Genetics - SAT Subject Test in Biology

A phenotype is a physical trait that can be observed. Thus, eye color (and hair color) are great examples of phenotypes. A genotype describes the genetic traits of a person (e.g. whether that person has the autosomal or recessive alleles of a gene). The genotype of an individual determines his or her phenotype.

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.

The correct answer is "X-linked." Mammalian females have two X chromosomes, with recessive alleles often not apparent unless there are two copies. Mammalian males have only one X chromosome, so any recessive alleles on it will be expressed.

The phenotypic ratio is the ratio of one phenotype to another (phenotype is the trait expressed, in this case color, while genotype is the allele combination (BB, bb, Bb, or bB) that produces that phenotype. The capital letters BB signify that the blue allele (B) is dominant to the white allele (b). As such, the only genotype that will produce white plants is bb. All other combinations (BB, Bb, bB) will produce a blue plant. If you cross a homozygous (both dominant or both recessive) dominant plant with a homozygous recessive plant, the dominant allele will be present in all of the offspring, as every possible allele the blue plant could contribute will be dominant to every possible allele the white plant could contribute, making all of the offspring blue.

A red flower must have the "R" allele, which conveys the red phenotype. Since this red allele is dominant, the flower will be red as long as one copy of this allele is present. Thus, the genotypes "Rr" and "RR" will both produce red flowers. If no red allele is present ("rr"), the flower will be white.

If other dingos are supplementing the population, we know that migration is occurring. Migration cannot occur if a population is to satisfy the conditions of Hardy-Weinberg equilibrium.

The Hardy-Weinberg principle asserts that allele frequencies in a population remain constant; for example, if 50% of people have blue eyes, then each successive generation will continue to have 50% of people with blue eyes as long as certain assumptions are met. The Hardy-Weinberg equilibrium assumes that within a population random mating occurs, no migration occurs, no mutations occur, no natural selection occurs, and the population is sufficiently large. If any of these assumptions are not met, the allele frequencies of the population will change, causing the population to evolve. For this question, all of the answer choices except "Random mutations" (remember there must be no mutations) are required assumptions for the Hardy-Weinberg equilibrium to be in effect.

Haploid cells contain one of each numbered chromosome, for a chromosome number of "n." Diploid cells contain two of each, so the number of chromosomes in a diploid cell is 2n, or twice the haploid number.

This question is essentially asking, "Which of the following is a point mutation of the original sequence?" A point mutation is a mutation in which one base pair of DNA is substituted for another. A point mutation will not change the reading frame of a DNA sequence because it does not involve adding or removing any bases. The only answer choice that is identical to the original sequence with the exception of one base pair being changed for another is 5'GATCACG3', so this is the correct answer.

The correct answer is the "S," or synthesis, phase. The cell cycle consists of six main parts: interphase, prophase, metaphase, anaphase, telophase, and cytokinesis. Interphase, the longest part of the cell cycle, is subdivided into three different phases: G1, S, and G2. G1 comes first; in this phase, the cell simply grows. In the S phase, the cell synthesizes (creates) a new set of DNA using its original genome as a template. In G2, the cell (which now contains two copies of DNA) continues to grow. These growth phases are an important part of interphase because the cell must grow enough to have enough cytoplasm to give to each of its daughter cells at the end of mitosis or meiosis.

Haploid cells contain only one of each numbered chromosome, whereas diploid cells contain two of each. Therefore, a haploid cell in a given organism would contain half as many chromosomes as would be in a diploid cell of the same organism.

A gamete will have n=23 chromosomes while a somatic cell is going to have a diploid number of cells, or 2n=46. This is the case because somatic cells are going to receive a copy of chromosomes from each parent, so 23 from the mother and 23 form the father to give 46 total chromosomes. If you are ever given the haploid number and asked for the diploid number, then you can simply double the haploid number.

In DNA, the four nucleic acid basis are adenine, guanine, cytosine, and thymine. RNA has uracil instead of thymine. Adenine pairs with thymine in DNA and uracil in RNA. Cytosine and Guanine pair together in both. Deoxyribose and ribose are the 5-carbon sugars in DNA and RNA nucleotides, respectively.

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". Children with genotypes "AA" and "AO" will have type A blood. Children with genotype "BO" will have type B blood. Children with genotype "BA" will have type AB blood. Genotype "O/O" is the only one that will result in type O blood. "O/O" is not a possible product of this punnet square. Both the mother and the father must have the "O" allele in order for a child to have genotype "O/O" and type O blood.

In RNA, the nucleic acid base adenine pairs with uracil. (In DNA, it pairs with thymine). In both RNA and DNA, cytosine pairs with guanine. Accordingly, RNA will have the same percent uracil as it does adenine—in this case, .

(In this case it will also contain guanine and percent cytosine, because the adenine and uracil together add up to and the remaining is divided evenly between the paired guanine and cytosine (and divided by two is ).

Gametes are germ cells used in sexual reproduction, such as eggs or sperm. They are haploid cells with a chromosome number of , while somatic (non-reproductive) cells are diploid with a chromosome number of . So, if an organism's gametes have chromosomes each, a somatic cell in that organism will have twice that number of chromosomes—in this case, .

As the disorder is X-linked and the subject is male, he only received an X-chromosome from his mother. Therefore, the allele for the disorder must have been inherited from his mother. However, this does not mean that the mother expressed the disorder herself, as she could have the dominant allele in addition to one recessive allele.

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.

Since the offspring receives one allele from each parent, crossing a purebred dominant organism with a purebred recessive organism (PPQQ x ppqq) will always result in a hybridized offspring (PpQq). Thus, Flower 1's genotype is PpQq. Also remember that a recessive phenotype always indicates double recessive alleles for that trait. And so, crossing Flower 1 with a white, short-stemmed flower will result in the cross PpQq x ppqq. The next step is to draw a 4x4 Punnett square, as seen in the diagram. The desired genotype for this question is ppqq (recessive phenotype), and from the Punnett square you will be able to see that 4/16 of the squares will carry this specific genotype.

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.