Question

Provide some examples of Dihybrid crosses that DO NOT show the classic Mendelian ratio of independent assortment.

Answer 1

Dihybrid cross : A cross that involves the analysis of two independent traits is termed as dihybid cross.

According to Mendals law of independent assortment ,every gene works in independent manner, this law states that assortment of genes of one pair is independent of the other pair at the time of gemetogenesis i.e each pair of contrasting character behaves independently. This law is applicable only when different genes are present on different chromosomes.

Limitation of law of independent assortment

• If two genes which encodes for different characters / or genes encode for the enzymes that works in same metabolic pathway ( i.e Gene interaction) , present on the same chromosome then law of independent assortment do not follow.

Example :

(1) Parental cross (seed shape and colour)

If alleles of 2 different genes present on the same chromosome ( R = Round shape, Y = yellow colour, y = green r= Wrinkled shape

Parent 1 Round shape and Yellow colour Parent 2 Wrinkled shape and green colour

RRYY X rryy

R Y   r y

R Y     r y

---------- Gamete formation by Meiosis-----------------

(Gamete )Parent 1 R Y     X   Parent 2 r y

After parental cross F1 generation R Y  

  r y

F1 generation 100% Round in shape and Yellow in colour

F1 cross R Y X   R Y

  r y r y

Round and Yellow Round and Yellow

------------Meiosis----------------

Gametes   R Y &  r y    X   R Y & r y   

  

F2 generation R Y R Y   r y       r y

  R Y r y r y     r y

3 : 1

Round and yellow in colour Wrinkled and green

Note : So the Mendelian dihybrid ratio 9:3:3:1 become 3:1 due to the presence of genes of two different traits on the same chromosome, As both genes present on the same chromosomes no recombination and no independent assortment occurs in gametogenesis, So only parental gametes forms.

(2) White substance Can be converted to the purple product only in the presence of enzyme A , enzyme A has high affinity for the white substrate.

White substance Can be converted to the Red product only in the presence of enzyme B, enzyme B has lower affinity for the white substrate.

• Due to the difference in the affinity for the white substrate, enzyme B can only work effectively when no enzyme A is present.

(Parent 1 ) AAbb X aaBB (Parent 2)

F1 AaBb (Purple)
  

F2 12 puple : 3 Red : 1 white

After cross, AABB (1), AABb (2), AaBB (2) AaBb (4), AAbb(1), Aabb (2) = One functional allele A convert all the substrate to purple product (12)

  aaBB (2), aaBb (1) = Red colour (3) due to the lack of enzyme A.

aabb (1) = white because no functional enzyme present and cannot synthesize any coloured product.

The 9:3:3:1 ratio becomes 12 purple : 3 Red : 1 white

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