Concept explainers
To analyze:
In a sample of
a. By calculating the chi-square analysis, check whether the number of children with CF in these families are reliable to the expected ratio?
b. Under expectations of binomial probability, what is the expected distribution of CF children in families with
c. Under expectations of binomial probability, is the above distribution of CF children in families with
Introduction:
In genetic crosses, to determine whether the data is significant or whether it fits in any of the Mendelian ratios, the Chi-square test
The test is performed in a series of steps- 1. Hypotheses (Null and Alternative), 2.Calculating chi-square test statistic, 3. Determine the Degree of Freedom df and p-value, 4. The decision about the null hypothesis, 5.Conclusion.
The formula for test statistics
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Genetic Analysis: An Integrated Approach (2nd Edition)
- If the frequency of those exhibiting a monogenic autosomal recessive phenotype caused by only one known clinically relevant allele is 1/8500 in a given population, what is the frequency of the wild type allele? Please give your answer as a percentage to 3 decimal places, do not include the % symbol.arrow_forwardIf the frequency of those exhibiting a monogenic autosomal recessive phenotype caused by a rare clinically relevant allele is 1/9250 in a given population, what is the carrier frequency? Please give your answer as a percentage to 3 decimal places, do not include the % symbol.arrow_forwardIf the frequency of those exhibiting a monogenic autosomal recessive phenotype caused by a rare clinically relevant allele is 1/8500 in a given population, what is the carrier frequency? Please give your answer as a percentage to 3 decimal places, do not include the % symbol. ANSWER: In a population where the frequency of those exhibiting a monogenic autosomal recessive phenotype caused by only one known mutation is 1/8500 the carrier frequency is a percent.arrow_forward
- The following genetic map describes three hypothetical human autosomal genes, each of which exhibits two alleles. Two-factor map distances are shown. A = Artistic (dominant) a = Inartistic (recessive) M = Moral (dominant) m = Immoral (recessive) G = Generous (dominant) g = Greedy (recessive) Assume that these traits exhibit simple Mendelian dominance/recessiveness. The coefficient of coincidence for this map is 0.4. An artistic, moral, generous heterozygous female of genotype AMG/amg marries an inartistic, immoral, greedy homozygous male of genotype amg/amg. What is the probability that their firstborn child will be inartistic, immoral and greedy? What is the probability that their firstborn child will be inartistic, moral and generous? What is the probability that their firstborn child will be artistic, immoral and generous?arrow_forwardAs it turned out, one of the tallest Potsdam Guards had an unquenchable attraction to short women. During his tenure as guard, he had numerous clandestine affairs. In each case, children resulted. Subsequently, some of the childrenwho had no way of knowing that they were relatedmarried and had children of their own. Assume that two pairs of genes determine height. The genotype of the 7-foot-tall Potsdam Guard was A9A9B9B9, and the genotype of all of his 5-foot clandestine lovers was AABB. An A9 or B9 allele in the offspring each adds 6 inches to the base height of 5 feet conferred by the AABB genotype. a. What were the genotypes and phenotypes of all the F1 children? b. Diagram the cross between the F1 offspring, and give all possible genotypes and phenotypes of the F2 progenyarrow_forwardHemophilia and color blindness are both recessive conditions caused by genes on the X chromosome. To calculate the recombination frequency between the two genes, you draw a large number of pedigrees that include grandfathers with both hemophilia and color blindness, their daughters (who presumably have one chromosome with two normal alleles and one chromosome with two mutant alleles), and the daughters sons. Analyzing all the pedigrees together shows that 25 grandsons have both color blindness and hemophilia, 24 have neither of the traits, 1 has color blindness only, and 1 has hemophilia only. How many centimorgans (map units) separate the hemophilia locus from the locus for color blindness?arrow_forward
- Cystic fibrosis is an autosomal disease that mainly affects the white population, and 1 in 20 whites are heterozygotes. Genetic testing can diagnose heterozygotes. Should a genetic screening program for cystic fibrosis be instituted? Should the federal government fund it? Should the program be voluntary or mandatory, and why?arrow_forwardIf the chi-square value came out to 14,224.221 and the p-value was said to be less than 0.01, what do you conclude about whether these traits follow a simple Mendelian inheritance pattern? Explain your answer.arrow_forwardFrom the pedigree shown here, answer the following questions with regard to individual VII-1. A. Who are the common ancestors of her parents? B. What is the inbreeding coefficient for this individual?arrow_forward
- One particularly useful feature of the Hardy-Weinberg equation is that it allows us to estimate the frequency of heterozygotes for recessive genetic diseases, assuming that Hardy-Weinberg equilibrium exists. As an example, let’s consider cystic fibrosis, which is a human genetic disease involving a gene that encodes a chloride transporter. Persons with this disorder have an irregularity in salt and water balance. One of the symptoms is thick mucus in the lungs that can contribute to repeated lung infections. In populations of Northern European descent, the frequency of affected individuals is approximately 1 in 2500. Because this is a recessive disorder, affected individuals are homozygotes. Assuming that the population is in Hardy-Weinberg equilibrium, what is the frequency of individuals who are heterozygous carriers?arrow_forwardConnection to Quantitative traits: SNPs are inherited in a Mendelian fashion and are often polygenic in nature. We can think of SNPs in terms of either contributing or non-contributing alleles. A study of SNPs correlated with heart disease has shown that heart problems are severe if 9 or more of the alleles at 6 loci are of the contributing variety. What is the probability the following parents will have a child that is susceptible heart disease? AaBbccDDEEFf x AaBbCCDdEeffarrow_forwardIt has been hypothesized that people who are heterozygous for the allele that causes the deadly genetic condition cystic fibrosis (which, among other symptoms, reduces fertility) are more resistant to the deadly disease tuberculosis. Question - if the cystic fibrosis allel protects against tuberculosis the same what the sickle cell allele protects against malaria, then which of the following should be true in comparison between regions with or with out TB. -cystic fibrosis deaths should be more common in regions with tb - CYSTIic fibrosis deaths should be less common in regions with tb - CYSTIic fibrosis deaths should be equally common in regions with tb -Regional differences in the cystic fibrosis death rate should be purly random and upredictable.arrow_forward
- Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage Learning