suppose you are monitoring the allelic and genotypic...

See similar textbooks

Similar questions

Tay-Sachs disease in humans is caused by a recessive allele. The disease is characterized by mental deficiency and blindness, with death occurring by 4 years of age. Among Ashkenazi Jews of central European ancestry, about 1 in 3600 children is born with the disease. Assume Hardy-Weinberg equilibrium and calculate the frequency of the Tay- Sachs allele in this population. Round your answer to the nearest 0.001. Type your answer...
in a village where hte proportion of individuals who are susceptible to malaria (genotype hbA/hbA) is 0.53 and hte population is asssumed to be at hardy weingberg equalibrium what is the expected freqency of the HbS alle? hint what is the genotype of people with sickle cell disease and how is that genotype represneted in the hardy weinberg equation
An allele of the G6PD gene acts in a recessive manner to cause sensitivity to fava beans, resulting in ahemolytic reaction (lysis of red blood cells) afteringestion of the beans. The same allele also confersdominant resistance to malaria. The heterozygote hasan advantage in a region where malaria is prevalent.Will the equilibrium frequency (qe) be the same foran African and a North American country? Whatfactors affect qe?
House mouse (Mus musculus) Gene of interest: B4galnt2 (encodes glycosyltransferase enzyme) • Allele R: Associated with prolonged bleeding due to issues with blood clotting; associated with resistance to bacterial infections due to absence of intestinal expression. A Allele C: Associated with increased susceptibility to bacterial infections; not associated with prolonged bleeding. The population is closed; there are no new mice coming in. The cages are setup to allow for random mating. The B4galnt2 gene is not associated with any mating preferences or non-random mating patterns. Studies have shown that mutations in this locus are extremely rare. The population is EXTREMELY large. The population is kept in well maintained environment, free of bacterial infections. The mouse food is supplemented with a low dose of blood coagulator, which helps the blood clot and eliminates any prolonged bleeding. None of the genotypes display any side effects to this medication. B The population is…
The S-s antigen system in humans is controlled by two codominant alleles S and s. In a group of 3146 individuals the following genotypic frequencies were found:188 SS, 717 Ss, and 2241 ss. The chi-square value of this test is?The degrees is freedom value is ?Using a significance level of 0.5 the chi-critical value is? from this chi- squar chi-square test is it significant or non significant is the nul hypothesis of HWE accepted or rejected
A study on blood types in a population found the following genotypic distribution among the people sampled: 1101 were MM, 1496 were MN and 503 were NN. i) calculate the allelic genotypic frequencies. ii) using x² test, determine whether or not this population is in Hardy-Weinberg equilibrium. Important information from the x² distribution is given in the picture attached.
The hemoglobin B gene (Hb) has a common allele (A) ofa SNP (rs334) that encodes the HbA form of (adult) hemoglobin and a rare allele (T ) that encodes the sicklingform of hemoglobin, HbS. Among 571 residents of a village in Nigeria, 440 were A/A and 129 were A/T, and 2were T/T individuals were observed. Use the χ2 test todetermine whether these observed genotypic frequencies fit Hardy–Weinberg expectations.
What would happen to the frequency of heterozygous carriers of sickle-cell anemia (with AS genotype) if mosquitoes were completely wiped out in a large region?
I asked this question before but the solution was not correct and I'm really trying to figure out how to do this to get the correct answer. Please show work!DNA typing is used to compare evidence DNA (E) left at a crime scene to two suspects(S1 and S2). Suspect 1 is excluded by the evidence, but suspect 2 remains included. What isthe frequency of suspect 2's genotype if the allelic frequencies in the population are f(A1) =0.1, f(A2) = 0.2, and f(A3) = 0.7, and the population is at Hardy–Weinberg equilibrium?The answer is 0.49.
8% of XY individuals are color blind in a population. Assume Hardy-Weinberg conditions. Submit your answer as it is. a) What is the percentage of color-blind XX individuals? b) What is the percentage of XX individuals who are carriers? c) If this population has 1000 individuals with 50% of male and 50% of female, how many carriers are present in this population? Submit your answer as it is. Do not round up.
Sickle cell anemia is a genetic disease most common in those of African ancestry that results from the mutation of a single basepair. This mutation results in a change to the hemoglobin protein. In this experiment, “S” will represent the normal allele and “s” thesickle allele. They are codominant alleles. Therefore, SS is normal, Ss is not fatal but mildly symptomatic, and ss is debilitating,painful, and often fatal. After hundreds or even thousands of generations, both alleles are still common in those of African ancestry. How would you explain this?
If the chi-square value is 0 under the degree of freedom 1. What could be the interpretation and does it fit in the population of Hardy-Weinberg equilibrium?