Genetics - Genetics

The correct answer here is uracil. Remember that in DNA replication, the nucleic acids present are TCGA. When we switch over to RNA production, the thyamine is replaced by uracil to form UCGA. If you chose any of the other options, remember that each one has a pair (AT and CG) and in RNA it switches over to UA and CG. If you remember that, you will always recall that uracil replaces thyamine in RNA sequencing

Nucleotide excision repair corrects DNA that turns into pyrimidine dimers. These are usually caused by ultraviolet radiation and result in sizable DNA adducts.

The correct answer here is meiosis. Some of the options look challenging because you could assume RNA or DNA replication would result in gametes. However, the term gamete singularly refers to a sex-germ cell that is the direct result of meiosis. We can cancel out mitosis because that is specifically to replicate non sex related cells.

CDK levels remain relatively constant, but activity varies. Cyclin levels vary to which the CDKs must bind to become active. Also, the activity depends on cyclin levels, CDK inhibitors, and phosphorylation of CDKs.

Viral RNA polymerases do not have the same proof-reading ability as DNA polymerases. This is a contributing factor to the difficulty of making vaccines from RNA viruses.

Retroviruses do have a single-stranded RNA genome, but they use DNA intermediates in replication. Retroviruses utilize reverse transcriptase to convert viral RNA into complementary DNA, which is then copied to produce double-stranded viral DNA.

Telomeres are repetitive DNA sequences at the end of chromosomes and they shorten which each division. Poly-a-tails are added to RNA at the end of transcription. 5' caps are nucleotides added to mRNA aiding in translation. The kinetichore is a protein structure on chromatids that allow spindle fiber attachment.

Gregor Mendel was a German friar who lived in the mid- to late-nineteenth century. He conducted a years-long series of experiments using pea plants in which he carefully tracked each plant's physical traits and determined how cross-fertilization (breeding the pea plant to another pea plant, rather than to itself) affected the plants' offspring. Seed color, flower shape, height, and pod shape were among the traits he studied. In the end, he summarized his work into three laws of heredity: the laws of segregation, independent assortment, and dominance. He also coined the terms "recessive" and "dominant," in reference to genetic traits.

Unfortunately, Mendel's work remained largely unacknowledged through his life. The importance of his studies was only recognized by other scientists decades after his death.

The r-selection strategy for reproduction is typically seen in environments that are very volatile and unpredicatable. It has a variety of characteristics including high brood sizes with a high mortality rate, and fast maturation with very little parental assistance. Low brood sizes are typically seen in the k-selection strategy for reproduction.

G1 checkpoint acts as the restriction point where the cell commits to cell cycle entry. This phase needs favorable environment in order to function properly. The G2/M checkpoint involves chromosome alignment on spindle in metaphase. This process requires adequate size for mitosis entry. The S checkpoint includes DNA quality control looking for proper duplication.

Failure to meet the checkpoint requirements may lead to delays from DNA damage, improper nutrients, or cell size. it may also lead to exiting the cell cycle and apoptosis (cell death). Necrosis is unprogrammed cell death, usually caused by inflammation, disease, or lack of oxygen/blood.

When dealing with two traits, it helps to approach each trait separately. The question asks for the fraction of plants in the second generation that are heterozygous for both traits.

First, we will look at the first generation. The parents are both pure breeding, meaning they are homozygous for each trait. One is purple (dominant) and wrinkled (recessive), while the other is white (recessive) and round (dominant). The result will be dihybrid offspring.

Parent cross: PPrr x ppRR

Offspring: All offspring will be PpRr and exhibit both dominant phenotypes (purple and round).

Now we will look at the first generation self-cross for each trait.

First generation: PpRr x PpRr

Offspring for color: 1 PP, 2 Pp, 1 pp; 3 purple and 1 white.

Offspring for seeds: 1 RR, 2 Rr, 1 rr; 3 round and 1 wrinkled.

We can see that half of the offspring will be heterozygous for color, and half of the offspring will be heterozygous for seed shape.

Since we are looking for plants that have both of these traits, we multiply these two probabilities together.

Mutations can lead to lack of regulation, which overall, leads to genomic instability providing opportunities for uncontrolled growth — for example, cancer from the loss of cell cycle control. Proto oncogenes are normal genes that promote and regulate cell growth. Mutations to the proto oncogene itself could lead to oncogenes, which are cancer promoting.

Proto oncogenes promote proteins via proliferation with regulation. Oncogenes promote cell proliferation without regulation, leading to genomic instability, which can lead to cancer. A mutagen is a physical or chemical substance that can increase the frequency of mutations. A carcinogen is a substance that is directly involved in causing cancer.

p53 is a famous tumor suppressor, which blocks cell cycle progression preventing damaged/mutated DNA from being duplicated. Proto oncogenes are normal genes that regulate cell growth and proliferation. Oncogenes are mutated proto oncogenes that lead to unregulated cell proliferation. The checkpoints are in the S, G1, G2/M phases.

In Hardy-Weinberg equilibrium, the sum of the dominant allele frequency (p) and the recessive allele frequency (q) is equal to 1.

The question says that 49% of the population consists of mice with the homozygous dominant gene, therefore, the dominant genotype frequency is equal to 0.49.

The question asks us to find the frequency of the recessive allele (q). In order to find the frequency of the recessive allele, we must first find the frequency of the dominant allele (p). According to the Hardy-Weinberg principle, the square root of the homozygous genotype frequency is equal to the allele frequency.

The dominant allele frequency is 0.7. Using this, we can solve for the recessive allele frequency.

A transition mutation represents a purine-to-purine mutation or a pyrimidine-to-pyrimidine mutation. This could be a change of A to T, or vice versa. It could also be a change of C to G, or vice versa. Remember, a transition mutation does not insert or delete any bases. It simply changes a base.

We can use the Hardy-Weinberg equilibrium formulas to calculate the allele frequencies.

We know that the frequency of homozygous recessive (pink) rabbits is . This is equal to in the Hardy-Weinberg calculation. We can use this information to solve for , the recessive allele frequency.

Now that we know the value of , we can solve for the value of .

The frequency of heterozygotes is equal to in the Hardy-Weinberg calculation. Now that we know the frequency of each allele, we can complete this calculation.

Since colorblindness is a sex-linked gene, female children need to have the sex-linked gene from both their mother and father in order to be colorblind. In this case, the mother will donate a colorblinded gene to half of all offspring, while the dad will give his colorblind gene to all his female children. This means that one out of every two female children will have this trait.