Answers

Crosses

Cross	Gametes First parent	Gametes 2nd parent	Genotype Ratio	Phenotype Ratio
AA X AA	all A	all A	all AA	all yellow
AA X Aa	all A	1/2 A, 1/2 a	1/2 AA, 1/2 Aa	all yellow
AA X aa	all A	all a	all Aa	all yellow
Aa X Aa	1/2 A, 1/2 a	1/2 A, 1/2 a	1/4 AA, 1/2 Aa, 1/4 aa	3/4 yellow, 1/4 green
Aa X aa	1/2 A, 1/2 a	all a	1/2 Aa, 1/2 aa	1/2 yellow, 1/2 green
aa X aa	all a	all a	all aa	all green

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Application

A = normal hemoglobin

a = sickle-cell hemoglobin

AA = normal

Aa = normal (called sickle-cell trait)

aa = sickle-cell anemia

A man with sickle-cell trait marries a normal woman. What is the probability that their children will have sickle-cell trait?

man = Aa

woman = AA

cross: Aa X AA

The first thing that needs to be done before analyzing any genetic cross is to determine what the gametes are that can be produced by each parent.

gametes, man = 1/2 A, 1/2 a

gametes, woman = all A

Once the gametes have been determined, a Punnett square analysis can be done.

The square below is used for this cross: AA X Aa.

One half of the offspring produced by this cross will be AA, the other half will be Aa.

The cross can also be written as shown below because one of the parents (AA) can produce only one kind of gamete (all A).

If both parents have sickle-cell trait, what percentage of their children will:

have a normal phenotype?

have sickle-cell trait?

have sickle-cell anemia?

 

man: Aa

woman: Aa

cross Aa X Aa

The first thing that needs to be done before analyzing any genetic cross is to determine what the gametes are that can be produced by each parent.

gametes, man = 1/2 A, 1/2 a

gametes, woman = 1/2 A, 1/2 a

Once the gametes have been determined, a Punnett square analysis can be done.

The square below is used for this cross: Aa X Aa.

Approximately 3/4 of the offspring will have the normal phenotype (AA and Aa in the table above).

Approximately 1/2 of the offspring will have sickle-cell trait (Aa),

Approximately 1/4 of the offspring will have sickle-cell anemia (aa).

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