SAT II Biology M : Inheritance Patterns

Study concepts, example questions & explanations for SAT II Biology M

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Example Questions

Example Question #1 : Inheritance Patterns

According to the law of independent assortment, what is the possible number of combinations that chromosomes can assort to independently in the gamete?

Possible Answers:

70,368,744,177,664

2,048

4,194,304

8,388,608

16,777,216

Correct answer:

8,388,608

Explanation:

According to the law of independent assortment, there are 2n combinations where chromosomes can assort into different gametes. So where n is the haploid number, you get 223=8,388,608. There are 8,388,608 possible combinations of chromosomes when assorting into gametes.

Example Question #2 : Inheritance Patterns

What contributes to genetic variation during human reproduction?

Possible Answers:

Random fertilization

Independent assortment

Nonrandom mating

All of these

Crossing over

Correct answer:

All of these

Explanation:

All three contribute to giving rise to genetic variation. Independent assortment allows for the chromosomes to assort in millions of random of combinations during fertilization. Crossing over between chromosomes produces recombinant chromosomes, or the combination of chromosomal DNA from two parents into one chromosome. Random fertilization allows aids with variation because it means any sperm can fertilize any egg. It is sometimes easy to overlook, but humans do not mate randomly. Lots of energy is put into choosing an optimal mate with whom to reproduce. 

Example Question #3 : Inheritance Patterns

Each person can have one of four possible blood types: A, B, AB, or O. Blood type A means that the "A" antigen is present on the red blood cell surface. Blood type B means that the "B" antigen is present on red blood cells' surfaces. Blood type "AB" means that both the "A" antigen and the "B" antigen are present on the red blood cells' surfaces. Blood type "O" means that no antigens are present on the red blood cells' surfaces.

A mother with blood type A and the genotype "A/O" and a father with blood type B and genotype "B/B" have a child with blood type AB. This is an example of what type of inheritance pattern?

Possible Answers:

Autosomal recessive

Incomplete dominance

Codominance

Complete autosomal dominance

X-linked autosomal dominance

Correct answer:

Codominance

Explanation:

The child is blood type AB, meaning that the child has both the "A" antigen and the "B" antigen on his or her red blood cells. The child is able to express the products of both genes simultaneously. The A antigen was inherited from mom, and the B antigen was inherited from dad. The "A" and "B" alleles are codominant because they can both be expressed in the same person at the same time if the person inherits both alleles, as is the case in this example.

Example Question #4 : Inheritance Patterns

Each person can have one of four possible blood types: A, B, AB, or O. Blood type A means that the "A" antigen is present on the red blood cell surface. Blood type B means that the "B" antigen is present on red blood cells' surfaces. Blood type "AB" means that both the "A" antigen and the "B" antigen are present on the red blood cells' surfaces. Blood type "O" means that no antigens are present on the red blood cells' surfaces.

Someone with genotype "A/A" or "A/O" will have type A blood. Someone with genotype "B/B" or "B/O" will have type B blood. Someone with genotype "A/B" will have AB blood, and someone with genotype "O/O" will have type O blood.

Assuming that blood type is not a sex-linked trait, what is the probability that a mother with genotype "A/O" and a father with genotype "A/B" will have a child with type B blood?

Possible Answers:

Correct answer:

Explanation:

The easiest way to solve this problem is to draw a punnet square. The genotypes of the parents are "AO" and "AB". The potential genotypes of their children are "AA", "AO", "BA", and "BO". Only genotype "BO" will produce type B blood. "BO" is one out of four results of this punnet square, so the probability of this outcome is .

Example Question #5 : Inheritance Patterns

In a certain flower, a blue petal phenotype is dominant to a white petal phenotype. If you cross a heterozygous flower with a homozygous recessive flower, what is the probability of inheritance for the white petal phenotype?

Possible Answers:

Correct answer:

Explanation:

Dominant alleles are referred to with capital letters, so let's call the dominant blue-petal allele B. Recessive alleles are referred to using lower case letters, so we will call the recessive white-petal allele b.

A heterozygous organism has one dominant and one recessive allele, so the heterozygous flower has one B allele and one b allele. Its genotype is Bb. Because B is dominant to b, its phenotype (the trait produced by its genotype) is blue petals.

A homozygous organism has two of the same allele. The homozygous flower will either have two BB alleles or two bb alleles. The question states that the flower with white petals is homozygous recessive, so its genotype is bb and its phenotype is white petals. The only genotype that produces a white phenotype is bb, because you need two recessive alleles in order to express the recessive trait.

When you cross the two flowers, each parent donates one of its two alleles for petal color to the offspring. Accounting for every possible combination of alleles from each parent, there are four possible outcomes from a cross between Bb and bb: Bb, Bb, bb, and bb. (It may also help to draw a punnet square to visualize the four possible combinations). As you can see, these outcomes lead to two possible genotypes: Bb and bb.  The Bb genotype produces flowers with blue petals, and the bb genotype leads to flowers with white petals. Because two of the four possible outcomes are genotype bb, two of the four possible outcomes are for flowers with white petals. Two out of four is equal to , so  is the correct answer.

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