Researchers studied the blood types of indigenous populations in the Andes. They used a two-allele typing system (M and N alleles). In one population the genotypic frequencies were: f (MM) = 0.70, f (MN) = 0.20, f (NN) = 0.10. What are the frequencies of M and N alleles for this population? Answer choices f(M) = 0.80 and f(N) = 0.20 f(M) = 0.70 and f(N) = 0.30 f(M) = 0.49 and f(N) = 0.01 f(M) = 0.90 and f(N) = 0.10 f(M) = 0.50 and f(N) = 0.50

You are studying a population of mice in a field that have black (genotype BB), dark brown, (Bb) or light brown (bb) fur.   In your population you have 49 black individuals, 274 dark brown and 206 light brown.   What is the frequency of the B allele in the population? (round to 2 decimal places)

A total of 6500 North American Caucasians were blood typed for the MN locus, which is determined by two codominant alleles LM and LN. The following data were obtained: Blood Type M MN N a. 0.52, b. 0.23 a. 0.52, b. 0.12 0.48, 0.12 Number a. What is the allele frequency of LM in the population (LM=p) b. What is the genotypic frequency of LNLN? a. 0.48, b. 0.23 1800 3200 1500

If a population has an allele A1 frequency of 0.59 and an allele A2 frequency of 0.41, what is the expected frequency of the A2A2 genotype?  Answers have 2 decimal places.

The MN blood group is a single-gene, two allele system in which each allele is codominant. Why are such codominant alleles ideal for studies of allele frequencies in a population.

A population of dragons or Varanus dragonus has two alleles B and P encoding for phenotypes blue and purple scales, is dominant over B. You have population of 345 individuals. PP = 100, BP; =90 , and BB = 155 . What is the frequency of the P allele? Please use and round to four decimal points.

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

How do I find the frequency of alleles in this population? Please explain how to solve this problem.

The UNF penguin lab decided to study the elusive polka dot penguin of Patagonia. They trapped a majority of the population using gummy worms as bait and collected the following information. Observed Genotype Frequencies (not in HWE) AA: 0.464 Aa: 0.427 aa: 0.109 If this population was in Hardy-Weinberg Equilibrium, what would be the expected genotype frequency for the heterozygotes? Retain 3 decimal places.