Help!! With Dihybrid Crosses!!!! (1 Viewer)

[Uknowuwantme]

i really really need help with dihybrid crosses...i don't know how to cross the following...and search engines aren't helping me at all!!!...if i don't turn this in by Monday imm DOOMED!! i don't know how to cross these:


Ttbb X ttbb

ttBb X Ttbb

Ttbb X TtBb


Help me please!!!

 

[rhia]

according to the syllabus, i don't think we need to know about dihybrid crosses, so i haven't bothered to burn that to memory.

http://en.wikipedia.org/wiki/Dihybrid_cross

 

[k_stroud22]

isn't dihybrid crosses to do with the option Genetics:the code broken? sry but i'm not doing that option. i think you have to do a heaps large punnet square. try hsc online

www.hsc.csu.edu.au/

it should have it.

 

[ubergeekery667]

ubergeekery667
Sorry it;s late, but I figured I'd try to close this out in case someone else searches the archives.

It’s no trouble at all. We’ll make the following assumptions:

A – This is a case of simple Mendelian inheritance
B – We’re dealing with complete dominance (T masks t and B masks b)
C – The alleles are not linked (independent assortment, T/t B/b on different chromosomes)
D – We can do single-trait crosses

The easiest way is to use punnett squares. When you say:

Ttbb X ttbb

That represents the mating of 2 individuals with genotypes Ttbb and ttbb

Each offspring should get one copy of each allele for both the T trait and the B trait.

Cross 1------------- Ttbb X ttbb


The first individual can only produce the following gametes:
Tb or tb

The second individual can only produce the following gametes:
tb

If you then do a punnett square you will produce the following:
Tb____tb
tb |Ttbb | ttbb|

Resulting phenotype breakdown:
Ttbb = 1/2 = T phenotype and b phenotype
ttbb = 1/2 = t phenotype and b phenotype

Cross 2-------------ttBb X Ttbb


For ttBb X Ttbb it is slightly more extensive, but no more complicated:

The only gametes that ttBb can produce are:
tB or tb

The only gametes that Ttbb can produce are:
Tb or tb

Doing the cross looks like this:
Tb___tb
tb |Ttbb | ttbb|
tB|TtBb | ttBb|

So, the genotypes that result are:
Ttbb
ttbb
TtBb
ttBb

Phenotype breakdown:
Ttbb = 1/4 = T and b phenotype
ttbb = 1/4 = t and b phenotype
TtBb = 1/4 = T and B phenotype
ttBb = 1/4 = t and B phenotype

Cross 3------------- Ttbb X TtBb
Gametes for Ttbb:
Tb, tb

Gametes for TtBb
TB,Tb,tB,tb (this is a lot like the distributive property of multiplication or the “Foil” method)

Doing the cross:
Tb tb
TB TTBb TtBb
Tb TTbb Ttbb
tB TtBb ttBb
tb Ttbb ttbb

Genotype Breakdown:
1 TTBb
2 TtBb
1 TTbb
2 Ttbb
1 ttBb
1 ttbb

Phenotype breakdown:
3 individuals with T and B phenotypes
3 individuals with T and b phenotypes
1 individual with t and B phenotypes
1 individual with t and b phenotypes

3:3:1:1

K, that’s about it. If you have q’s please post. There are some assumptions I have made, like the fact that you ought to know how to do monohybrid crosses.

Cheers,
Chris
Pittsburgh, PA, USA

 

[Dr_Doom]

Thanks! ubergeekery667! I fully get it now! U rule xD

 

[ari89]

Thanks! ubergeekery667! I fully get it now! U rule xD

umm...person who vanished after the hsc bumpin yr old threads?