Lab #1: Population Genetics by Kevin Gomez, Bishoy Galil, Tien Van, Hiu Nam (Tiger) Wong
Date: 1/9/2018
Bio 212 section AA
Winter 2018
Lab 1: post-lab assignment - Population Genetics
Lab #1: Population Genetics
by Kevin Gomez, Bishoy Galil, Tien Van, Hiu Nam Wong (Tiger)
POPGEN-AA
A. Distribution of Genes in a Population
It is hypothesised that the result genotypic ratio will follow the Mendelian's ratio. Hence, it is predicted that 1/2 of the individuals withdrawn will be heterozygous, while 1/4 will be homozygous white and remaining 1/4 will be homozygous speckled.
Null hypothesis: It is hypothesised that the result genotypic ratio will not follow the Mendelian's ratio.
I. Data Results
Frequency of Genotypes Counted in Gene Pool A
Genotypes
|
Frequency
|
Homozygous White
|
15
|
Homozygous Speckled
|
16
|
Heterozygous
|
19
|
Table 1: The table shows the frequency for each genotypes counted in Gene Pool A.
Frequency of Genotypes Counted in Gene Pool B
Genotype
|
Frequency
|
Homozygous White
|
16
|
Homozygous Speckled
|
15
|
Heterozygous
|
19
|
Table 2: The table shows the frequency for each genotypes counted in Gene Pool B.
Figure 1. Frequencies of homozygous white, homozygous speckled, and heterozygous individuals for population A and population B.
II. Conclusion and Chi Square Calculations
- Conclusion: The hypothesis says that we expect 1:2 of the product to be heterozygous, but we got an average percentage of 38% heterozygous as calculated based on the data in table 1 and table 2, and as shown in graph 1 and graph 2. According to the calculated Chi Square value, which is less than the critical value of 5.99 when p=0.05, we are not able to reject the null hypothesis. In other words, the observed genotypic ratio do not have significant difference comparing to the Mendelian's ratio. However, the data did not fully support our alternative hypothesis as well. According to Table 1 and Table 2, the percentages for homozygous white are 30% and 32% in gene pool A and B respectively, and homozygous speckled are 32% and 30%, while heterozygous accounted for 38% in both gene pools. Since the Mendelian's ratio suggests at least 50% will be heterozygous and only 38% was heterozygous, and only 25% should be homozygous dominant or recessive, this data suggests that the experimental results did not exactly followed the Mendelian's ratio. In addition, we can say that our hypothesis is not fully supported because our hypothesis is only supported by the ratio. The average percentage of white homozygous and speckled homozygous were both equal to 31%, which were expected to be equal. The difference was in the percentages ratio as the percentage of the heterozygous is supposed to be double of the homozygous ratio. The differences in the percentage is due to human errors, like miscounting and how we selected the beans.
- Chi Square Calculations:
Observed (O)
|
Expected (E)
|
O - E
|
(O-E)2
|
(O-E)2E
| |
Homozygous White
|
15
|
12.5
|
2.50
|
6.25
|
0.50
|
Homozygous Speckled
|
16
|
12.5
|
3.50
|
12.25
|
0.98
|
Heterozygous
|
19
|
25
|
-6.00
|
36.00
|
1.44
|
Total
|
50
|
50
|
2.92
|
Degrees of Freedom= 3-1 = 2
The critical value for 2 degrees of freedom at p=0.05 is equal 5.99 ( 2.92 < 5.99 )
As the chi square value is smaller than the critical value, therefore we can’t reject our null hypothesis. The observed genotypic ratio do not have a significant deviation from the expected Mendelian's genotypic ratio.
B. Genetic Drift or Natural Selection
Null hypothesis: It is hypothesised that there will be no selection thus the phenotypic ratio remains the same.
Table 6: Table above shows the Chi square value of each phenotype in Population A.
Table 7: Table above shows the Chi square value of each phenotype in Population B.
I. Data Results
Population A Generations
Alleles
|
1
|
2
|
3
|
4
|
5
|
6
|
7
|
8
|
9
|
10
|
Red
|
16
|
18
|
17
|
20
|
24
|
29
|
35
|
42
|
48
|
49
|
Speckled
|
12
|
15
|
13
|
9
|
12
|
9
|
5
|
2
|
0
|
0
|
Black
|
17
|
17
|
21
|
21
|
14
|
12
|
10
|
6
|
2
|
1
|
White
|
5
|
0
|
0
|
0
|
0
|
0
|
0
|
0
|
0
|
0
|
Table 4: The table shows the frequencies of each phenotype from generation 1 to generation 10 in Population A.
Population B Generations
Alleles
|
1
|
2
|
3
|
4
|
5
|
6
|
7
|
8
|
9
|
10
|
Red
|
9
|
10
|
10
|
10
|
15
|
21
|
22
|
20
|
21
|
22
|
Speckled
|
11
|
15
|
16
|
24
|
20
|
14
|
11
|
13
|
13
|
12
|
Black
|
8
|
11
|
17
|
13
|
15
|
15
|
17
|
17
|
16
|
16
|
White
|
22
|
14
|
7
|
3
|
0
|
0
|
0
|
0
|
0
|
0
|
Table 5: The table shows the frequencies of each phenotype from generation 1 to generation 10 in Population B.
Figure 2. Frequencies of red, speckled, black, and white beans counted over 10 generations in gene pool A.
Figure 3. Frequencies of red, speckled, black, and white beans counted over 10 generations in gene pool B.
Figure 4. Red beads, Black beads, white beans, and speckled bean used for the experiment.
II. Conclusion and Chi Square Calculations
- Conclusion: Based on our data and the graphs ( shown by /table 3, table 4, graph 1, and graph 2), we can reject our null hypothesis. The results are supporting our hypothesis that the change in phenotypic ratio from generation 1 to generation 10. This change is due to natural selection. For our experiment, refer to Figure 4., we used red and black beads, and white and speckled beans, which beads are larger in size when compared to beans. The reason for the natural selection was the different in sizes of beans, as the red and black beads were larger than other speckled and white beans. The larger sized beads had better opportunity, advantage, to survive than the smaller beans ( in this case, the larger the beans are the easier for them to be selected ). According to Figure 2, population A shows that the ability for red and black beads to survive is higher than small beans. The chi squares for both population shows that our value is higher than the critical value for degrees of freedom of 3. That means that there is a significant difference between the observed and the expected values, so we can reject our null hypothesis. So, we can say that the results we got are not due to chance, instead are due to natural selection.
- Chi Square Calculation:
Observed (O)
|
Expected (E)
|
O - E
|
(O-E)^2
|
((O-E)^2)/E
| |
Red
|
49
|
16
|
33
|
1089
|
68.06
|
Speckled
|
0
|
12
|
-12
|
144
|
12.00
|
Black
|
1
|
17
|
-16
|
256
|
15.06
|
White
|
0
|
5
|
-5
|
25
|
5.00
|
Total
|
50
|
50
|
100.12
|
Degrees of Freedom= 4-1 = 3
The critical value for 3 degrees of freedom at p=0.05 is equal 7.82 ( 100.12 > 7.82 )
As the chi square value is larger than the critical value, therefore we can reject our null hypothesis, the observed values show a significant difference from the Mendelian's ratio.
Population B
Observed (O)
|
Expected (E)
|
O - E
|
(O-E)^2
|
((O-E)^2)/E
| |
Red
|
22
|
9
|
13
|
169
|
18.78
|
Speckled
|
12
|
11
|
1
|
1
|
0.09
|
Black
|
16
|
8
|
8
|
64
|
8.00
|
White
|
0
|
22
|
-22
|
484
|
22.00
|
Total
|
50
|
50
|
48.87
|
Degrees of Freedom= 4-1 = 3
The critical value for 3 degrees of freedom at p=0.05 is equal 7.82 ( 48.87 > 7.82 )
As the chi square value is larger than the critical value, therefore we can reject our null hypothesis, the observed values show a significant difference from the Mendelian's ratio.
What startles me is how in both of your populations for the beads, that the white alleles went extinct pretty fast, especially population A. During the bean alleles, the reason why the white beads weren't picked as much while in the cup was because they were too small for our fingers, so the data turned out to be mostly bias towards homozygous speckled and homozygous white (in the end because that was all that was left). But it's interesting how your groups gene pool was high in numbers for heterozygous and our data was low for heterozygous, which shows that different ways of choosing the beans has a massive effect on the genotypes which I think shows a more directional mode of selection on a phenotype.
ReplyDeleteI found it interesting that you used half beads and half beans for experiment B. I didn't even think about that when completing my experiment. I had similar results (My group used 4 different types of beans) when it came to white beans. It seems that natural selection took part because of the size of the beans/beads, if 4 beads were used I believe that would be a case of genetic drift. It looks to me that white beans were the smallest in size out of the other 3 alleles causing it to be lost in both experiments. Its interesting to see that your black beads end result was lower than your red beads in both populations. I wonder if their is any significant differences in the beads that would lead to this result, or if its just a coincidence. Nice post.
ReplyDeleteHey everyone,
ReplyDeleteVery interesting data, I am wondering how you guys chose to select your beans/beads? It's also an interesting choice to use both beans and beads, seeing that I thought maybe none of your alleles would go extinct, but in part 2, you still had population A have white die out quickly and black almost died out as well, and then population A only had white die out, but black beads still had a significant amount of alleles. It was really nice seeing different data to think more of how the sizes or shapes and methods of pulling alleles could have really pushed one allele to strive more than another. Keep up the good work!
Hello guys,
ReplyDeleteFirst of all, I am really impressed by the organization of your post. It's so great that your post actually looks like a real procedure from the manual or the professor! I am glad to see that your group came to the same conclusion as my group concerning Part B: some kind of selection happened. Indeed, we used exclusively beans and we didn't know if doing the same experiment with half beads and half beans would have changed our conclusion. If we had to redo this expriment, you guys should try using only beads of the same size and reducing the ''human selection'' by using a spoon or tweezers. In this way, you could actually measure the real impact of randomness and not selection.
Great post! I found it interesting that the numbers were exactly the same for gene pool A and B for the first half of your experiment. Moving onto the second half I really liked how you explained the chi squared and what degrees of freedom was. Also I found it shocking that in your population A the red beads ended up having double the amount in generation 10 than it had in generation 10 of population B. Thank you for great work.
ReplyDeleteI noticed in your graphs of populations A and B over 10 generations that the red, black, and white beans either increased or decreased similarly in both graphs, but the speckled beans steadily decreased in population A and spiked during the fourth generation in population B. Did you notice anything during the experiment that might explain why the red, black, and white bean alleles were surprisingly consistent between the two populations but the speckled beans behaved differently in each population?
ReplyDelete