Please assist me with answering this Lab exercise question: Please demonstrate t

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Please assist me with answering this Lab exercise question: Please demonstrate the Hardy-Weinberg rule as shown in the images below. The first part posted are instructions of the experiment and the rest are the questions provided in response to the findings of the experiment to be accomplished on the tables as shown in the images.

(if possible, the beans should be of about the same size and shape Demonstration of Hardy-Weinberg Rule Collect 70 brown beans and 30 white beans. Let the brown beans represent the dominant allele for blue wings in a butterfly. Let the white beans represent the recessive allele for white wings. Mix the beans in a coffee cup and draw two beans at random from the cup. Note the color of the two beans and record your results on page one of this lab's report sheets (use B for brown bean and W for white bean). Put the beans back in the cup after each drawing and mix them before drawing another sample (you are trying to make the selection random). Repeat the process until you have recorded data from a total of 50 drawings. PART I Next, I'd like you to translate your "bean count" data into a real world predic- tion. Noting that a heterozygote butterfly has light blue wings (incomplete domi- nance), record the number of dark blue (BB), light blue (BW), and white (WW) butterflies that you expect to result from random mating. In other words, how many of the 50 offspring will have, for example, dark blue wings? The number of dark blue butterflies equals the number of samples (drawings of two beans) with two brown beans, right? How many butterflies will have light blue wings accord- ing to your model? How many white butterflies will you get? In this example, there are only two alleles that determine wing color. Let's let the letter p stand for the fraction of beans that are brown (p will also correspond to the frequency of the blue wing allele). Let's let q stand for the fraction of beans that are white (q is also the frequency of the white wing allele). We know that p plus q must equal one, since the sum of p plus q is equal to all possible alleles. p+q= Here's a formula to predict the frequency of combinations of two alleles: (p)(p) + 2pq + (q)(q) = 1 91

Explanation / Answer

Part I - Hardy- Weinberg principle

We have 70 brown beans and 30 white beans, so totally 100 beans.

We mix the beans and we make pairs of 50. We have to remember to keep two letters in a box.

Now, we made the drawing.

Questions:

How many drawings gave BB combinations? 30

How many drawings gave the WB result? 10

How many were WW? 10

In the 50 data boxes, 70 brown beans are recorded.

In the 50 data boxes, 30 white beans are recorded.

From this exercise one can understand the Hardy-weinberg principle.

At the generation 0, we have 50 diploid butterflies and the genotypic frequency of the wing colour could be represented by two beans, in various combinations, as follows:

Genotype

White beans

Brown beans

Homozygous dominant (BB)

Heterozygous (WB)

Homozygous recessive (WW)

Part II. Selection against the dominant phenotype

Now, we are going to do selection against the dominant phenotype (blue wing), hence we did not select the dark blue, BB (Homozygous) drawings but for the light blue, the BW/WB (Heterozygous) combinations and for the white wings (WW).

Table 1.

Questions:

How many brown beans are recorded in the boxes above? = 39 (37 from BW/WB and 2 from BB)

What is the frequency of the blue wing allele (brown bean) in the next generation of butterflies?

As said in the text, let us assume, “p” which stand for the fraction of that are brown i,e the frequency of blue wing allele and “q” which stand for the fraction of that are brown i,e the frequency of white wing allele. The calculation of allele frequency facilitates to identify the richness of particular allele among the individuals in the given population. A change in the allele frequency of a population is one of the reasons for evolution.

However, the frequency of allele B among 50 individuals can be calculated as follows:

There are 1 BB individual which make 2 B gametes

There are 37 WB/BW individual which would make 37 B and 37 W gametes

There are 12 WW individual which makes 24 W gametes.

So, the frequency of B among the 100 gametes is

B = 2+37/100 =0.39

How many white beans are recorded in the boxes above? = 61 (24 from WW and 37 from BW/WB)

What is the frequency of the white wing allele (white bean) in this next generation of butterflies? =

So, the frequency of W among the 100 gametes is

B = 24+37/100 =0.61

Inference: When compare to the starting population, the frequency of B allele decreases in the second generation and the selection against the dominant phenotype favours light blue and white wings.

Table 2.

How many brown beans are recorded in the boxes above? = 17 (from BW)

What is the frequency of the blue wing allele (brown bean) in the next generation of butterflies?

The frequency of allele B among 50 individuals can be calculated as follows:

There are 17 BW individual which would make 17 B and 17 W gametes

There are 33 WW individual which makes 66 W gametes.

So, the frequency of B among the 100 gametes is

B = 17/100 =0.17

How many white beans are recorded in the boxes above? = 83 (66beans from WW combinations and 17 beans from BW combinations)

What is the frequency of the white wing allele (white bean) in this next generation of butterflies?

So, the frequency of W among the 100 gametes is

B = 66+17/100 =0.83

Inference: When compare to the second generation, the frequency of B allele still decreases in the third generation and the selection against the dominant phenotype favours more light blue and white wings.

Which has the greater impact on allele frequencies, genetic drift or natural selection? Briefly, explain your answer and include some data as evidence.

Both, natural selection and genetic drift are important for genetic evolution.

Genetic drift occurs due to random sampling and it causes allele frequency changes by chance

On the other hand, natural selection is not a random process. It facilitates the selection of the trait contributing reproductive success of an individual in a given population over generations.

In the above example, we did selection against dominant phenotype. Eventually, the dominant phenotype (BB) did not get a chance/less chance to reproduce the offspring. Eventually, this type of selection facilitates the heterozygote (WB/BW) for light blue to decrease and the homozygote for white winged (WW) to increase in second and third generation.

Moreover, as said in the text, the small differences in the generated data and predicted data are due to genetic drift. Particularly, for the small population like here in this example genetic drift places also plays a major role in changes in allele frequency.

At genetic equilibrium, the allele frequencies follow Hardy-Weinberg law as follows:

P2(BB)+2pq(BW/WB)+q2 (WW)= 1

While using the equilibrium the allele frequency were changing the starting population,

The Hardy – Weinberg Equilibrium was,

(.7) (.7) + 2(.7)(.3)+(.3)(.3), and we expect to get 49% p (Dark blue) 42% heterozygotes BW/WB (Light blue) and 9% homozygous recessive (white)

But we got 60% (dark blue), 20% light blue and 20% white.

In the second generation,

It was (.39) (.39) + 2(.39)(0.61)+(0.61)(0.61)=1

i.e, 0.15 (p) +0.48 (pq) +0.37 (q)

But we got lesser numbers the calculated values. It is due to genetic drift.

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