Genetic Crosses

25%

$Hh \times Hh$ gives 25% HH, 50% Hh, and 25% hh. Only hh is homozygous for smooth hair: 25%.

0%: Each parent carries h, so hh offspring are possible: 1 of 4 boxes in the Punnett square.

50%: 50% is the heterozygous (Hh) offspring, which have wire hair.

75%: 75% is the offspring with wire hair (HH and Hh).

$\frac{4}{16}$

In the 16-box dihybrid Punnett square, 4 of the 16 combinations have aa. Those mice are albino whatever their B alleles: $\frac{4}{16}$.

$\frac{1}{16}$: 1/16 is only aabb. All aa mice are albino, including aaBB and aaBb, which makes 4/16.

$\frac{3}{16}$: All four aa combinations are albino, not three.

$\frac{9}{16}$: 9/16 is the A_B_ (agouti) group, not the albino group.

I-1: Aa; I-2: aa

I-2 is unshaded and therefore aa. I-1 expresses the dominant trait but has unshaded descendants, so I-1 must carry a and is Aa.

I-1: aa; I-2: Aa: This reverses the shaded dominant individual and unshaded recessive individual.

I-1: AA; I-2: Aa: I-2 is unshaded and must be aa; I-1 also has unshaded descendants and cannot be AA.

I-1: aa; I-2: AA: Both assignments conflict with the shading and the presence of unshaded descendants.

Punnett square A

Albino is recessive to all other alleles, so an albino offspring must inherit c from both parents. Square A includes c in each parent's gametes, allowing cc.

Punnett square B: This cross does not provide the c allele from both parents needed for cc.

Punnett square C: C is dominant to all other alleles, so offspring receiving C cannot be albino.

Punnett square D: Without c contributed by both parents, no cc albino offspring can be produced.

Aagg

The first parent is GG and can contribute only G, so every offspring receives at least one G. The genotype gg is impossible.

AaGG: This is possible when the first parent contributes AG and the second contributes aG.

aaGg: This is possible when the first parent contributes aG and the second contributes ag.

aaGG: This is possible when both parents contribute aG.

Table A

Black is BB and black lace is BL. A $BB \times BL$ cross produces 50% BB (black) and 50% BL (black lace), with no silver LL offspring.

Table B: This predicts silver offspring, but neither parent contributes an L-only pair without B.

Table C: This predicts 25% silver, but a BB parent contributes B to every offspring.

Table D: The expected probabilities are 50% black and 50% black lace, not these values.

100

The cross is AA × aa, so every child has genotype Aa. Because A is dominant, 100% of the children will have achondroplasia.

50%

There is a 50% chance of $RW$ offspring because the red bull always provides an $R$ allele and there is a 50% chance that the roan cow provides the $W$ allele.

0%: This is the probability of $WW$ offspring. The red bull always provides an $R$ allele, so every offspring is either red or roan.

25%: This would be the probability of red or white offspring from a cross between two roan cattle.

100%: This would be the probability of $RR$ from two red cattle or $WW$ from two white cattle.

All the sunflowers will be orange.

Every offspring is $Rr$. Incomplete dominance blends the phenotypes, so all the sunflowers will be orange.

One-quarter of the sunflowers will be yellow.: About one-quarter would be yellow from a cross between two hybrid red plants with dominant/recessive inheritance.

Half the sunflowers will be red.: About half would be red from a cross between a hybrid red plant and a recessive yellow plant with dominant/recessive inheritance.

Three-fourths of the sunflowers will have both red and yellow petals.: A mix of red and yellow petals would describe codominance, not incomplete dominance.

50%

Parent 1 ($aa$) does not carry the allele, and Parent 2 ($Aa$) carries one dominant allele. Child 1 has a 50% chance of inheriting the dominant allele from Parent 2 and having Marfan syndrome.

25%: A 25% probability would occur if the disorder were autosomal recessive and each parent were a carrier ($Aa$).

75%: A 75% probability would occur if each parent were heterozygous ($Aa$).

100%: A 100% probability would occur only if at least one parent were homozygous dominant ($AA$).