Question

The human genome contains approximately \(10^{6}\) copies of an \(A l u\) sequence, one of the best-studied classes of short interspersed elements (SINEs), per haploid genome. Individual Alus share a 282 -nucleotide consensus sequence followed by a 3 '-adeninerich tail region. Given that there are approximately \(3 \times 10^{9}\) bp per human haploid genome, about how many base pairs are spaced between each Alu sequence?

Short answer

Answer: The average distance between each Alu sequence in the human genome is approximately 3000 base pairs.

Step-by-step solution

Write down the given information

The human genome has approximately \(3 \times 10^9\) bp per haploid genome and contains \(10^6\) copies of Alu sequences per haploid genome.

Calculate the distance between Alu sequences

To find the average distance between each Alu sequence, divide the total number of base pairs by the number of Alu sequences: \(\text{Average distance} = \frac{\text{Total base pairs}}{\text{Number of Alu sequences}}\)

Substitute the values and solve

Now, substitute the given values of base pairs and Alu sequences into the formula: \(\text{Average distance} = \frac{3 \times 10^9 \ \text{bp}}{10^6 \ \text{Alu sequences}}\) Divide the values: \(\text{Average distance} = \frac{3 \times 10^9}{10^6}= 3000 \ \text{bp}\)

Present the result

The average distance between each Alu sequence in the human genome is approximately 3000 base pairs.