Question

1) The results you predict as a result of a controlled experiment can be described as an hypothesis, such as “selection of Wisconsin Fast Plants with the most trichomes in the first (parent) generation will result in an increase in trichome number in the plants of the second generation.” You are making a prediction based on scientific knowledge of selection, and are able to quantify the number of trichomes. This is your experimental hypothesis. A null hypothesis for your experiment would predict that there will be no difference between the groups as a result of the treatment. Your experimental goal would be to gather data to reject the null hypothesis. The data presented in Part D shows the results of artificial selection for hairy Wisconsin Fast Plants. Identify the null hypothesis for this investigation. The data presented in Part D shows the results of artificial selection for hairy Wisconsin Fast Plants. Identify the null hypothesis for this investigation.

a) There will be no difference between the mean number of trichomes in the second generation compared to the parent population.

b) If the mean number of trichomes is greater in the second generation than in the parent population, then selection has occurred. c) As a result of selection, the mean number of trichomes will be greater in the second generation.

d) If plants with the most trichomes in the first generation are selected as parents, then the second generation will have more trichomes.

2) In the preceding example, we calculated the probability of obtaining certain genotypes in the offspring based on allelic frequencies, but we can also use this method to determine the genetic makeup of a population. The Hardy-Weinberg equation is p 2 + 2pq + q 2 = 1. What do these variables represent, and how can this equation be used to describe an evolving population? In an earlier part of this investigation, we worked with a pair of alleles that were incompletely dominant to each other. Now let’s generalize and use terminology that can be applied to any genetic trait. You may want to print out the following instructions to use as a reference when working on Hardy-Weinberg problems. For a gene locus that exists in two allelic forms in a population, A and a: Let p = the frequency of A, the dominant allele Let q = the frequency of a, the recessive allele All the dominant alleles plus all the recessive alleles will equal 100% of the alleles for this gene, or, expressed mathematically, p + q = 1 for a population in genetic equilibrium. If this simple binomial is expanded we get the Hardy-Weinberg equation: p 2 + 2pq + q 2 = 1 The three terms of this binomial expansion indicate the frequencies of the three genotypes: p 2 = frequency of AA (homozygous dominant) 2pq = frequency of Aa (heterozygous) q 2 = frequency of aa (homozygous recessive) If we know the frequency of one of the alleles, we can calculate the frequency of the other allele: p + q = 1, so p = 1 – q q = 1 – p Let’s use this equation to solve the following problem: In pea plants, the allele for tall plants (T) is dominant to the allele for dwarf plants (t). If a population of 100 plants has 36 dwarf plants, what is the frequency of each allele? Here is a step-by-step guide: Let p = frequency of the dominant allele (R), and q = frequency of the recessive allele (r). q 2 = frequency of the homozygous recessive = 36%, or 0.36. Since q 2 = 0.36, what is q? Take the square root of 0.36, or q = 0.6. Now, p + q = 1, so subtract q from 1 to find the value of p, or 1 – 0.6 = 0.4; therefore, p = 0.4. That’s it! But let’s go a step further--how many of these plants are heterozygous tall (Tt) if the population is in Hardy-Weinberg equilibrium? Calculate 2pq = 2 × 0.4 × 0.6 = 0.48, or 48%. Since there are 100 plants, 48 are heterozygous tall. Suppose that green seeds (G) are dominant to yellow seeds (g) in peas. In a population of 500 individuals, 25% show the recessive phenotype. How many individuals would you expect to be homozygous dominant for this trait if the population is in Hardy-Weinberg equilibrium?

You need to list equations used and provide steps of problem solving. Providing an answer itself is not enough for full grade.

3. You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 36%. Calculate the frequency of the heterozygous genotype, homozygous dominant genotype and homozygous recessive genotype.

4. You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 49%. What is the frequency of the recessive allele, and the frequency of the dominant allele? What is the frequency of heterozygous genotypes?

Answer 1

Answer-

According to the given question-

Question 1-

Here we have a pea plant having different length of stem , tall pea plant is dominant over the dwarf plants.

The allele for tall plants = T

The allele for dwarf plants = t.

Total population = 100

Number of dwarf plants= 36

According to the Hardy Weinberg equilibrium-

p + q = 1 and p² + 2pq + q² = 1

where

p = frequency of dominant allele in the population

q = frequency of recessive allele in the population

p² = Frequency of individuals having dominant genotype in the population

q² =Frequency of individuals having recessive genotype in the population

2pq = Frequency of heterozygous individuals in the population

Frequency of individuals having recessive genotype = q² = 36 100 = 0.36

q = 0.36 = 0.6

frequency of the recessive allele = q = 0.6

p + 0.6 = 1 , p = 1 - 0.6 = 0.4

frequency of the dominant allele = p = 0.4

Frequency of individuals having dominant genotype =  p² = (0.4)² = 0.16

Frequency of heterozygous individuals in the population = 2pq = 2 0.4 0.6 = 0.48

The number of homozygous dominant = p² number of population = 0.16 100 = 16

The number of homozygous recessive = q² number of population = 0.36 100 = 36

The number of heterozygous = 2pq number of population = 0.48 100 = 48

Question 2-

Here the color of green seed is dominant over the yellow seed color.

The allele for Green color seed = G

The allele for yellow seeds = g

total population = 500

plants with recessive phenotype = 25% which means that 500 25 100 = 125 are having recessive phenotype

Frequency of individuals having recessive genotype = q² = 125 500 = 0.25

q = 0.25 = 0.5

frequency of the recessive allele = q = 0.5

p + 0.5 = 1 , p = 1 - 0.5 = 0.5

frequency of the dominant allele = p = 0.5

Frequency of individuals having dominant genotype =  p² = (0.5)² = 0.25

Frequency of heterozygous individuals in the population = 2pq = 2 0.5 0.5 = 0.50  

The number of homozygous dominant = p² number of population = 0.25 500 = 125

The number of homozygous recessive = q² number of population = 0.25 500 = 125

The number of heterozygous = 2pq number of population = 0.50 500 = 250

Thus the number of offspring which are homozygous dominant i.e. having green color seed = 125

Question 3.

Here we have a population , where the percentage of offspring which are having homozygous recessive genotype i.e. aa = 36%. which means that 36 out of 100 are having homozygous recessive genotype

Frequency of individuals having recessive genotype = q² = 36 100 = 0.36

q = 0.36 = 0.6

frequency of the recessive allele = q = 0.6

p + 0.6 = 1 , p = 1 - 0.6 = 0.4

frequency of the dominant allele = p = 0.4

Frequency of individuals having dominant genotype =  p² = (0.4)² = 0.16

Frequency of heterozygous individuals in the population = 2pq = 2 0.4 0.6 = 0.48

The number of homozygous dominant = p² number of population = 0.16 100 = 16

The number of homozygous recessive = q² number of population = 0.36 100 = 36

The number of heterozygous = 2pq number of population = 0.48 100 = 48

frequency of heterozygous genotype = 2pq = 0.48,

frequency of homozygous dominant genotype = p² = 0.16

frequency of homozygous recessive genotype = q²  =  0.36

Question 4.

Here we have a population in which the ,

percentage of individuals having homozygous recessive genotype = aa = 49%.   which means that 49 out of 100 are having homozygous recessive genotype

Frequency of individuals having recessive genotype = q² = 49 100 = 0.49

q = 0.49 = 0.7

frequency of the recessive allele = q = 0.7

p + 0.7 = 1 , p = 1 - 0.7 = 0.3

frequency of the dominant allele = p = 0.3

Frequency of individuals having dominant genotype =  p² = (0.3)² = 0.09

Frequency of heterozygous individuals in the population = 2pq = 2 0.3 0.7 = 0.42  

The number of homozygous dominant = p² number of population = 0.09 100 = 9

The number of homozygous recessive = q² number of population = 0.49 100 = 49

The number of heterozygous = 2pq number of population = 0.42 100 = 42

frequency of homozygous recessive genotype = aa =  q²  =  0.49

frequency of homozygous dominant genotype = AA = p² = 0.09

frequency of heterozygous genotype = Aa = 2pq = 0.42,