Question

Bacteria and other microbes can be used to clean up" an oil spill by breaking down oil into carbon dioxide and water. Two samples isolated from the Deepwater Horizon leak in the Gulf of Mexico were labeled A and B. The DNA of each was isolated and the percent thymine measured in each sample. Sample A contains 19.5% thymine and sample B contains 30.1 % thymine. Assume the organisms contain normal double-stranded DNA and predict the composition of the other bases. percent adenine in sample A: percent guanine in sample A: percent cytosine in sample A: percent adenine in sample B: percent guanine in sample B: percent cytosine in sample B: Both samples are then denatured to remove the secondary structure. Which will have the higher temperature to denature? OB OA

Answer 1

Sample A

It is given that sample A contains 19.5% thymine. In DNA, thymine and adenine form pairs. So, percentage of adenine in sample A  will be equal to percentage of thymine in sample A .

Percentage of adenine in sample A = 19.5%

Percentage of adenine and thymine in sample A =  19.5%+ 19.5%

Percentage of adenine and thymine in sample A = 39.0%

DNA contains four types of bases. They are adenine, thymine,guanine and cytosine. We have calculated that percentage of adenine and thymine in sample A will be 39.0%. Percentage of remaining bases( guanine and cytosine) in sample A can be calculated by subtracting  percentage of adenine and thymine in sample A from 100.

So, percentage of guanine and cytosine in sample A = 100 - percentage of adenine and thymine in sample A

Percentage of guanine and cytosine in sample A = 100 - 39.0%

Percentage of guanine and cytosine in sample A = 61 %

In DNA, guanine pairs with cytosine. So, percentage of guanine can be calculated as follows.

Percentageof guanineandcytosinein sample Percentageof guaninein sample A =

%

Percentageof guaninein sample A = * = 30.5

Percentage of guanine in sample A = 30.5 %

Percentage of cytosine in sample A= 30.5 %

Sample B

It is given that sample B contains 30.1% thymine.

So, percentage of adenine in sample B = 30.1%

Percentage of adenine and thymine in sample B = 30.1% + 30.1%

Percentage of adenine and thymine in sample B = 60.2 %

Percentage of guanine and cytosine in sample B = 100 - percentage of adenine and thymine in sample B

Percentage of guanine and cytosine in sample B = 100 - 60.2 %

Percentage of guanine and cytosine in sample B = 39.8 %

Percentageof guanineandcytosinein sample Percentageof guaninein sample B =

%

Percentageofguaninein sampleB = 39.8 = 19.9

Percentage of guanine in sample B = 19.9 %

Percentage of cytosine in sample B = 19.9 %

If both samples are denatured to remove the secondary structure, sample A will have the higher temperature to denature because percentage of guanine and cytosine is higher in sample A and they are paired by triple hydrogen bond. On the other hand, adenine and thymine are paired by double hydrogen bond.