I really need help with dihybrid crosses plz!
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 Garter
 Posts: 3
 Joined: Thu Jan 05, 2006 4:50 am
I really need help with dihybrid crosses plz!
In this problem ... it says "two pea plants heterozygous for both traits are crossed."
GGreen pea pod gYellow pea pod
TTall tshort
what are the chances of being homozygous for both traits in the F1 generation?
I'm confused on how to start it off ... like drawing the cross diagram. Thank you.
GGreen pea pod gYellow pea pod
TTall tshort
what are the chances of being homozygous for both traits in the F1 generation?
I'm confused on how to start it off ... like drawing the cross diagram. Thank you.
Dihybrid crosses involve 2 traits that will independently assort. Since the plants are heterozygous this means both plants will have the genotype of GgTt. The dihybrid cross looks a little like this.
GgTt x GgTt
______[GT]___[Gt]___[gT]___[gt]
[GT] _ GGTT _ GGTt _ GgTT _ GgTt
[Gt] _GGTt___ GGtt__ GgTt__ Ggtt
[gT] _ GgTT _GgTt _ ggTT __ ggTt
[gt] __GgTt___ Ggtt__ ggTt __ggtt
SO the chance of the F1 generation (first offspring) being homozygous (GGTT and ggtt) is 2:16.
Hope that helps.
GgTt x GgTt
______[GT]___[Gt]___[gT]___[gt]
[GT] _ GGTT _ GGTt _ GgTT _ GgTt
[Gt] _GGTt___ GGtt__ GgTt__ Ggtt
[gT] _ GgTT _GgTt _ ggTT __ ggTt
[gt] __GgTt___ Ggtt__ ggTt __ggtt
SO the chance of the F1 generation (first offspring) being homozygous (GGTT and ggtt) is 2:16.
Hope that helps.
Last edited by xnabizkox on Thu Jan 05, 2006 9:34 pm, edited 1 time in total.
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Jinasarangyoonhox3, the Punnet's square is very easy to draw. You only have to write the gametes from one parental individual in the first row and the gametes from the other parental individual in the first column. Then, only you have to combine them. xnabizkox has written it very well.
But, what were the question that problem says? If the problem ask for the probability of being homozygous for the two characters, you should answer 2/16 (because you have to take in account dominant homozygous and recessive homozygous). But if the problem ask for the probability of being dominant homozygous (GGTT) or recessive homozygous (ggtt), you should answer 1/16, as xnabizkox said.
See you!
But, what were the question that problem says? If the problem ask for the probability of being homozygous for the two characters, you should answer 2/16 (because you have to take in account dominant homozygous and recessive homozygous). But if the problem ask for the probability of being dominant homozygous (GGTT) or recessive homozygous (ggtt), you should answer 1/16, as xnabizkox said.
See you!
another one is just simply using mathematics :p > chanches. I'll explain:
If we cross GgTt with each other, we get:
_________________Gg______ x ________Gg
_______________/______________________\
___________GG(1/4)______Gg(1/2)______gg(1/4)
__________/______\_____/______\_____/______\ ____< (times Tt)
____GGTT_GGTt_GGtt_GgTT_GgTt_Ggtt_ggTT_ggTt_ggtt
Because TT also has a chance of 1/4 GGTT has a chance of 1/4 * 1/4 = 1/16
Tt has a chance of 1/2, thus GGTt has a chance of 1/4 * 1/2 = 1/8 = 2/16
And so on.
If we cross GgTt with each other, we get:
_________________Gg______ x ________Gg
_______________/______________________\
___________GG(1/4)______Gg(1/2)______gg(1/4)
__________/______\_____/______\_____/______\ ____< (times Tt)
____GGTT_GGTt_GGtt_GgTT_GgTt_Ggtt_ggTT_ggTt_ggtt
Because TT also has a chance of 1/4 GGTT has a chance of 1/4 * 1/4 = 1/16
Tt has a chance of 1/2, thus GGTt has a chance of 1/4 * 1/2 = 1/8 = 2/16
And so on.
sdekivit wrote:another one is just simply using mathematics :p > chanches. I'll explain:
If we cross GgTt with each other, we get:
_________________Gg______ x ________Gg
_______________/______________________\
___________GG(1/4)______Gg(1/2)______gg(1/4)
__________/______\_____/______\_____/______\ ____< (times Tt)
____GGTT_GGTt_GGtt_GgTT_GgTt_Ggtt_ggTT_ggTt_ggtt
Because TT also has a chance of 1/4 GGTT has a chance of 1/4 * 1/4 = 1/16
Tt has a chance of 1/2, thus GGTt has a chance of 1/4 * 1/2 = 1/8 = 2/16
And so on.
He he, cool example .

 Garter
 Posts: 3
 Joined: Thu Jan 05, 2006 4:50 am
Phenotypic ratio is 9:3:3:1
9 Green Tall
3 Green Short
3 Yellow Tall
1 Yellow Short
Do the punnett square and count how many different physical traits result from the cross.
9 Green Tall
3 Green Short
3 Yellow Tall
1 Yellow Short
Do the punnett square and count how many different physical traits result from the cross.
Changing the world one molecule at a time.
enzymatic clothing designs > http://enzymaticclothing.tripod.com
enzymatic blogspot > http://enzymaticclothing.blogspot.com/
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Jinasarangyoonhox3 wrote:thank you so much !!! also to get the phenotypic ratio of this problem ... im not too sure...
You only have to count all the different appearance traits in the Punnet's square, as xnabizkox said.
Take in account that it is a 9:3:3:1 segregation (because you'll see that 9/16 corresponds to G_T_, 1/16 to ggtt and the others are combinations). And seeing the genotypes of each result in Punnet's square, we can deduce the phenotypes. Take a look to xnabizkox post:
xnabizkox wrote:Phenotypic ratio is 9:3:3:1
9 Green Tall
3 Green Short
3 Yellow Tall
1 Yellow Short
Did you find all the phenotypes and their probabilities in the table of genotypes (Punnet's square)?

 Garter
 Posts: 3
 Joined: Thu Jan 05, 2006 4:50 am
xnabizkox wrote:GgTt x GgTt
______[GT]___[Gt]___[gT]___[gt]
[GT] _ GGTT _ GGTt _ GgTT _ GgTt
[Gt] _GGTt___ GGtt__ GgTt__ Ggtt
[gT] _ GgTT _GgTt _ ggTT __ ggTt
[gt] __GgTt___ Ggtt__ ggTt __ggtt
If you want to find the chances for having heterozygous for both treats, the answer would be 4/16 which we can simplify to 1/4. YES, IT IS CORRECT .
phenotypes can also be done in the same way:
________________Gg x Gg
_______________/________\
__________green(3/4)_yellow(1/4)
_________/___________________\
_____tall(3/4)_short(1/4)_tall(3/4)_short(1/4)
thus see here the phenotypes and their probabilities:
green and tall will be: (3/4 * 3/4) = 9/16
________________Gg x Gg
_______________/________\
__________green(3/4)_yellow(1/4)
_________/___________________\
_____tall(3/4)_short(1/4)_tall(3/4)_short(1/4)
thus see here the phenotypes and their probabilities:
green and tall will be: (3/4 * 3/4) = 9/16
Enzyme wrote:xnabizkox wrote:GgTt x GgTt
______[GT]___[Gt]___[gT]___[gt]
[GT] _ GGTT _ GGTt _ GgTT _ GgTt
[Gt] _GGTt___ GGtt__ GgTt__ Ggtt
[gT] _ GgTT _GgTt _ ggTT __ ggTt
[gt] __GgTt___ Ggtt__ ggTt __ggtt
If you want to find the chances for having heterozygous for both treats, the answer would be 4/16 which we can simplify to 1/4. YES, IT IS CORRECT .
with my 'probabilitytree': Gg = 1/2 and Tt = 1/2
> GgTt = 1/2 * 1/2 = 1/4 = 4/16
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