I really need help with dihybrid crosses plz!

[Jinasarangyoonhox3]

In this problem … it says "two pea plants heterozygous for both traits are crossed."

G-Green pea pod g-Yellow pea pod
T-Tall t-short

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. 😆

[xnabizkox]

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.

[Enzyme]

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! 😉

[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.

[Enzyme]

quote 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:.

[Jinasarangyoonhox3]

thank you so much !!! also to get the phenotypic ratio of this problem … im not too sure…

[xnabizkox]

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.

[Enzyme]

quote 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 9:3:3:1

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)?

[Jinasarangyoonhox3]

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?

[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:.

[sdekivit]

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

[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 ‘probability-tree’: Gg = 1/2 and Tt = 1/2

–> GgTt = 1/2 * 1/2 = 1/4 = 4/16

[Enzyme]

quote 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 ‘probability-tree’: 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.

[xnabizkox]

Thank you. It was actually a lot harder to make then I first imagined. But it’s all in the name of science 😀

[sdekivit]

ty enzyme ❗ 💡 😀

[Enzyme]

He he, you’re welcome ;).

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