Help pls answer: How many Diff. gametes cud b produced by a guinea pig wd d genes bbLIRrSs?pls help?

help me i've got more:

1.In horses, black is dependent upon a dominant gene B and chestnut id dependent upon its recesiv allele b. the trotting gait is due to a dominant gene T and the pacing gait is due to its recessive allele t. If a homozygous black pacer is mated to a homozygous chestnut trotter, what will b d possible phenotypes of the F1 generation?

2. If 2 F1 individuals from prob 1 were mated, what kind of offsprings cud dey have n in wat proportions?

3. If an F1 male from prob 1 were mated to a homozygous female black pacer, what kinds of offspring cud b produced and in wat proportions?

4. In cocker spaniels, black s due to a dominant gene B and red is due to d recessive allele B. Solid color is dpendent upon a dominant gene S and white spotting upon its recessive alllele s. A solid red male was mated 2 a black and white spotted fmale.They hav 5 spotted and 2 red and white spotted, what were d genotypes of the parents??? PlSSSSS ANSWEEERRR. pls biologists

Help pls answer: How many Diff. gametes cud b produced by a guinea pig wd d genes bbLIRrSs?pls help?
bbLlRrSs

There would be 8 different gametes (the first gene bb has only b to contribute, and the others have 2 alleles to contribute, so 1 x 2 x 2 x 2 = 8):

bLRS

bLRs

bLrS

bLrs

blRS

blRs

blrS

blrs



1. A homozygous black pacer would be genotye BBtt and the homozygous chestnut trotter would be genotype bbTT. The genotype of the possible F1 offspring would be: BbTt



2. BbTt x BbTt

F2:

BBTT

BbTT

BbTT

bbTT

BBTt

BbTt

BbTt

bbTt

BBTt

BbTt

BbTt

bbTt

BBtt

Bbtt

Bbtt

bbtt

The offspring in the F2 will have phenotypes in a 9:3:3:1 ratio. I'll let you figure out what the phenotypes are from the genotypes given above.



3. BbTt x BBtt



BBTt

BbTt

BBtt

Bbtt

All in equal proportions in terms of probability (25% for each)



4. S-bb x ssB-

Offspring:

5 spotted (black spotted?)

2 red and white spotted

If I assume the 5 are black spotted, then the genotypes of the parents have to have a recessive allele in each of the S and B genotypes, so:



Ssbb x ssBb