A population consists of 300 individuals with the following genotypes: AA – 100 Aa – 125 aa – 75 a. What are the values of p and q? b. Is the population in Hardy-Weinberg equilibrium? Explain.

In a population with two alleles at the R locus (R and r), the frequency of the genotype rr is 0.24. Assuming that the R locus is at Hardy-Weinberg equilibrium in this population, what is the frequency of heterozygotes (Rr)? Round and report your answer to the second decimal place (0.00). Your Answer:

The two-tailed p-value associated with this test is 0.860. Should the null hypothesis be rejected? Is there sufficient evidence to suggest that there is a statistically significant difference between the two means in question? If we now wanted to know whether there was a difference in weight between those with a low NYHA class (1 and 2) and those with a high NYHA class (3 and 4), what p-value would you report?

please tell us the expected average time to fixation of an allele at frequency p = 0.5 in a population of 100 individuals, we can safely assume that the allele does become fixed?

Cerebellar abiotrophy (CA) is a recessive, genetic neurological disease found in the Arabian horse breed. If the incidence of this disease is about 1 in 400 Arabian horses, estimate the frequency of carriers in a population in Hardy-Weinberg equilibrium. Value of q²: 0.0025 Value of q: 0.05 Value of p: 0.95 Carrier Frequency: 0.095 How many horses would you expect to be carriers for cerebellar abiotrophy if the population consists of 582 Arabian horses? (Round to the nearest whole number and enter 2 digits.)

What does it mean when your data/population conforms to Hardy-Weinberg Equilibrium?

Pretend that you are comparing the actual genotype distribution for a population with the distribution of genotypes predicted by the Hardy-Weinberg theorem.  So your hypothesis is that the population is in Hardy-Weinberg equilibrium (i.e. that actual population data fit the Hardy-Weinberg expectations).  If you carry out a chisquare goodness of fit test and calculate a total chisquare value of 0.03 with 1 degree of freedom (see table), what does this mean? (select all true statements)a) The data do NOT fit the hypothesized distribution.b) The data do fit the hypothesized distribution well enough, so we accept the hypothesis at this time (i.e. we cannot reject the hypothesis). c) The probability that the data came from a population in Hardy-Weinberg equilibrium is too small, so we reject the hypothesis.d) The probability that the data came from a population in Hardy-Weinberg equilibrium is too big, so we reject the hypothesis.e) The data support Hardy-Weinberg expectations – there is no…

Consider an autosomal locus with alleles A and a. If the the allele frequencies are as follows Freq(A) = 0.4, Freq(a) = 0.6. , then what is the predicted frequency of heterozygous Aa individuals, assuming the population is in Hardy-Weinberg equilibrium? Freq(Aa) = ? Enter a number between 0 and 1, inclusive, for example 0.33

In a population with two alleles at the R locus (R and r), the frequency of the genotype rr is 0.10. Assuming that the R locus is at Hardy-Weinberg equilibrium in this population, what is the frequency of heterozygotes (Rr)? Round and report your answer to the second decimal place (0.00).

For Wright- Fisher population with 500 of population size, it was found that frequency of allele b was 0.7 (B=0.3) in current population. Calculate the probability that B alleles will fix in the population? Show all workings