Genetic problems --

1. Chin fissure is controlled by a dominant allele and a smooth chin by a recessive allele. If the parents were heterozygous for the trait, the chance of producing a child with a chin fissure would be 75%.

Cc * Cc

	C	c
C	CC	Cc
c	Cc	cc

Chin fissure -- dominant, so expressed in heterozygous and homozygous dominant individuals -- 3/4 (or 75%) possible outcomes will have at least one dominant allele.

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2. (Assume a simple/dominant recessive relationship). If you cross two green pea plants. One plant is homozgyous and one is heterozygous. What fraction of their progeny (offspring) will be homozygous dominant, heterozygous or homozygous receissive? In other words, what is the genotypic ratio?

If both plants are the same phenotype (green) and one is heterozygous, then we have to assume that the homozygous one is homozygous dominant. If it was homozygous recssive it would have a different phenotype from the heterozygous individual.

So the cross is GG* Gg

	G	g
g	Gg	gg
g	Gg	gg

The genotypic ratio would be 2-Gg : 2 gg (or 1:1 homozygous dominant and heterozygous).

There would be no homozygous recessive.

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3. Albinism (lack of skin pigmentation) is caused by a recessive autosomal allele. A man and a woman, both normally pigmented, have an albino child. The mother is now pregnant again, and her doctor tells her she is carrying fraternal twins (resulting from two eggs between fertilized by separate sperm). What is the probability that both twins will have normal pigmentation

Since the parents have one albino child (who must be homozygous recessive since it has albinism), they both must be heterozgyous (both must give the child one copy of a recessive allele.

A a

A AA Aa

a Aa aa

If the mother is pregnant with faternal twins - it is the same as two independent births, as far as genetics. So the probability of have a child with normal pigemention if both parents are heterozygous is 3/4 (see punnett square above). Probablility of both twins having normal pigmentation would be 9/16 (3/4 * 3/4 = 9/16). Need to multiple the probability of both events together.

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4. In snapdragons plant color shows an incomplete dominance. Heterozygous plants have pink flowers, where as the two homozygotes have red flowers or white flowers. Two pink flowers are crossed. What proportion of their progeny would you predict would be pink

Since this is a case of incomplete dominance, you usually use superscripts to indicate the two alleles -- since neither allele is completely dominant over the other we can't use the upper and lower cases. C^RC^W * C^RC^W

C^R C^W

C^R C^RC^R C^RC^W

C^W C^RC^W C^WC^W

1/2 of the offspring would be predicted to be heterozygous -- so 50% of the offspring would be pink.

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5. If two heterozygous parents, CcDd, were crossed there are NINE possible combinations of GENOTYPES in the offspring? FOUR possible phenotypes?

	CD	Cd	cD	cd
CD	CCDD	CCDd	CcDD	CcDd
Cd	CCDd	CCdd	CcDd	Ccdd
cD	CcDD	CcDd	ccDD	ccDd
cd	CcDd	Ccdd	ccDd	ccdd

Nine possible Genotypes - (Four possible Phenotypes):
CCDD - 1 (dominant, dominant)
CCDd - 2 (dominant, dominant)
CcDD - 2 (dominant, dominant)
CcDd - 4 (dominant, dominant)
CCdd - 1 (dominant, recessive)
Ccdd - 2 (dominant, recessive)
ccDD - 1 (recessive, dominant)
ccDd - 2 (recessive, dominant)
ccdd - 1 (recessive, recessive)

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6. If tall is dominant to short, and purple flowers are dominant to white flowers, what is the phenotypic ratio of the potential offspring if you cross a short, white flower with a homozygous dominant tall, heterozygous purple flower? ttpp * TTPp

	tp	tp	tp	tp
TP	TtPp	TtPp	TtPp	TtPp
Tp	Ttpp	Ttpp	Ttpp	Ttpp
TP	TtPp	TtPp	TtPp	TtPp
Tp	Ttpp	Ttpp	Ttpp	Ttpp

Phenotypic ratio: 1:1 or 50% - Tall, Purple; 50% Tall, white

Or since one parent only have one possible gamete and the other only has two possible combinations, you can do a smaller punnett's square --

	tp
TP	TtPp
Tp	Ttpp

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7. Black fur in mice is dominant to brown fur (B – dominant allele, b-recessive allele). Short tails are dominant to long tails (T – dominant allele, t-recessive allele). What fraction of the progeny of the cross BbTt * BBTt will have black fur and long tails?

	BT	Bt	BT	Bt
BT	BBTT	BBTt	BBTT	BBTt
Bt	BBTt	BBtt	BBTt	BBtt
bT	BbTT	BbTt	BbTT	BbTt
bt	BbTt	Bbtt	BbTt	Bbtt

4/16 would be have black fur and long tails.

Alternatively you could do this using the multiplication method -- teating the two characters separately:

	B	b
B	BB	Bb
B	BB	Bb

All dogs would be black. (no homozygous recessive possible) 4/4

	T	t
T	Tt	Tt
t	Tt	tt

1/4 of the dogs would be predicted to have long tails (tt) So black & long tailed would be 4/4 * 1/4 = 4/16