Chapter 21, Problem 21.59E

Interpretation Introduction

(a)

Interpretation:

The time required by one DNA polymerase enzyme to replicate the genetic material in one cell is to be stated.

Concept introduction:

The biochemical process in which a DNA molecule produces two exact copies of itself is known as the process of DNA replication. During the process of DNA replication, the enzyme DNA helicase unwinds the double helical structure of original DNA molecule.

Answer to Problem 21.59E

The time required by one DNA polymerase enzyme to replicate the genetic material in one cell is $5.70\text{years}$.

Explanation of Solution

It is given that the single human cell consists of $3\times {10}^9$ nucleotides.

It is given that $1000$ nucleotides are added per minute.

The conversion of $\min$ to $\text{year}$ is done as shown below.

$1\min =1.902\times {10}^{-6}\text{year}$

Therefore, the time is $1.902\times {10}^{-6}\text{year}$.

The time is calculated by the following formula.

$\text{Time}\left(\text{in}\text{years}\right)=\left(\frac{\text{Number}\text{of}\text{nucleotide}\text{in}\text{single}\text{cell}\times \text{Time}\left(\text{years}\right)}{\text{Nucleotides}\text{added}\left(\text{per}\text{minute}\right)}\right)$ …(1)

Substitute the value of number of nucleotides in a single human cell and the activity of enzyme in equation (1).

$\begin{array}{rll}\text{Time}\left(\text{in}\text{years}\right)& =& \left(\frac{\text{Number}\text{of}\text{nucleotide}\text{in}\text{single}\text{cell}\times \text{Time}\left(\text{years}\right)}{\text{Nucleotides}\text{added}\left(\text{per}\text{minute}\right)}\right)\\ \text{Time}\left(\text{in}\text{years}\right)& =& \left(\frac{3\times {10}^9\times 1.902\times {10}^{-6}\text{year}}{1000}\right)\\ & =& 5.70\text{years}\end{array}$

Therefore, $5.70\text{years}$ are required by one DNA polymerase enzyme to replicate the genetic material in one cell.

Conclusion

The time required by one DNA polymerase enzyme to replicate the genetic material in one cell is $5.70\text{years}$.

Interpretation Introduction

(b)

Interpretation:

The number of DNA polymerase molecules needed to replicate a cell’s genetic material in $10\text{minutes}$ is to be stated.

Concept introduction:

The biochemical process in which a DNA molecule produces two exact copies of itself is known as the process of DNA replication. During the process of DNA replication, the enzyme DNA helicase unwinds the double helical structure of original DNA molecule.

Answer to Problem 21.59E

The number of DNA polymerase molecules needed to replicate a cell’s genetic material in $10\text{minutes}$ is $3\times {10}^5$.

Explanation of Solution

It is given that $1000$ nucleotides are added per minute.

The time is given as $10\text{minutes}$.

Therefore, one enzyme can add $10\times 1000$ nucleotides in $10\text{minutes}$.

It is given that the single human cell consists of $3\times {10}^9$ nucleotides.

The number of DNA polymerase molecules needed to replicate a cell’s genetic material is calculated by the following formula.

$\text{Number}\text{of}\text{DNA}\text{polymerase}=\frac{\text{Number}\text{of}\text{nucleotide}\text{in}\text{single}\text{cell}}{\text{Nucleotides}\text{added}\text{in}\text{given}\text{time}}$ …(2)

Substitute the value of nucleotides in single cell and nucleotides added in $\text{10}\text{minutes}$ in equation (2).

$\begin{array}{rll}\text{Number}\text{of}\text{DNA}\text{polymerase}& =& \frac{\text{Number}\text{of}\text{nucleotide}\text{in}\text{single}\text{cell}}{\text{Nucleotides}\text{added}\text{in}\text{given}\text{time}}\\ \text{Number}\text{of}\text{DNA}\text{polymerase}& =& \frac{3\times {10}^9}{10\times 1000}\\ & =& \frac{3\times {10}^9}{{10}^4}\\ & =& 3\times {10}^5\end{array}$

Therefore, $3\times {10}^5$ DNA polymerase molecules are needed to replicate a cell’s genetic material in $10\text{minutes}$

Conclusion

The number of DNA polymerase molecules needed to replicate a cell’s genetic material in $10\text{minutes}$ is $3\times {10}^5$.