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Chapter 13: 13-5TYU (page 268)

If we continue to follow the cell lineage from question 4, then the DNA content of a single cell at metaphase of meiosis II will be

  1. 0.25x
  2. 0.5x
  3. x
  4. 2x

Short Answer

Expert verified
  1. The option鈥�0.25x鈥� is false.
  2. The option鈥�0.5x鈥� is false.
  3. The option鈥渪鈥� is true.
  4. The option鈥�2x鈥� is false.

Step by step solution

01

Description of DNA

DNA is the genetic material present in the cell. The DNA gets replicated because the copies of DNA get infused into the daughter cells. It is produced via cell division.

02

Explanation of option “(A)”

The number of DNA does not get reduced by one-fourth. The combining of chromosomes does not result in a 0.25x number of DNA.

The number of DNA does not get reduced four times.

Therefore, the given option is false.

03

Explanation of option “(B)”

The cell lineage is considered as x from question 4. After meiosis I, the number of DNA gets reduced due to the chromosomes combining in pairs.

The DNA count that takes place does not reduce by half or 0.5x from 2x.

Therefore, the given option is false.

04

Explanation of option “(C)”

The metaphase of meiosis with the DNA content of the cell is 2x. This cell lineage follows in the metaphase of meiosis II. In the process of metaphase II, the sister chromatids are joined together. In a single cell, the DNA value tends to decrease from 2x to x.

Thus the cell with the DNA content as x is formed at the end of metaphase of meiosis II.

Therefore, the given option is true.

05

Explanation of option “(D)”

The sister chromatid joining in the metaphase of meiosis II causes a decrease in the DNA count of the single cell.

The value of 2x will not remain the same.

Therefore, the given option is false.

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Most popular questions from this chapter

A horticulturalist breeds orchids, trying to obtain a plant with a unique combination of desirable traits. After many years, she finally succeeds. To produce more plants like this one, she should crossbreed it with another plant or clone it. Why?

Because the variable on the x-axis varies continuously, it makes sense to plot the data on a line graph. (a) Plot each data point from the table onto the graph. (b) Connect the data points with line segments.

Given the fact that 1 fg of DNA = 9.78 脳 105 base pairs (on average), you can convert the amount of DNA per cell to the length of DNA in numbers of base pairs. (a) Calculate the number of base pairs of DNA in the haploid yeast genome. Express your answer in millions of base pairs (Mb), a standard unit for expressing genome size. Show your work. (b) How many base pairs per minute were synthesized during the S phase of these yeast cells?

What is the original source of variation among the different alleles of a gene?

Most of the yeast cells in the culture were in G1 of the cell cycle before being moved to the nutrient-poor medium. (a) How many femtograms of DNA are there in each yeast cell in G1? Estimate this value from the data in your graph. (b) How many femtograms of DNA should be present in each cell in G2? (See Concept 12.2 and Figure 12.6.) At the end of meiosis I (MI)? At the end of meiosis II (MII)? (See Figure 13.7.) (c) Using these values as a guideline, distinguish the different phases by inserting vertical dashed lines in the graph between phases and label each phase (G1, S, G2, MI, MII). You can figure out where to put the dividing lines based on what you know about the DNA content of each phase (see Figure 13.7). (d) Think carefully about the point where the line at the highest value begins to slope downward. What specific point of meiosis does this 鈥渃orner鈥� represent? What stage(s) correspond to the downward sloping line?
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