12 - Structure and Function of DNA

The three-dimensional structure of DNA is shown in the interactive (image) Enter the complementary sequence to the DNA strand: 5′- GTACGATCA -3′

3′- CATGCTAGT -5′ Explanation: To determine the complementary sequence to the DNA strand 5′- GTACGATCA -3′, match each base in the original DNA strand with its complementary base. In DNA base pairing, adenine (A) pairs with thymine (T) and guanine (G) pairs with cytosine (C). The original DNA strand and its complementary sequence are shown. 5′- GTACGATCA -3′ 3′- CATGCTAGT -5′

The bacteria Streptococcus pneumoniae has a virulent S strain and a nonvirulent R strain. The S strain is lethal to mice. The S strain contains a chemical factor that can transform the R strain to be virulent. The diagram shows a series of experiments conducted by injecting combinations of these strains into mice to identify the transforming factor. A R strain (nonvirulent) B S strain (virulent) C Heat‑treated S strain D R strain and heat‑treated S strain E R strain and heat‑treated S strain with polysaccharides, lipids, RNA, and proteins destroyed F R strain and heat‑treated S strain with polysaccharides, lipids, RNA, proteins, and DNA destroyed If DNA is the transforming factor, match the expected results of each experiment by placing the appropriate mouse image. (dead mouse or alive mouse)

A - alive B - dead :( C - alive D - dead :( E - dead :( F - alive Explanation: This series of experiments demonstrates that DNA contains biological information that can be passed from one organism to another. Griffith discovered a transforming factor that held information to change the nonvirulent Streptococcus pneumoniae R strain into the virulent S strain. Avery, MacLeod, and McCarty built on the early work of Griffith to identify DNA as the molecule that stores biological information. Mice injected with the R strain are not infected and will survive, whereas mice injected with the S strain are infected and will die. Heat treatment kills live bacteria by breaking them open, rupturing, and releasing the cellular components. The heat‑treated S strain alone is not infectious, and mice will survive because the injected bacteria are all dead. When live R strain is combined with the heat‑treated S strain, the infected mice will die. An undetermined transforming factor in the cellular components of the heat‑treated S strain causes the live R strain to become virulent. This process is called transformation. To determine the identity of the transforming factor, Avery, MacLeod, and McCarty isolated the transforming factor by destroying each major component of the cell in turn. Then, they tested for the ability of the remaining cellular components of the S strain to transform the R strain to a virulent state. When lipids, polysaccharides, RNA, and proteins from the heat‑treated S strain were destroyed and then combined with the R strain, the mice still died because the transforming factor had not been destroyed. However, when DNA was additionally destroyed, the mice injected with this combination lived. This indicated that the transforming factor had been destroyed and the R strain had not been transformed, confirming DNA as the genetic molecule.

Nucleotides are molecules that make up the structures of RNA and DNA. Modify the compounds to create adenosine monophosphate (AMP). (See image) Complete the sentence about AMP. AMP is comprised of the base, __________, the sugar, ___________ and one phosphate.

AMP is comprised of the base, adenine, the sugar, ribose, and one phosphate. A nucleotide is composed of a phosphate group, a five‑carbon sugar, either ribose or deoxyribose, and a nucleobase (nitrogenous base). The phosphate group forms a bond with either the 2‑, 3‑, or 5‑carbon of the sugar (generally with the 5‑carbon). The sugar also bonds with the nucleobase by a glycosidic bond. In this case, an N‑glycosidic bond would be formed. Also formed in this reaction are two water molecules. There are two types of nucleobases, pyrimidines and purines. Pyrimidines consist of a single pyrimidine ring whereas purines consist of a pyrimidine ring fused to an imidazole ring. AMP contains the purine adenine. The ribose sugar contains a hydroxyl group on the 2' carbon, whereas deoxyribose does not. Therefore, AMP contains the sugar ribose.

Identify the correct name or abbreviation for the given nucleoside or nucleotide. dGDP guanosine GDP dADP ADP A structural formula shows two fused aromatic, heterocyclic rings attached to a five carbon sugar. The fused rings include a six membered ring and a five membered ring. The six membered ring has an amine group attached to the carbon at position 2 and a carbonyl group at position 6. The five membered ring is attached to carbon 1 of the sugar. The sugar has free hydroxyl groups at carbon 2 and carbon 3. Carbon 5 of the sugar is attached to a pyrophosphate group, also called a diphosphate group. Identify the correct name or abbreviation for the given nucleoside or nucleotide. adenosine deoxyadenosine guanosine dGDP deoxyguanosine

First image = GDP Second image = Deoxyguanosine Explanation: A pyrimidine has a single heterocyclic ring. Cytosine, thymine and uracil are pyrimidines (ending become "-idine"). A purine has two fused heterocyclic rings. Adenine and guanine are purines ("-ine" -> "-osine"). A nucleoside is a purine or pyrimidine base attached to a sugar via an N-glycosidic bond. The sugar is typically either ribose or deoxyribose. Compared to ribose, deoxyribose is missing the 2′ hydroxyl group. The chart describes the names of each nucleoside. Base: adenine With ribose : adenosine With deoxyribose: deoxyadenosine Base: thymine With ribose : ribothymidine With deoxyribose: thymidine or deoxythymidine Base: cytosine With ribose: cytidine With deoxyribose: deoxycytidine Base: guanine With ribose : guanosine With deoxyribose: deoxyguanosine Base: uracil With ribose : uridine With deoxyribose: deoxyuridine Note that thymidine is normally found only on deoxyribose and not ribose, so the deoxy- prefix is not necessary, though may be included. Similarly, uracil is typically found only on ribose and not deoxyribose, though deoxyuridine does exist. A nucleotide is a nucleoside that is attached to one or more phosphates. Nucleotides attached to one, two, or three phosphates are called 5′‑monophosphates (MP), 5′‑diphosphates (DP) or 5′‑triphosphates (TP), respectively. The chart gives the abbreviations for the different nucleotides. (Deoxyribose adds "d" to beginning) Base: adenine (A) With ribose (1, 2 or 3 phosphates): AMP, ADP, ATP With deoxyribose (1, 2, or 3 phosphates): dAMP, dADP, dATP Base: thymine With ribose (1, 2 or 3 phosphates): TMP, TDP, TTP With deoxyribose (1, 2, or 3 phosphates): dTMP, dTDP, dTTP Base: cytosine With ribose (1, 2 or 3 phosphates): CMP, CDP, CTP With deoxyribose (1, 2, or 3 phosphates): dCMP, dCDP, dCTP Base: guanine With ribose (1, 2 or 3 phosphates): GMP, GDP, GTP With deoxyribose (1, 2, or 3 phosphates): dGMP, dGDP, dGTP Base: uracil With ribose (1, 2 or 3 phosphates): UMP, UDP, UTP With deoxyribose (1, 2, or 3 phosphates): dUMP, dUDP, dUTP (LOL.. dUMP)

Place an asterisks (*) next to the 3' carbon atoms in the polynucleotide shown. Answer Bank: * (multi-use) What is the name of this polynucleotide? CGT GTC TCG CTG

GTC Explanation: A polynucleotide is a molecule composed of several nucleotide units joined together by phosphodiester linkages. The backbone of the molecule consists of alternating sugars and phosphate groups. Polynucleotides are formed by joining the 3' hydroxyl group of one nucleotide with the 5' phosphate group of another. Each carbon atom bonded to the oxygen of the hydroxyl group is a 3' carbon atom; each carbon atom bonded to the oxygen of a phosphate group is a 5' carbon atom. The 5' end of the polynucleotide has a free phosphate group, while the 3' end has a free hydroxyl group. A polynucleotide is named from the 5' end to the 3' end, using the one letter abbreviations for each base. The bases in this polynucleotide are guanine, thymine, and cytosine (from the 5' end to the 3' end). The name of this polynucleotide is GTC.

The nucleotides adenosine‑5'‑monophosphate (A) and thymidine‑5'‑monophosphate (T) form a dinucleotide. Modify the structures to give the dinucleotide AT.

I hate these. Explanation: Two nucleotides are joined at the 5' end of one nucleotide to the 3' end of the other nucleotide. In this reaction, the hydroxyl group on carbon 3 of adenosine‑5'‑monophosphate will react with the phosphate on carbon 5 of thymidine‑5'‑monophosphate. A new phosphodiester bond is formed between the two nucleotides. The order of the dinucleotide indicates that adenosine‑5'‑monophosphate is at the 5' end and the thymidine‑5'‑monophosphate is at the 3' end. The dinucleotide TA would give the reverse order.

Select each of the characteristics that are key features of genetic material. It must have observable characteristics that are heritable. It must be able to replicate and transmit to progeny. It must differentiate in each cell type. It must encode the blueprint to form proteins and other structures. It must periodically mutate to generate variation.

It must be able to replicate and transmit to progeny. It must encode the blueprint to form proteins and other structures. It must periodically mutate to generate variation. Explanation: DNA is the genetic material for most life on earth. Genetic material encodes all the information necessary for biological processes that occur in an organism. This includes all the genes present in the genome of a cell or organism, as well as introns, enhancers, promoters, and other DNA elements. The key properties of genetic material are storage and expression of information, replication and transmission to the next generation, and variation by mutation. The first key property is represented in the central dogma. Information is encoded in DNA and expressed as proteins. The second key property ensures that replication of the parental genome provides an almost identical copy to progeny cells. Offspring can only survive if a reliable working copy of the parental genome is passed down. Mutations are the third key property and the source of variability within a population. A mutation is defined as a change in chemical composition of DNA. Differentiation is a process of physical specialization at the cellular level. A cell undergoes differentiation, but this almost never involves directly modifying the genome. This process occurs throughout the life of a multi‑cellular organism, as tissues and cell types develop and are maintained, and as stem cells divide and specialize. Traits are the observable characteristics of gene expression. A phenotype is the expressed trait for an organism or cell. Neither traits nor phenotypes are heritable. Alleles are defined as the unit of inheritance. They are encoded in a linear sequence of DNA. A genome is all of the genetic material contained in a cell or organism. Confusion arises because the genotype of two alleles of a gene encodes a protein that ultimately may determine a trait, but not all genes are expressed.

Identify the true statement about the structure of the DNA double helix. The phosphate groups are on the outside of the double helix because they are charged and interact with the aqueous environment surrounding the DNA. The nitrogenous bases are on the outside of the double helix because they are hydrophilic and interact with the aqueous environment surrounding the DNA. The nitrogenous bases are on the inside of the double helix because they are hydrophilic and do not interact with the aqueous environment surrounding the DNA. The phosphate groups are on the inside of the double helix because they are nonpolar and do not interact with the aqueous environment surrounding the DNA.

The phosphate groups are on the outside of the double helix because they are charged and interact with the aqueous environment surrounding the DNA. Explanation: The arrangement of the phosphate and nitrogenous base groups within the DNA double helix was proposed by Watson and Crick. The nitrogenous bases are nonpolar and hydrophobic. Thus, they avoid the aqueous environment surrounding the DNA and instead interact with one another on the inside of the DNA double helix. In contrast, the phosphate groups are charged. Thus, they are on the outside of the double helix where they interact with the aqueous environment surrounding the DNA.

Identify the phosphate ester (phosphoester) bond and the N‑glycosidic bond in the nucleotide shown. Answer Bank: phosphoester bond N‑glycosidic bond

The single bond on the left between P-O is the phosphoester bond On the right side, the single bond between the N-C is the N-glycosidic bond See image for explanation

What contribution did James Watson and Francis Crick make to our understanding of DNA? They found that proteins are not used as the genetic code, but that DNA is the genetic code. They determined that adenine and thymine occur in equal amounts in DNA. They produced a clear X‑ray diffraction picture of the three‑dimensional structure of DNA. They pieced together the available evidence and modeled the structure of DNA.

They pieced together the available evidence and modeled the structure of DNA. Explanation: James Watson and Francis Crick developed an accurate model of the DNA molecule that accounted for all the evidence scientists had accumulated about DNA. Work in several labs had already led scientists to conclude that DNA, and not protein, was the genetic material used by living organisms. However, how DNA acted as a code and how DNA was structured remained unknown. Two scientists provided critical pieces of evidence about DNA structure that Watson and Crick used to construct their model. Erwin Chargaff had noticed that the amount of thymine in DNA was always very close to the amount of adenine in DNA. Likewise, the amounts of cytosine and guanine were approximately equal in DNA. Rosalind Franklin produced an X‑ray diffraction picture of DNA that allowed her to determine that the DNA was a helix and to measure distances between parts of the DNA molecule. Watson and Crick tested models until they developed one that was consistent with all of this evidence.