The gene for rough coat is dominant to the gene for a smooth and black is domina

Question

The gene for rough coat is dominant to the gene for a smooth and black is dominant to white. Cross-and animal homozygous for both rough and black with a smooth white one.
a) What will be the appearance of the F1?
b) What will be the appearance of F2?
c) What will be the appearance of the offspring of a cross of the F1 with the rough black parent?
d) What will be the appearance of the offspring of a black or test cross of the F1 with the smooth white parent?

How many different types of gametes could be produced by a guinea pig of the genotype bbLlRrSsTt?

Explanation / Answer

taking R for rough and r for smooth B for black and b for white a) RRBB * rrbb now RRBB gives two gametes both (RB) rrbb gives two gametes both (rb) so the cross is between RB and rb and we get 100% RrBb appearance will be rough and black b) doing the self cross of RrBb will give simply crossing of Rr *Rr gives 1/4 RR 1/2 Rr 1/4 rr Bb * Bb gives 1/4 BB 1/2 Bb 1/4 bb so RRBB= 1/4 * 1/4 =1/16 RRBb= 1/8 RRbb= 1/16 RrBB= 1/8 RrBb=1/4 Rrbb =1/8 rrBB = 1/16 rrBb=1/8 rrbb= 1/16 rough and black= 9/16 rough and white = 3/16 smooth and black = 3/16 smooth and white = 1/16 c)F1 = RrBb rough black parent = RRBB RRBB * RrBb in any case of crossing we do not get bb as one gamete has dominant and we dont get rr as one parent has dominant RB RB RRBB Rb RRBb rB RrBB rb RrBb all rough black d)F1 and smooth white(rrbb) RrBb * rrbb rb RB RrBb Rb RRbb rB rrBb rb rrbb 1/4 rough black 1/4 rough white 1/4 smooth black 1/4 smooth white bb give only b Ll give L and l (2 types) Rr gives R and r Ss gives S and s Tt gives T and t hence types of gametes = 2*2*2*2 =16 types

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