Describe gene expression

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Kiwigirl
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Describe gene expression

Post by Kiwigirl » Mon Oct 09, 2006 10:16 pm

The following is from a table (I don't know how to use html to make a table, so I have listed the contents in bold) which shows the effect of two genes in providing resistance to powdery mildew disease in peach trees.

Note that the observed phenotypic ratio did not agree with the expected phenotypic ratio.

F2 Class: T_H_
F2 Expected Phenotypic ratio: 9
Reaction to powdery mildew disease: Highly resistant (HR).

F2 Class: T_hh
F2 Expected Phenotypic ratio: 3
Reaction to powdery mildew disease: Highly resistant (HR).

F2 Class: ttH
F2 Expected Phenotypic ratio: 3
Reaction to powdery mildew disease: Fairly resistant (FR).

F2 Class: tthh
F2 Expected Phenotypic ratio: 1
Reaction to powdery mildew disease: Susceptible (S).

Observed phenotypic ratio: 12 HR : 3 FR : 1 S


Question:
If plants were the genotypes Tthh and ttHH were crossed, what proportion of their offspring would be expected to be failry resistant to powdery mildew disease?

I have two answers for this problem, and I don't know which one is correct, or if either one is correct at all. Please help me answer this so I can compare it with my own answers and understand it better. Thank you!

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canalon
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Post by canalon » Mon Oct 09, 2006 10:30 pm

RAther give us your answer and ow you arrive to each one, and we will help you.

Note also that the expected and observed phenotypic ratio are in perfect agreement (9+3=12 in my books...)
Patrick

Science has proof without any certainty. Creationists have certainty without
any proof. (Ashley Montague)

Kiwigirl
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Post by Kiwigirl » Mon Oct 09, 2006 10:48 pm

Well in my book it says that "the observed phenotypic ratio did not agree with the expected phenotypic ratio." So I am just going by what it says.

The first answer was:

From the data it appears that a single T allele alone confers high resistance and a single H allele in a tt background confers fair resistance. If we cross Tthh and ttHH then all offspring will have one H (so we will have no susceptibles). On top of this Hh background we expect a 50:50 mix of Tt and tt genotypes, so we end up with 50% high resistance (Tt) and 50% fair resistance (ttHh) in the crossed generation.

The second answer I have is:

Observed Ratio (12:3:1) is deviating from dihybrid ratio of 9:3:3:1. It is a case of gene of interaction – Dominant Epistasis. Gene T is dominant over gene H.

Parent : TThh x ttHH

Gametes : Th tH

F1 : TtHh ( Highly Resistant)

F2 : T_H_ : 9 (Highly Resistant) T > H
T_hh : 3 (Highly Resistant)
ttH_ : 3 (Fairly resistant)
tthh : 1 (Susceptible)

So, 0 % in F1 generation and 18.75 % in F2 generation are fairly resistant to powdery mildew.

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Post by Kiwigirl » Mon Oct 09, 2006 10:52 pm

What I am confused with is... for the first answer, it basically ignores that "the observed phenotypic ratio did not agree with the expected phenotypic ratio" which it can't do, because I have to go by what it says in the book. The second answer, however, uses TThh × ttHH, when the question specifically states that the genotypes are Tthh × ttHH. Help please :)

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