The amount of water vapor in the air is known as the _______.
1 answer:
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Answer:
a) Chi-Sq = 14.08. Degrees of freedom = 3. Genes are linked.
b) Distance= 40.5 map units.
Explanation:
All the offspring from the dihybrid cross of plants of different phenotype for both traits had green pods and normal leaves, so these two are the dominant traits.
<u>The alleles for the two genes are:</u>
- Y_= green pods, yy= yellow pods
- C_=normal leaves, cc=curling leaves
The parental plant with yellow pods and curling leaves (yc/yc) was crossed with the wild-type plant (YC/YC). The F1 plants were heterozygous YC/yc.
The F1 plants were then test crossed: YC/yc X yc/yc.
<u>The resulting F2 was:</u>
- 117 green pods, normal leaves (YC/yc)
- 115 yellow pods, curling leaves (yc/yc)
- 78 green pods, curling leaves (Yc/yc)
- 80 yellow pods, normal leaves (yC/yc)
Total: 390
a) If the genes were independent, the four gametes produced by the F1 individual would have the same frequency: 1/4. The test cross individual can only produce 1 type of gamete: <em>yc</em>.
The expected offspring would be:
- 1/4 x 390 = 97.5 green pods, normal leaves
- 1/4 x 390 = 97.5 yellow pods, curling leaves
- 1/4 x 390 = 97.5 green pods, curling leaves
- 1/4 x 390 = 97.5 yellow pods, normal leaves
Null hypothesis: genes are not linked.
The Chi square can be calculated as follows:
Where O is the observed number of offspring of a particuar phenotype and E is the expected number of offspring showing the same phenotype.
For dihybrid crosses, the degrees of freedom (DF) can be calulated as number of phenotypes - 1. In this case there are 4 possible phenotypes, so 3 DF. If you take a look at a Chi-Square Table, for 3 DF the Chi-Sq must be greater than 7.815 (p < 0.05) in order be statistically significant.
In our case, Chi-Sq was 14.08, so we can reject the null hypothesis. The genes are linked.
b) Distance = Frequency of Recombination x 100
The recombinant gametes are Yc and yC, so there are a total of 78+80=158 recombinant individuals in the offspring. Frequency of Recombination= 158/390=0.405
Distance=0.405 x 100 = 40.5 map units.