Question

Given the fact that 1 fg of DNA = 9.78 × 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?

Short answer

(a) 12 Mb (12´10⁶ base pairs).

(b)2.0´10⁵ base pairs per minute.

Step-by-step solution

Haploid and diploid organisms

DNA is the genetic material that contains instructions for the genotype and phenotype of an organism. Organisms with a complete set of paired chromosomes, one inherited from each parent, are called diploid organisms.

On the other hand,organisms with half the set of paired chromosomes are called haploid organisms.

Calculation of (a)

Given,\({\rm{1fg of DNA = 9}}{\rm{.78 \times 1}}{{\rm{0}}^{\rm{5}}}{\rm{ bp}}\). Let us assume then,

\(\begin{aligned}{l}{\rm{1fg of DNA = 10 \times 1}}{{\rm{0}}^{\rm{5}}}{\rm{ bps}}\\{\rm{1fg of DNA = 1 }} \times {\rm{ 1}}{{\rm{0}}^6}\,{\rm{bps}}\end{aligned}\)

The haploid genome of yeast has 12 chromosome numbers.Thus, the genome in base pairs is calculated as:

\(\begin{aligned}{l}{\rm{12 fgs of DNA = 1 }} \times {\rm{ 1}}{{\rm{0}}^6}\,{\rm{b}}\,{\rm{ps }} \times \,{\rm{12}}\\{\rm{12 fgs of DNA}}\, = \,12\, \times \,{\rm{1}}{{\rm{0}}^6}\,{\rm{b}}\,{\rm{ps}}\end{aligned}\)

The haploid yeast genome is 12 Mbp.

Calculation of (b)

The graph shows that the S phase occurred from the 1-hour mark to the 3-hour mark. Thus,the amount of genetic material doubled during this time is the difference between the DNA amount at the 3-hour and 1-hour mark.

According to the data, DNA amount doubled is 2 hours,\({\rm{47 fg}}\,{\rm{ - }}\,{\rm{24 fg = 23 fg}}\)

Thus, the rate of fg synthesized per minute is calculated by dividing DNA content by 120 minutes:

\(\frac{{{\rm{23 fg}}}}{{{\rm{120}}}}{\rm{ = }}\,{\rm{0}}{\rm{.192}}\,{\rm{fg/min}}\)

Now, the genetic contentfrom fg/min to base pairs/min is calculated as:

\({\rm{0}}{\rm{.192 fg/min\; \times \;}}\left( {{\rm{9}}{\rm{.78\; \times \;1}}{{\rm{0}}^{\rm{5}}}{\rm{\;base pairs/fg}}} \right){\rm{ = 187,80}}0\).

Thus, approximately 2 ×10⁵base pairs/min were synthesized during the S phase of yeast cells.