Biology Forum › Genetics › I really need help with dihybrid crosses plz!

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January 5, 2006 at 4:53 am #3094Jinasarangyoonhox3Participant
In this problem … it says "two pea plants heterozygous for both traits are crossed."
GGreen pea pod gYellow pea pod
TTall tshortwhat 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. 😆

January 5, 2006 at 9:05 am #36232xnabizkoxParticipant
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.

January 5, 2006 at 11:01 am #36238EnzymeParticipant
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! 😉

January 5, 2006 at 11:34 am #36244sdekivitParticipant
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_ggttBecause 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.

January 5, 2006 at 12:16 pm #36251EnzymeParticipantquote sdekivit: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 :wink:.

January 5, 2006 at 8:42 pm #36283Jinasarangyoonhox3Participant
thank you so much !!! also to get the phenotypic ratio of this problem … im not too sure…

January 5, 2006 at 9:33 pm #36287xnabizkoxParticipant
Phenotypic ratio is 9:3:3:1
9 Green Tall
3 Green Short
3 Yellow Tall
1 Yellow ShortDo the punnett square and count how many different physical traits result from the cross.

January 5, 2006 at 10:25 pm #36298EnzymeParticipantquote Jinasarangyoonhox3: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:
quote xnabizkox:Phenotypic ratio is 9:3:3:19 Green Tall
3 Green Short
3 Yellow Tall
1 Yellow ShortDid you find all the phenotypes and their probabilities in the table of genotypes (Punnet’s square)?

January 5, 2006 at 11:15 pm #36303Jinasarangyoonhox3Participant
okay … i understand … thanks …
also my final question …. if i want to find the chances of having offspring heterozygous for both traits for the pea plants …. would the answer be 4:16 … which reduces to 1/4. so 1/4 would be the answer? is that correct?

January 6, 2006 at 9:25 am #36317EnzymeParticipantquote xnabizkox: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 :wink:.

January 6, 2006 at 1:07 pm #36332sdekivitParticipant
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

January 6, 2006 at 1:09 pm #36334sdekivitParticipantquote Enzyme:quote xnabizkox: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 :wink:.
with my ‘probabilitytree’: Gg = 1/2 and Tt = 1/2
–> GgTt = 1/2 * 1/2 = 1/4 = 4/16

January 6, 2006 at 4:50 pm #36341EnzymeParticipantquote sdekivit:quote Enzyme:quote xnabizkox: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 :wink:.
with my ‘probabilitytree’: Gg = 1/2 and Tt = 1/2
–> GgTt = 1/2 * 1/2 = 1/4 = 4/16
He he, your ‘probability tree’ is very nice :D.

January 6, 2006 at 6:19 pm #36345xnabizkoxParticipant
Thank you. It was actually a lot harder to make then I first imagined. But it’s all in the name of science 😀

January 6, 2006 at 7:42 pm #36358sdekivitParticipant
ty enzyme ❗ 💡 😀

January 21, 2006 at 10:29 pm #38021EnzymeParticipant
He he, you’re welcome ;).


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