|
Data Run |
BB Baby Bug Count |
Bb Baby Bug Count |
bb Baby Bug Count |
|
1 |
5 |
3 |
2 |
|
2 |
1 |
6 |
3 |
|
3 |
2 |
5 |
3 |
|
4 |
3 |
6 |
1 |
|
5 |
2 |
7 |
1 |
|
6 |
3 |
5 |
2 |
|
7 |
3 |
4 |
3 |
|
8 |
4 |
4 |
2 |
|
9 |
5 |
4 |
1 |
|
10 |
2 |
6 |
2 |
Procedure II - Part A - Bug Population changes when there is a breeding preference for blue rimmed bugs
Data Table - Enter your Final Bug Counts
|
BB Bug Count |
Bb Bug Count |
bb Bug Count |
|
30 |
50 |
20 |
I am trying to work on the questions below; But I am wondering if my count numbers are incorrect 30 , 50 and 20?
The Question that I need help on are these:
[17] Do your results suggest anything about what the composition of this population might be at some distant point in the future? Defend your answer.
[18] Based on the initial starting population, use the Hardy-Weinberg equation to predict the future bug population phenotype composition. Hint: Under the Background tab, go to the Summary of Formulas Needed for Calculations section, see the example titled Using the Hardy-Weinberg Equation, then do Step 1 and Step 2 using the initial starting population for this data run
Here ,the different alleles are as follows,
BB = homozygous dominant
Bb = heterozygous
bb = homozygous recessive
now according to Hardy wienberg equation,
p^2 + 2pq +q^2 = 1
now ,this equation tells us that in genetic equillibrium ,the overall frequency remains constant i.e
BB + Bb + bb = 100%
as we can clearly see that count is 30 ,50 and 20. so as per assumption of equation,we will get offspring count or frequency same as parent.
in future generation ,at any time heterozygous will have greater frequency.
Data Run BB Baby Bug Count Bb Baby Bug Count bb Baby Bug Count 1 5...
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