You discover two segregating genes (A and B) that affect resistance in a population of snails. You are interested in determining if the two loci are in LD. You analyzed 100 snails for allelic and haplotype states, and observed the frequencies reported in the table below. a. Calculate D for genes A and B. (D = gAB9ab - GAB9AB) b. Is this population in linkage equilibrium or linkage disequilibrium with respect to genes A and B? Frequency Allele A Allele a 0.6 0.4 Allele B 0.8 Allele b 0.2 0.48 Haplotype AB Haplotype Ab 0.12 0.32 Haplotype aB Haplotype ab 0.08

Last year you sampled a population of asters in the Rochester area and the allele frequencies for one locus (region of the genome) were p=0.7, q=0.3. Assuming that the population is in Hardy-Weinberg Equilibrium with respect to that locus, which of the following would you expect when sampling this year?   a. 42% of the asters are expected to be homozygous dominant.   b. 91% of the asters are expected to be heterozygous.   c. 9% of the asters are expected to be homozygous recessive.   d. 53% of the asters are expected to be heterozygous.

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…

Albinism is due to a recessive allele of an autosomal gene. Let a represent the albino allele and A represent the normal non-albino allele. If one out of every 10,000 people were albino and assuming that the population is in Hardy-Weinberg equilibrium, A. what would be the phenotype frequencies of albino and non-albino ? B. what would be the allele frequencies of a and of A? C. what fraction of people can be expected to be of the genotype Aa? D. what fraction of people can be expected to be of the genotype AA?

Snow geese (Chen caerulescens) come in two color types, white “snows” and “blues” with dark bodies. A single gene controls coloration, where the dark (“blue”) allele (D) is dominant. A population of 30,012 geese includes 9236 dark individuals. Genetic testing reveals that 7636 of the 9236 dark individuals are heterozygous (Dd). According to the Hardy-Weinberg equilibrium, what is the expected number of individuals of genotype Dd? Group of answer choices A.) 14752.40 B.) 13596.09 C.) 8898.56 D.) 3608.26