1. A pair of genes AB are completely linked together, so they are tightly linked. Give the genotypic ratio of the cross between a heterozygous and recessive parents. 2. Why is coefficient of coincidence a basis for strength of linkage?
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1. A pair of genes AB are completely linked together, so they are tightly linked. Give
the genotypic ratio of the cross between a heterozygous and recessive parents.
2. Why is coefficient of coincidence a basis for strength of linkage?
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- 2. Juvenile free-living ground squirrels utilize responses known as "alarm calls" in response to predators. These calls can alert others to danger. Newly weaned squirrels emerge from the nest at about 25 days of age. Do the squirrels emerge with automatic responses to alarm signals, or are the responses learned? To answer this question, a biologist played recordings of alarm calls for three random samples from populations (by age) of squirrels and observed the squirrels' responses. She reasoned that if the responses were not different for the different age groups, the squirrels must be responding by instinct. On the other hand, if the responses differ with the age of the squirrels the young squirrels must be learning to respond to the alarm calls. Her data for the responses, categorized by age, is presented below: Response to alarm call Enter or Stand up or run to burrow freeze Totals Days since weaning 1-5 21 18 39 6-10 16 24 40 10-15 12 19 Totals 49 49 98 Is there convincing evidence…-. Thanks to Gregor Mendel's famous experiments in the 1800s, we know that in a hybridization event between a homozygous plant carrying two yellow-pod genes and a heterozygous plant carrying one yellow-pod and one green-pod gene, there is a 75% chance that the offspring plant has green pods, and a 25% chance the offspring plant has yellow pods. If one of these hybridization experiments results in 12 offspring, what is the probability that exactly 5 of them have yellow pods? Make sure to define a random variable, state its distribution, and translate the question into a probability statement before solving. plants w/ yellow ands at of 12 ՈՐThe coat color of black bears depends on a gene with two alleles. Allele B is dominant and confers black coat, whereas allele b is recessive and homozygous bears have white coats (albino black bears, I know it sounds weird but it is true). As we know from genetics each bear has two alleles and the coat colors are as follows: BB-black; Bb-black; bb-white. A researcher obtained the following data for the genotypes (BB, Bb or bb) of black bears: Genotype Number of bears BB 42 Bb 24 bb 21 You are performing the following hypothesis test: H0: The genotypes of the bears follow a binomial distribution, with a probability of receiving the b allele equal to the observed proportion of the b allele in the population. HA: The genotypes of the bears do not follow a binomial distribution What is the expected frequency of bears with genotype BB, Bb, and bb? Give the absolute expected frequency (number of bears) not the fraction/proportion of BB bears
- Consider a situation where you have a parental cross with the mother and father phenotypes listed below. Remember that the genotype for the wild-type parent is always homozygous. The counts of the F1offspring are listed in Table 1. Two randomly selected individuals are selected and mated to produce a set of F2individuals. If you test the F2counts to determine whether they are consistent with an autosomal dominant mode of inheritance, what is your decision? Use a 0.05 significance level. PARENTAL CROSS Parental cross: Mother with disease phenotype, Father with wild-type phenotype. Table 1. F1 DATA Gender Phenotype Disease Wild-type Male 0 23 Female 0 34 Table 2. F2 DATA Gender Phenotype Disease Wild-type Male 7 25 Female 0 25 A. Do not reject the null hypothesis that the F2 data are consistent with an autosomal dominant mode of inheritance; chi-square goodness of fit test p-value is greater than 0.05. B. Reject the null hypothesis that the F2…A consumer products testing group is evaluating two competing brands of tires, Brand 1 and Brand 2. Though the two brands have been comparable in the past, some technological advances were recently made in the Brand 1 manufacturing process, and the consumer group is testing to see if Brand1 will outperform Brand 2. Tread wear can vary considerably depending on the type of car, and the group is trying to eliminate this effect by installing the two brands on the same 10 cars, chosen at random. In particular, each car has one tire of each brand on its front wheels, with half of the cars chosen at random to have Brand 1 on the Español left front wheel, and the rest to have Brand 2 there. After all of the cars are driven over the standard test course for 20,000 miles, the amount of tread wear (in inches) is recorded, as shown in the table below. Car 1 3 4 6. 8 69 10 Brand 1 0.23 0.29 0.28 0.23 0.17 0.26 0.17 0.29 0.30 0.21 Brand 2 0.22 0.27 0.27 0.29 0.31 0.24 0.27 0.33 0.22 0.25 Difference…a researcher in a large organisation wishes to study sickness absence among its employees. the organisation has a large number of branches. each keeps its records of its sickness leave.after a sample was collected it became apparent that the branches fell into three natural groups in terms of sales small, medium and large.from the data of all the branches the researcher found that 210 randomly selected staff 90 worked in small branches 36 in medium branches and the rest worked in large branches. in total 96 of the selected staff had no sickness day off of which 52 worked in small branches and 29 worked in large branches. form a table showing this information clearlya researcher in a large organisation wishes to study sickness absence among its employees. the organisation has a large number of branches. each keeps its records of its sickness leave.after a sample was collected it became apparent that the branches fell into three natural groups in terms of sales small, medium and…
- 4. To determine whether dogs prefer a specific brand of dog food to that of a competitor, market researchers are working with a local dog shelter to randomly select dogs for participation in a study. All randomly selected dogs are put into a large room with two bowls of food — one with their brand and another with the competitor's brand. The first two dogs immediately go for the dog food containing their ingredient, so the researchers stop the experiment, concluding this is enough evidence that their product will sell the most. Is this a good experiment and conclusion? State reasons for your explanation, and what you would do differently?Assortative mating is a nonrandom mating pattern where individuals with similar genotypes and/or phenotypes mate with one another more frequently than what would be expected under a random mating pattern. Researchers studied this topic and collected data on Scandinavian men and their female partners. They found out 60% of Scandinavian men have blue eyes. Among those men with blue eyes, 80% of them choose female partners with blue eyes. Among those Scandinavian men whose eye color is not blue, 20% of them have partners with blue eyes. Given that a man’s partner has blue eyes, what’s that probability this man has blue eyes? If 60% of Scandinavian men have blue eyes and you randomly select 8 men, what’s the probability that none of them has blue eyes?A crop researcher investigated the phenotypes that resulted from crossing two different types of cucumber plants. There are 4 possible resulting phenotypes (numbered 1, 2, 3, 4 in the data table below). Mendel's laws of genetic inheritance ('the model') suggest that the proportions should be 9/16, 3/16, 3/16 and 1/16 for phenotypes 1, 2, 3, 4 respectively. The classification of a sample of 1400 observations is given below together with the proportions suggested by Mendel's laws of inheritance (the model) Give the degrees of freedom that would be used in a test of whether or not the crossing results in a population that follows Mendel's laws of inheritance. Degrees of freedom = The test is carried out at significance level α = 0.05 using the method described in the lecture notes. If the decision is to not reject the null hypothesis, what is the appropriate conclusion for the test? Select the appropriate conclusion from the list below. There is sufficient evidence at α = 0.05 that…
- According to the CCD, 1 in 25 adults have mutation A in their blood. Test TT for that mutation will give a positive result 95% of the time when the adult actually has the mutation and a negative result the other 5% of the time. If the adult does not have the mutation, then the TT test gives a positive result 1% of the time and a negative result the other 99% of the time. Suppose an adult is randomly selected and given the TT test. Need help with e. a.What is the probability that the selected adult has mutation A? b. What is the probability that the selected adult has mutation A and test TT gives a positive result? c. What is the probability that test TT gives a positive result for the selected adult? d. What is the probability that test TT gives a negative result for the selected adult? e. What is the probability that the selected adult has the mutation given that test TT gives a positive result? f.If 1000 random adults are selected, about how many would be expected to give a positive…2. Suppose AB Drug Company develops a new drug, designed to prevent colds. The company states that the drug is equally effective for men and women. To test this claim, they choose a simple random sample of 100 women and 160 men from a population of 100,000 volunteers. At the end of the study, 38% of the women caught a cold; and 51% of the men caught a cold. Based on these findings, can we reject the company's claim that the drug is equally effective for men and women? Use a = 0.05 level of significance.5. Two processes A and B, were used to produce stainless steel shafts in an industrial company. Process A and Process B produce 9% and 12% nonconforming stainless steel shafts, respectively. a. A random sample of 100 stainless steel shafts were taken from Process A. What is the UTM probability that more than 95% of the stainless steel shafts are conforming? b. A random sample of 90 stainless steel shafts from Process A and 80 stainless steel shafts UTM UT UTM 8U from Process B were taken randomly. Find the probability that the difference between proportion of the nonconforming stainless steel shafts from Process A and Process B UTM UTM is at most 0.1. O UTM UT UTM 6 UTM U O UTM UT