Bio Question...please help! :(

[whale86]

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quick question, i forgot what the "trick" or shortcut is for figure out the number of possible offspring from for instance BbBB x TTtt without actually drawing out a punnett square.....anyone remember it? 😕 i think its something raised to the power of something 😛 I just dont remember what that "something" is!! 🙁

 

[system7]

are you talking about how many different gametes can be produced?

for that it's 2^n where n= #of heterzygotes.

for example, an organism with genotype HhGgLLKk, there are three heterozygotes (LL is homozygous), therefore you will get 2^3 gametes which equals 8.

 

[spritezhang]

are you talking about how many different gametes can be produced?

for that it's 2^n where n= #of heterzygotes.

for example, an organism with genotype HhGgLLKk, there are three heterozygotes (LL is homozygous), therefore you will get 2^3 gametes which equals 8.

Correct, and make sure you know both genotype and phenotype ratios, my DAT has those question and it is dihybird.

 

[whale86]

thanks guys!! 🙂

spritezhang- what do u mean genotype and phenotype ratios? u mean know how to solve the 9:3:3:1 kinda problems? if you could clarify or give an example it would be great....

 

[spritezhang]

thanks guys!! 🙂

spritezhang- what do u mean genotype and phenotype ratios? u mean know how to solve the 9:3:3:1 kinda problems? if you could clarify or give an example it would be great....

Yes, that's what i am talking about. But it will not be 9:3:3:1 that easy, the parental genotype is given, for example, BbFf and bbff, what's the phenotype ratio of the offsprings?

 

[whale86]

oh ok! Thanks 🙂

 

[sfoksn]

I know what you are asking now... Yeah there is a way to calculate the phenotypic ratio and genotypic ratio by being given the genotypes of parents.

Give me an example and I can show you how to do it, if you would like.

 

[RCT PC CRN]

I know what you are asking now... Yeah there is a way to calculate the phenotypic ratio and genotypic ratio by being given the genotypes of parents.

Give me an example and I can show you how to do it, if you would like.

We can take spritezhang example to try it out. BbFf and bbff - I'm assuming capital letter as dominant.

I got 1:1:1:1 - genotype 1:1:1:1 - phenotype (DD: Dr:rD:rr) D=dominant r=recessive

Can you confirm?

 

[whale86]

We can take spritezhang example to try it out. BbFf and bbff - I'm assuming capital letter as dominant.

I got 1:1:1:1 - genotype 1:1:1:1 - phenotype (DD: Dr:rD:rr) D=dominant r=recessive

Can you confirm?

could you please explain how you got that answer?

 

[whale86]

We can take spritezhang example to try it out. BbFf and bbff - I'm assuming capital letter as dominant.

I got 1:1:1:1 - genotype 1:1:1:1 - phenotype (DD: Dr:rD:rr) D=dominant r=recessive

Can you confirm?

I got: 1:1:2 being that...

1 was DD
1 was DR
and 2 were rr

D-dominant
R-recessive


correct me if im wrong please...

 

[sfoksn]

We can take spritezhang example to try it out. BbFf and bbff - I'm assuming capital letter as dominant.

I got 1:1:1:1 - genotype 1:1:1:1 - phenotype (DD: Dr:rD:rr) D=dominant r=recessive

Can you confirm?

Yes, that is 1:1:1:1.

I think we learn those stuff in intro genetics class.

So this is how you would do it:

You have BbFf x bbff.

Chance of B- is 1/2, because 1/2 chance of getting B from the first parent. Once you get the first parent's dominant gene, it doesn't matter which gene you get from the second parent in regards to B gene.

Same logic, chance of being bb is 1/2 as well.

You use the same logic for F gene, and chance of F- is 1/2, and ff is 1/2.

So chance of B-F- = chance of B- x chance of F-, which would be 1/2 x 1/2 = 1/4.

It's like statistics 🙂

 

[whale86]

i guess you can also break the dihybrid cross into 2 small crosses and get the 1:1:1:1 ratio fast!

thanks guys! 🙂

 

[MoElyassi]

i guess you can also break the dihybrid cross into 2 small crosses and get the 1:1:1:1 ratio fast!

thanks guys! 🙂

this way it will be faster..
interesting. thanks.