Question

Can someone please help me with the following questions. Please provide an answer/explanation especially for the first three.

?. Suppose a population of flour beetles has 1,000 individuals. Normally the beetles are red; however, this population is polymorphic for a mutant autosomal body color, black, designated by bb. Red is dominant to black, so BB and Bb genotypes are red. Assume the population is at Hardy–Weinberg equilibrium, with equal frequencies of the two alleles. What would be the allele frequencies if a population bottleneck occurred and only four individuals survived: one female red heterozygote and three black males?           A) 0.125 red; 0.875 black B)0.875 red; 0.125 black C) 0.25 red; 0.75 black D) 0.75 red; 0.25 black    E) 0.5 red; 0.5 black

?. What would be the expected frequencies of the red and black phenotypes in the original population of beetles? A) 0.5 red; 0.5 black    B) 0.75 red; 0.25 black C) 0.25 red; 0.75 black D) All red E) Insufficient information is given to answer the question.

?. Given the original population data and a Hardy-Weinberg equilibrium, what are the frequencies of the homozygous dominant, heterozygous, and homozygous recessive (in that order) after 100 generations, assuming that no selection or other evolutionary processes are operating?      A) 0.25; 0.5; 0.25      B) 0.75; 0.20; 0.05     C) 0; 0.5; 0.5      D.) 0; 0; 1   E) None of the above

?. The chart below shows the performance of different genotypes at the A locus for three different populations of lizards in two different environments. The first number is their running speed (in m/s) and the second is their relative strength. In which population is there evidence of a trade-off between running speed and relative strength? A) Population 1 B) Population 2 C) Population 3 D) Both a and b E) None of the above

Population

1 (AA) 6.2/10 (Aa)7.6/12 (aa)8.2/15

    

2 (AA)5.1/18 (Aa)6.2/11 (aa)7.4/8

  

3 (AA)6.9/14 (Aa)6.5/11 (aa)4.7/9

?. The half-life of carbon-14 is 5,700 years. If 10 percent of the original carbon-14 is left in a structure, that structure is _______ years old.    A )   less than 2,850   B) about 5,700   C) between 11,400 and 17,100   D) between 17,100 and 22,800   E) about 57,000

?. Which is the best way to estimate the length of time that two species have been separated evolutionarily? A) compare pieces of DNA coding for a structural protein that is present in both species B) studying the many nucleotide sites in both species and taking an average differences between DNA and RNA C) comparing pieces of DNA for a functional enzyme that is present in both species D) comparing the DNA differences between two species in non-coding regions E) Comparing fitness levels

?. Incomplete dominance could be considered as supporting evidence for A) Fleming Jenkins blending hypothesis argument B) Alfred Wallace’s theory of Biogeography C) Darwin’s theory of Natural Selection D) Cuvier’s catastrophe hypothesis E) None of the above.

?. Penicllium is a mold that grows on bread. It produces penicillin, a toxin which destroys or inhibits bacteria and other fungi from growing on the bread. This is best labelled A) commensalism B) amensalism C) neutralism D) obligate mutualism C) contest competition

Answer 1

1. After bottle neck, four individuals remain. Out of which, three are black, means people with homozygous recessive phenotype (bb).

The frequency of people with homozygous recessive phenotype (bb or q2) = 3/4 = 0.75

q = 0.87 or approximately 0.875

Now, the frequency of p allele = 1- 0.875 = 0.125

Thus, the correct option is A) 0.125 red; 0.875 black.

2. In the original population, the frequency of both the alleles is equal. Means, p = 0.5 and q = 0.5

Now, the frequency of homozygous dominant individuals, p2 = 0.25

The frequency of heterozygotes = 2pq = 0.5

the frequency of homozygous recessive individuals, q2 = 0.25

Given that, Red is dominant to black, so BB and Bb genotypes are red.

The frequency of individulas with red phenotype = the frequency of homozygous dominant individuals + the frequency of heterozygous individuals = 0.25 +0.5 = 0.75

The frequency of individulas with black phenotype = the frequency of homozygous recessive individuals= 0.25

Thus, the correct option is B) 0.75 red; 0.25 black