The F
1
generation is all Gg.
The F
2
generation is Gg x Gg, resulting in a
3:1 phenotype (3 G_ : 1 gg).
If any of this is unclear, do the Punnett square.
Detailed answer for part a)
There are a total of 8023 plants.
Of these 8023, there’s an expected 3:1 ratio of yellow:green plants.
Therefore, we expect:
8023 x 0.75 = 6017.25 yellow plants
8023 x 0.25 = 2005.75 green plants
We observe 6000 yellow plants, and thus also observe 2023 green plants.
So our “e” is 6017.25 yellow plants and 2005.75 green plants.
Our “o” in part a) is 6000 yellow plants
and 2023 green plants.
Do the chisquared: Σ (oe)
2
/e
(60006017.25)
2
/6017.25 + (20232005.75)
2
/2005.75 = 0.049 + 0.148 = 0.197 = chisquared statistic
Use the chisquared table to find the pvalue where the calculated chisquared statistic falls.
Remember,
degrees of freedom (df) = # categories – 1; since there are two categories (phenotypes), the df = 1.
You’ll
find that the pvalue for calculated chisquared statistic for part a) is somewhere between 0.75 and 0.5.
(Note: you do NOT need to calculate the exact pvalue for the corresponding chisquared statistic for this
class.)
Since the pvalue is between 0.75 and 0.5, the null hypothesis (H
o
) is accepted.
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View Full DocumentPart a)
Chisquared value = 0.197; pvalue between 0.75 & 0.5; H
o
accepted
Part b)
Chisquared value = 0.922; pvalue between 0.5 & 0.25; H
o
accepted
Part c)
Chisquared value = 1.484; pvalue between 0.25 & 0.1; H
o
accepted
Part d)
Chisquared value = 5.060; pvalue between 0.025 & 0.01; H
o
rejected
Part e)
Chisquared value = 6.287; pvalue between 0.025 & 0.01; H
o
rejected
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 Winter '11
 Kumar
 Genetics, Plants, Legume

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