What this deck covers
This deck focuses on Describe Dna Structure And Components, giving you a quick way to review the definitions, rules, and examples that matter most for Biology.
Study Describe Dna Structure And Components in Biology with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.
0% Complete
What does the abbreviation DNA stand for in biology?
Tap card or press Space to flip
Deoxyribonucleic acid. The full name of the molecule that stores genetic information.
How well did you know it?
Card 1 / 100
Space to flip · ← / → to move · once flipped, → Got it · ← Still learning
This deck focuses on Describe Dna Structure And Components, giving you a quick way to review the definitions, rules, and examples that matter most for Biology.
Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.
Answer: Deoxyribonucleic acid. The full name of the molecule that stores genetic information.
Answer: Deoxyribose. The five-carbon sugar component of DNA nucleotides.
Answer: G-C-C-T. Apply base-pairing rules: C pairs with G, A pairs with T.
Answer: 40%. Base-pairing rules mean C = G, so both are 40%.
Answer: Hydrogen bonds. Weak bonds that hold complementary base pairs together.
Answer: Nucleotide. The building block that polymerizes to form the DNA chain.
Answer: The amount of C equals the amount of G. Chargaff's rule: C always equals G in double-stranded DNA.
Answer: G-C-C-T. Apply base-pairing rules: C pairs with G, A pairs with T.
Answer: A–T. A-T forms two hydrogen bonds for weaker pairing.
Answer: Double helix. Two antiparallel strands twisted into this characteristic spiral structure.
Answer: Deoxyribose. The five-carbon sugar component of DNA nucleotides.
Answer: A chain of nucleotides linked by phosphodiester bonds. Multiple nucleotides joined together form a DNA strand.
Answer: 5′ end and 3′ end. Named for the carbon atoms in the deoxyribose sugar ring.
Answer: Purine-pyrimidine. This pairing maintains uniform helix diameter and stability.
Answer: Phosphodiester bond. Covalent bond connecting nucleotides to form the DNA backbone.
Answer: 30%. Base-pairing rules mean A = T, so both are 30%.
Answer: A pairs with T; C pairs with G. Watson-Crick base pairing ensures complementary strand formation.
Answer: Cytosine and thymine. Single-ring structures that are smaller than purines.
Answer: The amount of A equals the amount of T. Chargaff's rule: A always equals T in double-stranded DNA.
Answer: Thymine. DNA contains thymine while RNA contains uracil instead.
Answer: Phosphate group, deoxyribose sugar, nitrogenous base. Three essential parts that together form each DNA building block.
Answer: A chain of nucleotides linked by phosphodiester bonds. Multiple nucleotides joined together form a DNA strand.
Answer: Purine-pyrimidine pairing maintains uniform helix width. Large purine paired with small pyrimidine maintains constant width.
Answer: 3 hydrogen bonds. G-C pairs have more hydrogen bonds than A-T pairs.
Answer: On the interior, paired across the two strands. Bases stack in the center and pair across the double helix.
Answer: Opposite directions (antiparallel). Antiparallel orientation is essential for proper base pairing.
Answer: The amount of A equals the amount of T. Chargaff's rule: A always equals T in double-stranded DNA.
Answer: Base sequence (nucleotide sequence). The linear arrangement of bases that encodes genetic information.
Answer: Double helix. Two antiparallel strands twisted into this characteristic spiral structure.
Answer: 18%. Base-pairing rules mean C = G, so both are 18%.
Answer: Deoxyribose has H; ribose has OH at the 2′ carbon. DNA's sugar lacks the hydroxyl group that RNA's sugar has.
Answer: Hydrogen bonding between bases. Weak bonds allow strand separation during DNA processes.
Answer: Phosphodiester linkage. Strong covalent bonds provide structural stability to DNA.
Answer: 3 hydrogen bonds. G-C pairs have more hydrogen bonds than A-T pairs.
Answer: Glycosidic bond. Links the base to the 1′ carbon of the sugar.
Answer: C–G. C-G forms three hydrogen bonds for stronger pairing.
Answer: Adenine (A). T forms two hydrogen bonds with A in Watson-Crick pairing.
Answer: Thymine (T). A purine pairs with a pyrimidine following base-pairing rules.
Answer: Cytosine (C). C forms three hydrogen bonds with G in Watson-Crick pairing.
Answer: Antiparallel. Strands run in opposite directions for proper base pairing.
Answer: The 3′ end. This end has an unattached OH group on the sugar.
Answer: 38%. T = A = 12%, so C = G = (100%−24%)/2=38%.
Answer: Adenine and guanine. Double-ring structures that are larger than pyrimidines.
Answer: The amount of C equals the amount of G. Chargaff's rule: C always equals G in double-stranded DNA.
Answer: The base sequence. Variable base sequence carries genetic information, not backbone.
Answer: The 5′ end. This end has an unattached phosphate group on the sugar.
Answer: The base sequence. Variable base sequence carries genetic information, not backbone.
Answer: Sugar-phosphate backbone. Forms the structural framework on the outside of the helix.
Answer: Adenine, thymine, cytosine, guanine. The four nitrogen-containing bases that encode genetic information.
Answer: Each base sequence determines the other by pairing rules. Base-pairing rules ensure one strand determines the other's sequence.
Answer: A pairs with T; C pairs with G. Watson-Crick base pairing ensures complementary strand formation.
Answer: 18%. Base-pairing rules mean C = G, so both are 18%.
Answer: Nucleoside. Base attached to sugar without the phosphate group.
Answer: Base sequence (nucleotide sequence). The linear arrangement of bases that encodes genetic information.
Answer: Phosphate group, deoxyribose sugar, nitrogenous base. Three essential parts that together form each DNA building block.
Answer: Adenine and guanine. Double-ring structures that are larger than pyrimidines.
Answer: 2 hydrogen bonds. A-T pairs have fewer hydrogen bonds than G-C pairs.
Answer: Between the 3′ OH of one sugar and the 5′ phosphate of the next. Links the 3′ OH of one sugar to the 5′ phosphate of another.
Answer: Glycosidic bond. Links the base to the 1′ carbon of the sugar.
Answer: The 5′ end. This end has an unattached phosphate group on the sugar.
Answer: Guanine (G). G forms three hydrogen bonds with C in Watson-Crick pairing.
Answer: Thymine. DNA contains thymine while RNA contains uracil instead.
Answer: A–T. A-T forms two hydrogen bonds for weaker pairing.
Answer: T-G-C. Apply base-pairing rules: A pairs with T, C pairs with G.
Answer: Adenine (A). T forms two hydrogen bonds with A in Watson-Crick pairing.
Answer: Antiparallel. Strands run in opposite directions for proper base pairing.
Answer: 40%. Base-pairing rules mean C = G, so both are 40%.
Answer: 2 hydrogen bonds. A-T pairs have fewer hydrogen bonds than G-C pairs.
Answer: 30%. Base-pairing rules mean A = T, so both are 30%.
Answer: Nucleoside. Base attached to sugar without the phosphate group.
Answer: A-A-T-G. Apply base-pairing rules: T pairs with A, C pairs with G.
Answer: Opposite directions (antiparallel). Antiparallel orientation is essential for proper base pairing.
Answer: 38%. T = A = 12%, so C = G = (100%−24%)/2=38%.
Answer: One strand runs 5′→3′ and the other runs 3′→5′. Strands have opposite orientations for proper base pairing geometry.
Answer: On the interior, paired across the two strands. Bases stack in the center and pair across the double helix.
Answer: T-G-C. Apply base-pairing rules: A pairs with T, C pairs with G.
Answer: One strand runs 5′→3′ and the other runs 3′→5′. Strands have opposite orientations for proper base pairing geometry.
Answer: Between the 3′ OH of one sugar and the 5′ phosphate of the next. Links the 3′ OH of one sugar to the 5′ phosphate of another.
Answer: A-A-T-G. Apply base-pairing rules: T pairs with A, C pairs with G.
Answer: Deoxyribonucleic acid. The full name of the molecule that stores genetic information.
Answer: Thymine (T). A purine pairs with a pyrimidine following base-pairing rules.
Answer: The 3′ end. This end has an unattached OH group on the sugar.
Answer: 5′ end and 3′ end. Named for the carbon atoms in the deoxyribose sugar ring.
Answer: Deoxyribose has H; ribose has OH at the 2′ carbon. DNA's sugar lacks the hydroxyl group that RNA's sugar has.
Answer: Cytosine and thymine. Single-ring structures that are smaller than purines.
Answer: Purine-pyrimidine pairing maintains uniform helix width. Large purine paired with small pyrimidine maintains constant width.
Answer: Adenine, thymine, cytosine, guanine. The four nitrogen-containing bases that encode genetic information.
Answer: Guanine (G). G forms three hydrogen bonds with C in Watson-Crick pairing.
Answer: 30%. A = T = 20%, so C = G = (100%−40%)/2=30%.
Answer: Hydrogen bonds. Weak bonds that hold complementary base pairs together.
Answer: Cytosine (C). C forms three hydrogen bonds with G in Watson-Crick pairing.
Answer: Each base sequence determines the other by pairing rules. Base-pairing rules ensure one strand determines the other's sequence.
Answer: Purine-pyrimidine. This pairing maintains uniform helix diameter and stability.
Answer: C–G. C-G forms three hydrogen bonds for stronger pairing.
Answer: Hydrogen bonding between bases. Weak bonds allow strand separation during DNA processes.
Answer: Phosphodiester bond. Covalent bond connecting nucleotides to form the DNA backbone.
Answer: Nucleotide. The building block that polymerizes to form the DNA chain.
Answer: Sugar-phosphate backbone. Forms the structural framework on the outside of the helix.
Answer: Phosphodiester linkage. Strong covalent bonds provide structural stability to DNA.
Answer: 30%. A = T = 20%, so C = G = (100%−40%)/2=30%.