Dihybrid Genetics problems

1 . Black fur in mice (B) is dominant to brown fur (b).  Short tails (T) is dominant to long tails (t).
         What proportion of the progeny of the cross BbTt x BBtt will have black fur and long tails?

Answer:  BbTt   x   BBtt     

x

BT

Bt

bT

bt

Bt

BBTt

BBtt

BbTt

Bbtt

 

Black

Black

Black

Black

 

Short tail

Long

Short

Long

50%  Black Long  

2. A couple has three children, all of whom have brown eyes and blond hair.  Both parents are
         homozygous for brown eyes (BB), and one is blond (rr) while the other is a redhead (Rr).  What
         is the probability that the next child will be a brown-eyed redhead? 

P1   Brown Eye  Blond      x   Brown Eye Redhead

            BB rr                  x          BB Rr

x

BR

Br

Br

BBRR

BBrr

  phenotype

 

Brown eye

Brown Eye

Redhead

Blond

Assuming that RR is redhead then the probability is 50%  or P= 0.5, or a ½

3. In mice, black is dominant over tan and short tails are dominant over long.

x

BS

Bs

bS

bs

BS

BBSS

BBSs

BbSs

BbSs

Bs

BBSs

BBss

BbSs

Bbss

bS

BbSS

BbSs

bbSS

bbSs

bs

BbSs

Bbss

bbSs

bbss

 

9 Black Short; 3 Black Long; 3 Tan Short; 1 Tan Long

     

   

4. In certain bacteria, an oval shape is dominant over round and thick cell walls are dominant over thin. Cross a heterozygous oval, thick cell walled bacteria with a round, thin cell walled bacteria. Describe the phenotype of the offspring.

Allele Key  Q= oval  q= round     T= thick   t= thin

Oval Thickwall (QqTt)   x    Round  thin wall (qqtt)

x

qt

QT

QqTt

 Qt

Qqtt

qT

qqTt

qt

qqtt

Oval Thickwall :  Oval thin  wall; round Thickwall; round thin wall

1:1:1:1

5.
In guinea pigs, black coat colour is dominant over white, short hair is dominant over long.. Show the Punnett Square for a cross between a homozygous black, short-haired guinea pig and a homozygous white, long-haired guinea pig. What do the offspring look like?

Black (B) > White (b)           Short (S)> Long (s)

Homozygous black Short tailed ( BBSS)    x     Homozygous white, long haired ( bbss)

x

bs

Bs

BbSs

All Heterozygous   Black Short

6.  In humans, the presence of freckles is due to a dominant gene (F) and the non-freckled condition is
         due to its recessive allele (f).   Dimpled cheeks (D) is dominant to non-dimpled cheeks (d).  Two
         persons with freckles and dimpled cheeks  have two children.  One has freckles but no dimples and
         one has dimples but no freckles.

        a.  What are the genotypes of the parents?

          Parents     F_D_   x    F_D_           Children  F_ dd    ,   ffD_

Therefore  Parents must be heterozygous at each gene loci

                          FfDd     x    FfDd

         b.What are the possible phenotypes and genotypes of the children which they could produce?

x

FD

Fd

fD

fd

FD

FFDD

FFDd

FfDD

FfDd

Fd

FFDd

FFdd

FfDd

Ffdd

fD

FfDD

FfDd

ffDD

ffDd

fd

FfDd

Ffdd

ffDd

ffdd

Phenotypic Ration: 9 Freckled Dimpled;3 Freckled Non Dimple; 3 Non-freckle Dimple; 1 Non Freckle Non Dimple

9:3;3;1

 

c.     What are the chances that they would have a child which lacks both freckles and dimples?

The probability is 1 in 16 (this is a correction thank you Mavila Marina Miller)

   d.  A person with freckles and dimples whose mother lacked both freckles and dimples marries a
        person with freckles but no dimples (whose father did not have freckles or dimples).  What are
        the chances that they would have a child which lacks both freckles and dimples?

               FfDd  X  Ffdd

 

Fd

fd

FD

FFDd

FfDd

Fd

FFdd

Ffdd

fD

FfDd

ffDd

fd

Ffdd

ffdd