Question

In the Rocky Mountains, a particular mouse species has three fur colors (brown, gray, and white) that are controlled by two alleles (B and b) that display incomplete dominance. Scientists studying these mice conducted a survey to determine what proportion of the population displays each fur color. During the first study year, the area that was sample was largely composed of rocky outcroppings covered in a gray-brown lichen. One year later the scientists resampled this population. In this second survey, the scientists found that the lichens had died, leaving behind light gray remains. Is the initial mouse population (i.e. Generation 1) in Hardy-Weinberg equilibrium? If the population is not in equilibrium, explain how you know this by making reference to your mathematical calculations. If it is in equilibrium, what does this mean? Show ALL work! Based on the actual versus expected genotype frequencies for Generation 1 (calculated in #1), are any of the phenotypes better adapted for the initial environment? Are any maladaptive? Use the background information provided to hypothesize why these genotypes(s) have a higher or lower fitness. Did evolution (i.e. change in allele frequency) occur between the first and second generation of this population? If so, use the background information to determine which genotypes are more fit and which are maladaptive in Generation 2. Show ALL work!

Answer 1

According to Hardy-Weinberg equilibrium,

p² + 2pq + q² = 1

here p is frequency of allele "B" & q is frequency of allele "b".

Allele frequency = No. of copies of an allele in a population / Total no. of all alleles for that gene in a population

Genotype frequency = No. of individuals with a particular genotype in a population / Total no. of individuals in a population

ANS 1. A population will be in Hardy-Weinberg equilibrium if it fits the equation p² + 2pq + q² = 1

Here p² = genotype frequency of BB

q² = genotype frequency of bb

2pq = genotype frequency of Bb

CALCULATIONS :

Allele frequency of "p" = 90 + 2(55) / 2(55) + 2(90) + 2(10) = 0.65

Allele frequency of "q" = 90+ 2(10) / 2(55) + 2(90) + 2(10) = 0.35

EXPECTED GENOTYPE FREUENCIES -

"BB" = p² = (0.65)² = 0.4225 ; expected number of BB rats = 0.4225*155 = 65.48 or 65 approx.

actual number of BB rats = 55

"Bb" = 2pq = 2*0.65*0.35 = 0.455 ; expected number of Bb rats = 0.455*155 = 70.525 or 71 approx.

  actual number of Bb rats = 90

"bb" = q² = (0.35)² = 0.1225 ; expected number of bb rats = 0.1225*155 = 18.98 or 19 approx.

actual number of bb rats = 10

ACTUAL GENOTYPE FREQUENCIES -

p² = Individuals with BB genotype / Total population i.e. 55/155 = 0.354 or 35.4%

q² = Individuals with bb genotype / Total population i.e. 10/155 = 0.064 or 6.4%

2pq = Individuals with Bb genotype / Total population i.e. 90/155 = 0.580 or 58%

p² + 2pq + q² = 0.354 + 0.580 + 0.064 = 0.998 (approx. 1)

Yes, the population is in Hardy-Weinberg equilibrium. This means under a given set of assumptions, the allele and genotype frequencies will not change over the course of many generations. The dominant traits do not necessarily increase from one generation to another.

ANS 2. From the above results, population of gray (Bb) rats is better adapted and population of white (bb) rats is maladaptive. This could be possibly due to the natural selection of gray and brown rats because of gray-brown lichen which protected them from being hunted. White rats could be easily spotted against gray-brown lichen and hence, their population declined than expected.

ANS 3. Evolution is a slow and gradual process. One year is very short time span. So, the answer will be NO. Change in first and second generation individuals are due to natural selection. This is beacuse in second year, lichens were wiped off. So now, white rats were selected naturally against brown and gray rats.

The more fit genotypes in 2nd generation are Bb and bb and maladaptive genotype is BB.

In generation 2,

allele frequency of p = 2(15) + 60/ 2(15) + 2(60) + 2(45) = 0.375

allele frequency of q = 2(45) + 60/ 2(15) + 2(60) + 2(45) = 0.625

Expected genotype frequency of BB = p² = 0.140

Actual Genotype frequency of BB = 15/120 = 0.125

Expected genotype frequency of bb = q² = 0.390

Actual genotype frequency of bb = 45/120 = 0.375

Expected genotype frequency of Bb = 2pq = 2*0.375*0.625 = 0.468

Actual genotype frequency of Bb = 60/120 = 0.5