Mastering Cell Division: Mitosis and Meiosis Quiz

Mitosis is the process of cell division where a parent cell divides into two genetically identical daughter cells with the same number of chromosomes (2n) as the parent cell. This ensures the growth, development, and repair of multicellular organisms.

Metaphase is the stage of cell division where chromosomes line up in the middle of the cell. Region III is the only one where you would expect to find cells at metaphase.

When colchicine is added to cells in G2, the chromosomes would coil and shorten but have no spindle to which to attach, leading to disrupted cell division.

The major difference between mitosis and meiosis I lies in the separation of genetic material. In mitosis, sister chromatids separate, ensuring identical genetic information in daughter cells, whereas in meiosis I, homologous chromosomes separate, leading to genetic diversity in resulting cells.

In meiosis, by the end of meiosis I, the cell is haploid (has half the number of chromosomes). After meiosis II, the DNA amount is reduced further resulting in cells with half the amount of DNA compared to the cell that began meiosis.

The error in either anaphase I or II could lead to an egg with an extra chromosome (25) since these stages involve the separation of chromosomes during meiosis, potentially resulting in an extra chromosome in the resulting egg.

Genetic recombination specifically occurs during crossing over, a process where homologous chromosomes exchange genetic material. This results in genetic variation among offspring. The other options do not involve the direct recombination of genetic material.

In the given genotype Rrtt, the possible gamete options are rt (which is option D). This is because during gamete formation, the alleles segregate randomly, leading to the possibility of different combinations of alleles in the gametes.

The probability of plants with white axial flowers in the F2 generation can be determined using a Punnett square. Since the genes for flower color and location assort independently, the probability of a plant having white flowers (regardless of the location) is 1/4. Additionally, the probability of a plant having axial flowers is 1/2. Therefore, the combined probability of plants with white axial flowers is 1/4 x 1/2 = 1/8. Since there are two ways this combination can occur (white axial flowers with red terminal flowers or white terminal flowers with red axial flowers), we multiply the probability by 2, resulting in 1/8 x 2 = 1/4. Finally, the total probability of white axial flowers in the F2 generation is 1/4 x 1/4 = 1/16. Thus, the correct probability of plants with white axial flowers in the F2 generation is 3/16.

Pleiotropy is the ability of a single gene to have multiple effects on an organism's phenotype. This phenomenon can result in a wide range of seemingly unrelated traits being influenced by the same gene.

The probability of the third child having extra digits is 1/2 because the man has the dominant trait for extra digits, which means he has two copies of the gene responsible for extra digits. Since the second child already has extra digits, the man's extra digit gene must have been passed on. Therefore, the probability of the gene being passed on to the third child is 1/2.

In this scenario, the woman is a carrier of the recessive allele on one of her X chromosomes. Since color blindness is a recessive trait and she has one normal X chromosome, the probability of passing on the color-blindness allele to her son is 50%. The son will inherit one X chromosome from his mother and one Y chromosome from his father, making the probability of him being color blind 50%.

When a parent with type AB blood (genotype AB) marries a parent with type O blood (genotype OO), their children can only inherit the A antigen from the AB parent since the O parent does not have any antigens.

The correct answer is D as indicated by the presence of affected individuals in multiple generations, which is characteristic of an autosomal recessive trait.

Sexual reproduction results in many new gene combinations through genetic recombination during meiosis, leading to genetic diversity among offspring. This genetic diversity can result in some individuals having traits that give them a reproductive advantage, which is a key component of natural selection.

Transcriptional control, chromatin condensation control, and alternative splicing are all stages in gene expression that can affect the amount of final active gene product besides translational control. This method focuses only on mRNA levels, which may not accurately represent the actual levels of the gene product.

If cells primarily controlled gene expression at the post-translation level, it would require more energy as the processes involved in post-translational modification and regulation are generally more energy-intensive compared to transcriptional control.

RNA polymerase moves in the 3' -> 5' direction along the DNA template strand to synthesize RNA, following the complementary base pairing rules. This process ensures the formation of a complementary RNA strand to the template DNA strand.

The correct tRNA anticodon loop should be complementary to the mRNA codon. In this case, since the mRNA sequence is CCG-ACG, the tRNA anticodon loop sequence should be GGC for the complementary pairing. Therefore, option B) 3' GGC 5' is the correct answer.

The genetic code being the same across organisms means that the genetic information is universally transferable, allowing genes from one organism to potentially be expressed in another.

The correct sequence of nucleotides is as follows: 3' AAA-GAA-TAA-CAA 5'. Each nucleotide triplet codes for an amino acid, with AAA coding for phenylalanine (phe), GAA coding for leucine (leu), TAA coding for isoleucine (ile), and CAA coding for valine (val). Option B has a different sequence that does not correspond to the correct polypeptide sequence. Options C and D also have sequences that do not match the specified polypeptide sequence.

When plasmids are used to produce a desired protein, the desired gene is inserted into the plasmid outside of the bacterium

Taq polymerase is heat stable and can withstand the temperature changes of the cycler, making it ideal for PCR reactions.

In gel electrophoresis, altering the nucleotide sequence of the DNA fragment would not impact the rate at which it moves through the gel, as the movement of DNA fragments is primarily determined by size and charge, not nucleotide sequence.

DNA ligase is used to join or ligate DNA fragments together, not to cut DNA and create sticky ends.