Spectrum Biology - September 2016

Gametes KL Kl kL kl

F 2 -generation

Chocolate : 9

Khaki : 3

Buff : 4

KL KKLL

(Chocolate)

KKLl

(Chocolate)

KkLL

(Chocolate)

KkLl

(Chocolate)

Kl KKLl

(Chocolate)

KKll

(Khaki)

KkLl

(Chocolate)

Kkll

(Khaki)

kL KkLL

(Chocolate)

KkLl

(Chocolate)

kkLL

(Buff)

kkLl

(Buff)

kl KkLl

(Chocolate)

Kkll

(Khaki)

kkLl

(Buff)

kkll

(Buff)

Con nec tivity with the Laws

Lablab has two genes, K and L. In the recessive state, the second

or supplementary gene (ll) has an effect on seed coat colour.

Dominant K independently produces khaki colour while its

recessive allele gives rise to buff colour irrespective of the

supplementary gene being dominant or recessive. In the

dominant state the supplementary gene ()L− changes the effect

of dominant allele of pigment forming gene (K) into chocolate

colour. F 2 phenotypic ratio is 9 : 3 : 4. It is also considered as

recessive epistasis.

When an agouti (CCAA) mice is crossed with an albino (ccaa)

mice, in F 1 -generation all agouti mice are obtained. On

inbreeding, in F 2 -generation agouti, black and albino mice are

produced. How would you justify the presence of black mice in

F 2 -generation? With the help of Punnett square predict the

phenotypic ratio of F 2 -generation.

Sol.If we consider a cross between mice of two different coat colour.

Connectivity with the Laws

The coat colour in mice is governed by two dominant genes

A and C. The presence of gene C produces black colour which

along with gene A changes its expression to agouti colour. Thus,

all combinations with atleast one C and one A give agouti colour.

The black mice possesses factor for black colour (C), but not the

gene A for agouti colour. Thus, all combinations with at least

one C, but with A produce black individuals. If gene for black

colour is absent agouti is unable to express itself hence, albinos

are seen. Therefore, in all combinations, where C is absent in the

presence or absence of A, the individuals are albinos. Thus,

presence of black mice is F 2 -generation can be explained on the

basis of supplementary genes or recessive epistasis.

When a brown dog (bbii) was crossed with a white dog (BBII), in

F 1 -generation all white dogs were obtained. Which type of dogs

will be obtained in F 2 -generation? Explain.

Sol.From the facts given in the problem, we can conclude following

results.

Connectivity with the Laws

It can be explained on the basis of dominant epistasis. In dogs

white coat colour appears to be dominant. It develops due to a

gene I which prevents the formation of pigment responsible for

colour of coat. The hypostatic gene B produces black coat while

its counter part b produces brown colour only when gene I is

homozygous recessive. Thus, all those cases where I is present

are white in colour. Similarly, all those cases with at least one

B, but absence of I are black while the case in which both B and I

are absent produces brown individual.

A white leghorn (IICC) variety of chicken is crossed with white

polymouth (iicc) variety. What would be the phenotype of F 1 and

F 2 -generation? Explain with the help of Punnett square. On the

basis of which law of genetics it can be explained?

Sol.Let us consider a cross between white leghorn (IICC) and white

polymouth (iicc) variety of chicken.

Connectivity with the Laws

It can be explained on the basis of law of dominant epistasis.

C gene is responsible for the coat colour of chickens. I prevents

the development of pigment even when pigment is present.

Colourless forms are produced when

Both colour (C) and inhibiting form (I) are present in their

dominant form.

Both colour and inhibiting factors are absent as their

dominant allele, i.e. when genetic constitution is ccii.

Only dominant gene for colour is absent. The inhibiting factor

may or may not be present, i.e. either ccII or ccIi or ccii.

ccaa

(Albino)

Ccaa

(Black)

ccAa

(Albino)

CcAa

ca (Agouti)

Ccaa

(Black)

CCaa

(Black)

CcAa

(Agouti)

CCAa

Ca (Agouti)

ccAa

(Albino)

CcAa

(Agouti)

ccAA

(Albino)

CcAA

cA (Agouti)

CcAa

(Agouti)

CCAa

(Agouti)

CcAA

(Agouti)

CCAA

CA (Agouti)

% CA cA Ca ca

&

Agouti

(CcAa)

Agouti

(CCAA)

Albino

(ccaa)

×

Gametes

Gametes

F -generation 2

Agouti : 9

Black : 3

Albino : 4

F -generation 1

bbii

(Brown)

Bbii

(Black)

bbIi

(White)

BbIi

bi (White)

Bbii

(Black)

BBii

(Black)

BbIi

(White)

BBIi

Bi (White)

bbIi

(White)

BbIi

(White)

bbII

(White)

BbII

bI (White)

BbIi

(White)

BBIi

(White)

BbII

(White)

BBII

BI (White)

% BI bI Bi bi

&

All white

(BbIi)

Brown dog

(bbii)

White dog

(BBII)

×

Gametes

Gametes

F -generation 2

White : 12

Black : 3

Brown : 1

F -generation 1

iicc

(White)

Iicc

(White)

iiCc

(Coloured)

IiCc

ic (White)

Iicc

(White)

IIcc

(White)

IiCc

(White)

IICc

Ic (White)

iiCc

(Coloured)

IiCc

(White)

iiCC

(Coloured)

IiCC

iC (White)

IiCc

(White)

IICc

(White)

IiCC

(White)

IICC

IC (White)

% IC iC Ic ic

&

All white

(IiCc)

White leghorn

(IICC)

White polymouth

(iicc)

×

Gametes

Gametes

F -generation 2

White : 13

Coloured : 3

F -generation 1