Genetics Answer Sheet

1. a) parents: BB, and bb; gametes B; b; offspring 100% heterozygous Bb (genotype); 100% Black (phenotype)

    b) parents Bb and Bb; gametes (parent 1; B and b) parent 2; B and b; offspring 25% BB, 50% Bb, 25% bb (genotypes); 75% Black, 25% white (phenotypes)

    c) parents Bb and bb; gametes (parent 1; B and b) parent 2; b only; offspring 50% Bb, 50% bb (genotype); 50% Black; 50% white (phenotypes)





Parents must be heterozygous (Ee)  to produce two yellow offspring (ee)  all others are purple (the purple offspring would be E _)

3. L = long wings; l = short wings

Parents L _ x L _; produce 49 short winged (ll) offspring; and 148 L _ offspring

Probable genotypes of parents would have to be heterozygous:  Ll x Ll to produce a 3:1 ratio of long to short wings

The genotypes of the offspring would be 25% LL; 50% Ll, 25% ll; which means that 2/3 of the 148 long winged offspring or 99 should be heterozygous.

4. The Father is B_ the mom is bb; all the offspring are B_; can't tell if dad is Bb because there is a 50:50 chance of him always giving the one B.                                                                                                     

5. B = black B'= white; shows incomplete dominance so BB' = blue

a) BB x BB'; offspring would be 1/2 BB; 12 BB' (genotypes); 1/2 black, 1/2 blue (phenotypes)

b) BB' x BB'; offspring would be 1/4 BB; 1/2 BB'; 1/4 B'B' (genotypes); 1/4 Black; 1/2 Blue; 1/4 White (phenotypes)

c) BB' x B'B'; offspring would be 1/2 BB'; 1/2 B'B' (genotypes); 1/2 blue; 1/2 white (phenotypes)

6. RR = red  Rr= pink, rr = white (incomplete dominance)

a) parents RR x rr; all offspring would be Rr so expect 100% pink no red expected AND 100% of the offspring are carrying one R allele

b) if cross two F1 plants (pink Rr); get 1/4 RR; 1/2 Rr; 1/4 rr (genotypes); get 1/4 Red; 1/2 Pink; 1/4 white (phenotypes).

7. XX female would yield gametes that all have one X;  an XY male would yield 1/2 of his gametes carrying an X and 1/2 carrying a Y.