Structure of nucleic acids and replication of DNA — practice questions

1. Recall complementary base pairing: In DNA, guanine (G) always pairs with cytosine (C), and adenine (A) always pairs with thymine (T).
2. Calculate cytosine percentage: Since G pairs with C, if guanine (G) is 32%, then cytosine (C) must also be 32%.
   • G = 32%
   • C = 32%
3. Calculate total G+C percentage: Total percentage of G and C bases is 32% + 32% = 64%.
4. Calculate total A+T percentage: The remaining percentage must be A and T. Total bases are 100%, so A + T = 100% - 64% = 36%.
5. Calculate thymine percentage: Since A pairs with T and their combined percentage is 36%, then adenine (A) = 18% and thymine (T) = 18%.

Therefore, the percentage of thymine bases in this segment is 18%.

1. Identify the given information:
   • Number of base pairs = 4,500
   • Distance per base pair = 0.34 nm
2. Calculate the total length in nanometres (nm): The total length is the number of base pairs multiplied by the distance between each pair.
   • Formula: Total Length = Number of base pairs × Distance per base pair
   • Calculation: Total Length = 4,500 × 0.34 nm = 1530 nm
3. Convert the length from nanometres (nm) to micrometres (µm): We know that 1 µm = 1000 nm. To convert from nm to µm, we divide by 1000.
   • Formula: Length in µm = Length in nm / 1000
   • Calculation: Length in µm = 1530 / 1000 = 1.53 µm

Therefore, the total length of the DNA molecule is 1.53 µm.