BIOL3190 Lecture Notes - Lecture 15: Dihybrid Cross, Null Hypothesis, Centimorgan
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Problem A: Shipwreck Founder Effect Genetics
In the year 1782, a small passenger schooner (the HMS Leaky) was blown off course, crashing into the rocks of a then-uncharted mid-Atlantic island. The 30 survivors built shelter and found plentiful food on the island. Within a year, it was clear that there was little chance of rescue and no means of leaving. Soon, the warm tropical nights led couples to succumb to the romantic tendencies that moonlight walks on the beach can induce. Before long, babies were born, and the ĂąÂÂnewù civilization of the island now known as Gilligania was off to the races.
A genetic analysis of the remains of those original islanders showed the following ABO blood type phenotypes:
Blood Type | Number of People |
A* | 9 |
B** | 7 |
AB | 4 |
O | 10 |
*Four heterozygotes (iA iO) and five homozygotes for iA
** Five heterozygotes (iB iO) and two homozygotes for iB
Keep in mind that the blood type alleles iA and iB are codominant over each other, and that the iO allele is recessive to both iA and iB. If you need to do so, review the genetics of this system in CampbellĂąÂÂs Biology. Based on this information, please answer the following questions (show your work):
1. What were the frequencies of the three alleles in the original survivor population on the island?
2. Determine the genetic structure (allele and genotype/phenotype frequencies) for this gene system in the island population in 1932, when the island was ĂąÂÂrediscoveredù by cartographers. The population at that time had grown to 185 people.
THis was all the information we were given
Introduction: A Chi-square test is used to compare observed data with expected data according to a hypothesis. For instance, if you were crossbreeding 2 heterozygous pea plants, you would expect to see a 3:1 phenotypic ratio in the offspring. In this case, if you were to breed 400 pea plants, you would expect to see 300 plants showing the dominant trait and 100 showing the recessive trait. But what happens if you observe only 260 plants with the dominant trait and 140 plants with the recessive trait? Does this mean something is wrong with Mendelian genetics or is this difference in expected results just due to chance (random sampling error)? These are the questions that can be answered using Chi-square statistics. The results of this statistical test is used to either reject or accept (fail to reject) the null hypothesis. The null hypothesis states there is no significant difference between the observed results and the expected results. This means that if the null hypothesis is accepted, the difference in observed and expected results was just a matter of chance and so the observed results basically "fit" with what was expected. Degrees of freedom (df) = number of independent outcomes (Y) being compared less 1 df = Y-1 At the 95% confidence interval we are 95% confident that there is a significant difference between the observed and expected results, therefore rejecting the null hypothesis. Probability Value - Is the decimal value determined from the X2 table and is the probability of accepting the null hypothesis. A 0.05 probability value equates to a 95% confidence interval.
The Chi-squared test formula is: Example: If we cross two pea plants that are heterozygous yellow pods, we would expect a 3:1 phenotypic ratio. So let's say we actually did the cross and got 280 plants with green pods and 120 plants with yellow pods. Question: Is this a 3:1 phenotypic ratio? This is the value of Chi-squared Test. We have a total of 400 plants and we expect a 300 green:100 yellow phenotypic ratio If the calculated Chi-squared value is less than the critical value listed in the Chi-squared table, then we accept the null hypothesis. This means that there is no significant difference between the observed and the expected values. Our degrees of freedom (df) = 2 outcomes - 1, or df = 1. Now we go the X2 table below and using the df = 1 and probability value of 0.05, our critical value is 3.84. Since our calculated X2 value is 5.33, and is larger than the critical value, we reject the null hypothesis and can say (at 95% confidence) that there is a significant difference between our observed and expected values.
The parent generation is yellowed podded and green podded pea plants. You cross a yellow podded pea plant with a green podded pea plant and you get 100% yellow podded plants in the F1 Generation (Phenotypic ratio 4 : 0, yellow to green). What will be the expected phenotypic ratio when you allow the F1 generation to reproduce?
Fill out the Punnett square.
If we actually did the cross and got 1150 yellow and 350 green. Would this be a consistent with what was expected?
Learning Outcomes Questions
1. Why would you run a Chi-squared test?
To determine if our data is consistent with expected results. | ||
a To determine if our data is consistent with expected results. b To determine if our data exactly matches the expected results. | ||
c To determine the expected results. | ||
d | To compare the phenotypic ratios to the genotypic ratios. |
2. Determine the degrees of Freedom of the phenotypic ratio for this genetic cross.
a. 1
b. 2
c. 3
d. 4
e. 5
3. Using the data given, what is the result of your Chi-squared analysis? x2= ___.
a. | 2.22 | |
b | 2.71 | |
c | 4.36 | |
d | 187.78 | |
e | 448.27 |
4. Using the results of your Chi-squared analysis, do we fail to reject or reject the null hypothesis?
a. | Fail to reject the null | |
b. | Reject the null | |
c. | It cannot be determined from the data given |