January 5, 2006 at 4:53 am #3094
In this problem … it says "two pea plants heterozygous for both traits are crossed."
G-Green pea pod g-Yellow pea pod
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. 😆
January 5, 2006 at 9:05 am #36232
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] _ 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 #36238
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 #36244
another one is just simply using mathematics :p –> chanches. I’ll explain:
If we cross GgTt with each other, we get:
_________________Gg______ x ________Gg
__________/___|___\_____/___|___\_____/___|___\ ____<– (times Tt)
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.
January 5, 2006 at 12:16 pm #36251quote sdekivit:
He he, cool example :wink:.
January 5, 2006 at 8:42 pm #36283
thank you so much !!! also to get the phenotypic ratio of this problem … im not too sure…
January 5, 2006 at 9:33 pm #36287
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.
January 5, 2006 at 10:25 pm #36298quote Jinasarangyoonhox3:
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:
Did you find all the phenotypes and their probabilities in the table of genotypes (Punnet’s square)?
January 5, 2006 at 11:15 pm #36303
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 #36317quote xnabizkox:
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 #36332
phenotypes can also be done in the same way:
________________Gg x Gg
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 #36334quote Enzyme:
with my ‘probability-tree’: Gg = 1/2 and Tt = 1/2
–> GgTt = 1/2 * 1/2 = 1/4 = 4/16
January 6, 2006 at 4:50 pm #36341quote sdekivit:
He he, your ‘probability tree’ is very nice :D.
January 6, 2006 at 6:19 pm #36345
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 #36358
ty enzyme ❗ 💡 😀
January 21, 2006 at 10:29 pm #38021
He he, you’re welcome ;).
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