BIOL 222 Chapter Notes - Chapter 2.5: Binomial Theorem, Null Hypothesis, Statistical Hypothesis Testing
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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 |
Please show detailed work. Thanks in advance.
16. Syntenic genes can assort independently when
A) they are very close together on a chromosome.
B) they are located on different chromosomes.
C) crossing over occurs rarely between the genes.
D) they are far apart on a chromosome and crossing over occurs frequently between the genes.
E) they are far apart on a chromosome and crossing over occurs very rarely between the genes.
17. The alleles of linked genes tend to
A) segregate together more often than expected by random assortment
B) assort independently.
C) be mutated more often than unlinked genes.
D) experience a higher rate of crossing over.
E) assort independently and show a higher rate of crossing over.
18. If you know that the frequency of recombination between genes X and Y is 34% and between X and Z is 25%, can you predict the order of the three genes?
A) Yes; the order is X-Z-Y.
B) Yes; the order is X-Y-Z.
C) Yes; the order is Z-X-Y.
D) No; based on this data alone, the order could be Z-Y-X or X-Y-Z.
E) No; based on this data alone, the order could be X-Z-Y or Z-X-Y.
Question 19 - 20. You have performed the following dihybrid cross in Drosophila using the black body color (b) and vestigial wing (vg) mutations. The b+ (grey body) and vg+ (normal wing) are dominant wild type alleles. These genes are autosomal.
Female â b+ vg+/b vg à male â b vg/b vg
Progeny:
Phenotype # of Progeny
Grey body normal wing 965
Black body vestigial wing 944
Grey body vestigial wing 208
Black body normal wing 195
19. Assuming linkage between black and vestigial, the estimated recombination frequency would be:
0.17
0.09
0.82
1.00
0.50
20. What key test could you use to determine whether the observed offspring frequencies deviate from those expected by chance alone?
A) Pascal's triangle
B) The product rule
C) The Chi-square (Ï2) test
D) The law of random assortment
E) The sum rule
21. In a genome wide association study (GWAS) designed to map the gene(s) that control height you divide subjects into a group of 1000 who are all more than seven feet tall and a control group of 1000 people of average height. You find the following associations between two genetic markers and the height trait:
| Tall group | Control group | ||
| Marker 1 | Allele A | 20% | 50% |
| Allele T | 80% | 50% | |
| Marker 2 | Allele G | 15% | 15% |
| Allele C | 85% | 85% |
What is your best guess for which marker is more closely linked to a gene that influences height?
A) Marker 1
B) Marker 2
22. Two pure breeding parents produce red and white flowers. They are crossed and the F1 produces pink flowers. When the F1 are selfed to produce the F2, nine distinct classes of pigmentation are present among F2 individuals. What is your best guess of the minimum number of genes that underlie flower pigmentation in this species?
A. 2
B. 3
C. 4
D. 5
E. 6
23. In a quantitative genetic experiment you identify two genes that confer bands of color on the back of a fly. At each gene, a dominant allele causes one band of color. If flies that are heterozygous at both loci are crossed, what ratio of offspring do you expect in each phenotype (i.e., number of color bands) class? (answer options are given from lowest to highest band number)
A) 1:1:1:1:1
B) 1:2:2:2:1
C) 1:4:6:4:1
D) 4:4:4:4:4
