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Chapter 14: Problem 60

Explain how the components of DNA fit together. a. DNA is composed of nucleotides, consisting of a 5 carbon sugar, a phosphate, and a nitrogenous base. DNA is a double helical structure in which complementary base pairing occurs. Adenine pairs with thymine and guanine pairs with cytosine. Adenine and thymine form two hydrogen bonds and cytosine and guanine form three hydrogen bonds. The two individual strands of DNA are held together by covalent bonds between the phosphate of one nucleotide and sugar of the next. The two strands run anti parallel to each other. b. DNA is composed of nucleotides, consisting of a 5 carbon sugar, a phosphate, and a nitrogenous base. DNA is a double helical structure in which complementary base pairing occurs. Adenine pairs with cytosine and guanine pairs with thymine. Adenine and cytosine form two hydrogen bonds and guanine and thymine form three hydrogen bonds. The two individual strands of DNA are held together by covalent bonds between the phosphate of one nucleotide and sugar of the next. The two strands run anti parallel to each other. c. DNA is composed of nucleotides, consisting of a 5 carbon sugar, a phosphate, and a nitrogenous base. DNA is a double helical structure in which complementary base pairing occurs. Adenine pairs with cytosine and guanine pairs with thymine. Adenine and cytosine form three hydrogen bonds and guanine and thymine form two hydrogen bonds. The two individual strands of DNA are held together by covalent bonds between the phosphate of one nucleotide and sugar of the next. The two strands run antiparallel to each other. d. DNA is composed of nucleotides, consisting of a 5 carbon sugar, a phosphate, and a nitrogenous base. DNA is a double helical structure in which complementary base pairing occurs. Adenine pairs with cytosine and guanine pairs with thymine. Adenine and cytosine form three hydrogen bonds and guanine and thymine form two hydrogen bonds. The two individual strands of DNA are held together by covalent bonds between the phosphate of one nucleotide and sugar of the next. The two strands run parallel to each other.

Short Answer

Expert verified
Option (a) is correct. Adenine pairs with Thymine (2 hydrogen bonds), Guanine pairs with Cytosine (3 hydrogen bonds), and the strands are antiparallel.

Step by step solution

01

- Identify the Correct Base Pairing

Understand that in DNA, the correct complementary base pairing is Adenine (A) pairs with Thymine (T), and Guanine (G) pairs with Cytosine (C). This is according to Watson-Crick base pairing rules.
02

- Verify Hydrogen Bonds

Confirm that Adenine (A) and Thymine (T) form two hydrogen bonds, and Guanine (G) and Cytosine (C) form three hydrogen bonds.
03

- Analyze DNA Strand Orientation

Check if the DNA strands run antiparallel (opposite directions). DNA strands are antiparallel, meaning one strand runs 5' to 3' and the other runs 3' to 5'.
04

- Identify Incorrect Statements

Analyze each option. Options (b), (c), and (d) suggest incorrect base pairings of Adenine with Cytosine and Guanine with Thymine, and/or incorrect hydrogen bond counts, and/or incorrect strand orientation (parallel in (d)).
05

- Choose the Correct Answer

Based on the steps above, option (a) is the only one that correctly describes the components of DNA with proper base pairing, hydrogen bonds, and antiparallel orientation of DNA strands.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

nucleotides
Nucleotides are the fundamental building blocks of DNA. Each nucleotide consists of three components: a 5-carbon sugar (deoxyribose in DNA), a phosphate group, and a nitrogenous base. There are four types of nitrogenous bases in DNA: Adenine (A), Thymine (T), Guanine (G), and Cytosine (C). The backbone of the DNA strand is formed by alternating sugar and phosphate groups, with the nitrogenous bases extending from the sugar. Each nucleotide is linked to the next by a covalent bond, creating the stable structure of the DNA strand.
complementary base pairing
Complementary base pairing is crucial for the structure and function of DNA. In DNA, the bases pair in a specific manner: Adenine (A) pairs with Thymine (T) through two hydrogen bonds, and Guanine (G) pairs with Cytosine (C) through three hydrogen bonds. This specific pairing is known as Watson-Crick base pairing, and it ensures that the DNA strands are complementary to each other. Each base pair is a rung on the DNA ladder, contributing to the precise replication and transmission of genetic information.
antiparallel strands
The two strands of the DNA double helix are antiparallel, meaning they run in opposite directions. One strand runs from the 5' end to the 3' end, while the other runs from the 3' end to the 5' end. This orientation is essential for the complementary base pairing and the function of DNA during replication and transcription. The antiparallel structure allows the enzymes involved in these processes to properly read and interpret the genetic code.
hydrogen bonds
Hydrogen bonds play a key role in the stability and integrity of the DNA double helix. These bonds form between the nitrogenous bases of the two DNA strands. Adenine (A) pairs with Thymine (T) via two hydrogen bonds, while Guanine (G) pairs with Cytosine (C) via three hydrogen bonds. These hydrogen bonds are strong enough to hold the two strands together but weak enough to allow the strands to separate during replication and transcription, making them both stable and flexible.
covalent bonds
Covalent bonds are essential for the construction of the DNA molecule. Within each DNA strand, covalent bonds connect the phosphate group of one nucleotide to the 3' carbon of the sugar of the next nucleotide. This creates a continuous sugar-phosphate backbone that provides structural stability to the DNA molecule. The covalent bonds are incredibly strong, ensuring the DNA strand does not break easily under normal physiological conditions.

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Most popular questions from this chapter

What type of body cell does not exhibit telomerase activity? a. adult stem cells b. embryonic cells c. germ cells d. liver cells

Discuss how mutations can increase variation within a population. a. Substitution mutations may cause a different amino acid to be placed at a specific location, causing small changes in the protein. Frame shift mutations usually cause multiple amino acid changes, increasing chances that a new protein will form, leading to radically different characteristics in the offspring. b. Substitution mutations may cause multiple amino acid changes, increasing chances that a new protein will form, leading to radically different characteristics in the offspring. Frame shift mutations may cause a different amino acid to be placed at a specific location, causing small changes in a protein. c. Substitution mutations may cause a different amino acid to be placed at a specific location, resulting in major changes to the protein and leading to radically different characteristics in the offspring. Frame shift mutations cause multiple amino acid differences in a protein, leading to small changes in the protein. d. Substitution mutations result in a different amino acid being placed at a specific position in a protein, causing small changes. Silent mutations could result in new characteristics possessed by an offspring when a stop codon is substituted for an amino acid.

Explain the events taking place at the replication fork. If the gene for helicase is mutated, what part of replication will be affected? a. Helicase separates the DNA strands at the origin of replication. Topoisomerase breaks and reforms DNA鈥檚 phosphate backbone ahead of the replication fork, thereby relieving the pressure. Single-stranded binding proteins prevent reforming of DNA. Primase synthesizes RNA primer which is used by DNA polymerase to form a daughter strand. If helicase is mutated, the DNA strands will not be separated at the beginning of replication. b. Helicase joins the DNA strands together at the origin of replication. Topoisomerase breaks and reforms DNA鈥檚 phosphate backbone after the replication fork, thereby relieving the pressure. Single-stranded binding proteins prevent reforming of DNA. Primase synthesizes RNA primer which is used by DNA polymerase to form a daughter strand. If helicase is mutated, the DNA strands will not be joined together at the beginning of replication. c. Helicase separates the DNA strands at the origin of replication. Topoisomerase breaks and reforms DNA鈥檚 sugar backbone ahead of the replication fork, thereby increasing the pressure. Single-stranded binding proteins prevent reforming of DNA. Primase synthesizes DNA primer which is used by DNA polymerase to form a daughter strand. If helicase is mutated, the DNA strands will be separated at the beginning of replication. d. Helicase separates the DNA strands at the origin of replication. Topoisomerase breaks and reforms DNA鈥檚 sugar backbone ahead of the replication fork, thereby relieving the pressure. Single-stranded binding proteins prevent reforming of DNA. Primase synthesizes DNA primer which is used by RNA polymerase to form a parent strand. If helicase is mutated, the DNA strands will be separated at the beginning of replication.

Discuss the significance of mutations in tRNA and rRNA. a. Mutations in tRNA and rRNA would lead to the production of defective proteins or no protein production. b. Mutations in tRNA and rRNA would lead to changes in the semi-conservative mode of replication of DNA. c. Mutations in tRNA and rRNA would lead to production of a DNA strand with a mutated single strand and normal other strand. d. Mutations in tRNA and rRNA would lead to skin cancer in patients of xeroderma pigmentosa

Compare and contrast the roles of DNA polymerase I and DNA ligase in DNA replication. a. DNA polymerase I removes the RNA primers from the developing copy of DNA. DNA ligase seals the ends of the new segment, especially the Okazaki fragments. b. DNA polymerase I adds the RNA primers to the already developing copy of DNA. DNA ligase separates the ends of the new segment, especially the Okazaki fragments. c. DNA polymerase I seals the ends of the new segment, especially the Okazaki fragments. DNA ligase removes the RNA primers from the developing copy of DNA. d. DNA polymerase I removes the enzyme primase from the developing copy of DNA. DNA ligase seals the ends of the old segment, especially the Okazaki fragments.
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