Overview, DNA, and Genes

Question 1

the first promising candidates for genetic materials were

Answer 1

proteins, because they were abundant, diverse and complex molecules

Question 2

in the Griffith’s Transformation Experiment (1928), Microbiologists identified two strains of the bacterium _______________

Answer 2

Microbiologists identified two strains of the bacterium Streptococcus pneumoniae

Question 3

in the Griffith’s Transformation Experiment (1928), the two strains Streptococcus pneumoniae were

Answer 3

R- strain and S- strain

Question 4

difference between the R- and S- strains of Streptococcus pneumoniae

Answer 4

The R-strain produced rough colonies on a
bacterial plate, while the other S-strain was smooth (Figure ). More importantly, the S-strain bacteria caused fatal infections
when injected into mice, while the R-strain did not (top, Figure ).

Question 5

in the Griffith’s Transformation Experiment (1928), Griffith in 1929
noticed that upon mixing “heat-treated” S-strain cells together with some R-type bacteria (neither should kill the mice), what happened to the mice ?

Answer 5

the mice
died and there were S-strain, pathogenic cells recoverable. Thus, some non-living component from the S-type strains contained
genetic information that could be transferred to and transform the living R-type strain cells into S-type cells.

Question 6

how does the Avery, MacLeod and McCarty’s Experiment (1944) explain the transform of living R-type strain cells into S-type

Answer 6

Avery, MacLeod and McCarty separated the S-type cells into various components, such as proteins, polysaccharides, lipids, and
nucleic acids. Only the nucleic acids from S-type cells were able to make the R-strains smooth and fatal. Furthermore, when
cellular extracts of S-type cells were treated with DNase (an enzyme that digests DNA), the transformation ability was lost. The
researchers therefore concluded that DNA was the genetic material, which in this case controlled the appearance (smooth or rough)
and pathogenicity of the bacteria

Question 7

Hershey and chase studied the transmission of genetic information in a virus called the

Answer 7

T2 bacteriophage,

Question 8

in the Hershey and chase experiments, the T2 bacteriophage used which host bacterium

Answer 8

used Escherichia coli as its host
bacterium (Figure ). Like all viruses, T2 hijacks the cellular machinery of its host to manufacture more viruses

Question 9

T/F the T2 bacteriophage contains many classes of genetic material

Answer 9

false, The T2 phage
itself only contains both protein and DNA, but no other class of potential genetic material.

Question 10

how did Hershey and chase determine which of aDNA or protein contained. genetic material

Answer 10

To determine which of these two types of molecules contained the genetic blueprint for the virus, Hershey and Chase grew viral
cultures in the presence of radioactive isotopes of either phosphorus ( P) or sulphur ( S). The phage incorporated these isotopes
into their DNA and proteins, respectively (Fig 1.5). The researchers then infected E. coli with the radiolabeled viruses, and looked
to see whether P or S entered the bacteria. After ensuring that all viruses had been removed from the surface of the cells, the
researchers observed that infection with P labeled viruses (but not the S labeled viruses) resulted in radioactive bacteria. This
demonstrated that DNA was the material that contained genetic instructions.

Question 11

From the complementary strands model of DNA, proposed by Watson and Crick in 1953, there were three straightforward possible
mechanisms for DNA replication:

Answer 11

(1) semi-conservative, (2) conservative, and (3) dispersive

Question 12

semi conservative model of DNA replication

Answer 12

. The semi-conservative model proposes the two strands of a DNA molecule separate during replication and then strand acts as a
template for synthesis of a new, complementary strand

Question 13

conservative model of DNA replication

Answer 13

The conservative model proposes that the entire DNA duplex acts as a single template for the synthesis of an entirely new
duplex

Question 14

The dispersive model of DNA replication

Answer 14

The dispersive model has the two strands of the double helix breaking into units that which are then replicated and
reassembled, with the new duplexes containing alternating segments from one strand to the other.

Question 15

Meselson and Stahl experiment (1958)

Answer 15

Meselson and Stahl used different isotopes of Nitrogen, which is a major component
in DNA. Nitrogen-14 ( N) is the most abundant natural isotope, while Nitrogen-15 ( N) is rare, but also denser. Neither is
radioactive; each can be followed by a difference in density – “light” 14 vs “heavy”15 atomic weight in a CsCl density gradient
ultra-centrifugation of DNA.
The experiment starts with E. coli grown for several generations on medium containing only N. It will have denser DNA. When
extracted and separated in a CsCl density gradient tube, this “heavy” DNA will move to a position nearer the bottom of the tube in
the more dense solution of CsCl (left side in Figure ). DNA extracted from E. coli grown on normal, N containing medium
will migrate more towards the less dense top of the tube.
If these E. coli cells are transferred to a medium containing only N, the “light” isotope, and grown for one generation, then their
DNA will be composed of one-half N and one-half N. If the this DNA is extracted and applied to a CsCl gradient, the
observed result is that one band appears at the point midway between the locations predicted for wholly N DNA and wholly N
DNA (Figure ). This “single-band” observation is inconsistent with the predicted outcome from the conservative model of
DNA replication (disproves this model), but is consistent with both that expected for the semi-conservative and dispersive models.
If the E. coli is permitted to go through another round of replication in the N medium, and the DNA extracted and separated on a
CsCl gradient tube, then two bands were seen by Meselson and Shahl: one at the N- N intermediate position and one at the
wholly N position (Figure ). This result is inconsistent with the dispersive model (a single band between the N- N
position and the wholly N position) and thus disproves this model. The two band observation is consistent with the semiconservative model which predicts one wholly N duplex and one N- N duplex. Additional rounds of replication also support
the semi-conservative model/hypothesis of DNA replication. Thus, the semi-conservative model is the currently accepted
mechanism for DNA replication. Note however, that we now also know from more recent experiments that whole chromosomes,
which can be millions of bases in length, are also semi-conservatively replicated.

Question 16

T/F DNA is the genetic material for all organisms

Answer 16

FALSE, While DNA is the genetic material for the vast majority of organisms, there are some viruses that use RNA as their genetic
material. These viruses can be either single or double stranded and include SARS, influenza, hepatitis C and polio, as well as the
retroviruses like HIV-AIDS. Typically there is DNA used at some stage in their life cycle to replicate their RNA genome.

Also, the case of Prion infections agents transmit characteristics via only a protein (no nucleic acid present).

Question 17

_________infect by
transmitting a misfolded protein state from one aberrant protein molecule to a normally folded molecule.

Answer 17

prions

Question 18

__________responsible for bovine spongiform encephalopathy (BSE, also known as “mad cow disease”) in cattle and deer and Creutzfeldt–
Jakob disease (CJD) in humans.

Answer 18

prions are responsible for bovine spongiform encephalopathy (BSE, also known as “mad cow disease”) in cattle and deer and Creutzfeldt–
Jakob disease (CJD) in humans.

Question 19

T/F All known prion diseases act by altering the structure of the brain or other neural tissue and all are
currently untreatable and ultimately fatal.

Answer 19

TRUE

Question 20

T/F Watson and crick determined that DNA is made up of a
series four different types of molecules, called bases or nucleotides: adenine (A), cytosine (C), thymine (T), guanine (G).

Answer 20

FALSE When Watson and Crick set out in the 1940’s to determine the structure of DNA, it was already known that DNA is made up of a
series four different types of molecules, called bases or nucleotides: adenine (A), cytosine (C), thymine (T), guanine (G). Watson
and Crick also knew of Chargaff’s Rules, which were a set of observations about the relative amount of each nucleotide that was
present in almost any extract of DNA. Chargaff had observed that for any given species, the abundance of A was the same as T, and
G was the same as C. This was essential to Watson & Crick’s model.

Question 21

Chargaff’s Rules

Answer 21

, which were a set of observations about the relative amount of each nucleotide that was
present in almost any extract of DNA. Chargaff had observed that for any given species, the abundance of A was the same as T, and
G was the same as C. This was essential to Watson & Crick’s model.

Question 22

The Double Helix

Answer 22

Using proportional metal models of the individual nucleotides, Watson and Crick deduced a structure for DNA that was consistent
with Chargaff’s Rules and with x-ray crystallography data that was obtained (with some controversy) from another researcher
named Rosalind Franklin. In Watson and Crick’s famous double helix, each of the two strands contains DNA bases connected
through covalent bonds to a sugar-phosphate backbone (Fig 1.8, 1.9). Because one side of each sugar molecule is always connected
to the opposite side of the next sugar molecule, each strand of DNA has polarity: these are called the 5’ (5-prime) end and the 3’ (3-
prime) end, in accordance with the nomenclature of the carbons in the sugars. The two strands of the double helix run in antiparallel (i.e. opposite) directions, with the 5’ end of one strand adjacent to the 3’ end of the other strand. The double helix has a
right-handed twist, (rather than the left-handed twist that is often represented incorrectly in popular media).

Question 23

blending inheritance

Answer 23

blending inheritance proposed that some undefined essence, in its entirety,
contained all of the heritable information for an individual. It was thought that mating combined the essences from each parent,
much like the mixing of two colors of paint. Once blended together, the individual characteristics of the parents could not be
separated again.

Question 24

how did Mendel disprove blending inheritance

Answer 24

He started with well-characterized strains, repeated his experiments many times, and kept careful records of his observations.
Working with peas, Mendel showed that white-flowered plants could be produced by crossing two purple-flowered plants, but only
if the purple-flowered plants themselves had at least one white-flowered parent (Fig 1.11). This was evidence that the genetic factor
that produced white-flowers had not blended irreversibly with the factor for purple-flowers. Mendel’s observations disprove
blending inheritance and favor an alternative concept, called particulate inheritance, in which heredity is the product of discrete
factors that control independent traits.

Question 25

T/F Mendel experiment prove blending inheritance

Answer 25

False He started with well-characterized strains, repeated his experiments many times, and kept careful records of his observations. Working with peas, Mendel showed that white-flowered plants could be produced by crossing two purple-flowered plants, but only if the purple-flowered plants themselves had at least one white-flowered parent (Fig 1.11). This was evidence that the genetic factor that produced white-flowers had not blended irreversibly with the factor for purple-flowers. Mendel’s observations disprove blending inheritance and favor an alternative concept, called particulate inheritance, in which heredity is the product of discrete factors that control independent traits.

Question 26

“one gene, one enzyme (or protein)”

Answer 26

Life depends on (bio)chemistry to supply energy and to produce the molecules to construct and regulate cells. In 1908, A. Garrod described “in born errors of metabolism” in humans using the congenital disorder, alkaptonuria (black urine disease), as an example of how “genetic defects” led to the lack of an enzyme in a biochemical pathway and caused a disease (phenotype). Over 40 years later, in 1941, Beadle and Tatum built on this connection between genes and metabolic pathways. Their research led to the “one gene, one enzyme (or protein)” hypothesis, which states that each of the enzymes that act in a biochemical pathway is encoded by a different gene. Although we now know of many exceptions to the “one gene, one enzyme (or protein)” principle, it is generally true that each different gene produces a protein that has a distinct catalytic, regulatory, or structural function.

Question 27

Beadle and Tatum used the fungus _______________(a mold) for their studies because it had practical advantages as a laboratory organism

Answer 27

Beadle and Tatum used the fungus Neurospora crassa (a mold) for their studies because it had practical advantages as a laboratory organism

Question 28

neurospora crassa advantages as laboratory organism

Answer 28

They knew that Neurospora was prototrophic, meaning that it could synthesize its own amino acids when grown on minimal medium, which lacked most nutrients except for a few minerals, simple sugars, and one vitamin (biotin). They also knew that by exposing Neurospora spores to X-rays, they could randomly damage its DNA to create mutations in genes. Each different spore exposed to X-rays potentially contained a mutation in a different gene

Question 29

Srb and Horowitz in 1944

Answer 29

Srb and Horowitz in 1944 tested the ability of the amino acids to rescue auxotrophic strains. They added one of each of the amino acids to minimal medium and recorded which of these restored growth to independent mutants. For example, if the progeny of a mutagenized spore could grow on minimal medium only when it was supplemented with arginine (Arg), then the auxotroph must bear a mutation in the Arg biosynthetic pathway and was called an “arginineless” strain (arg-).

Question 30

The complete set of DNA within the nucleus of any organism is called its

Answer 30

nuclear genome

Question 30

The complete set of DNA within the nucleus of any organism is called its nuclear genome and is measured as the

Answer 30

C-value in units of either the number of base pairs or picograms of DNA. There is a general correlation between the nuclear DNA content of a genome (i.e. the C-value) and the physical size or complexity of an organism.

Question 30

The C-value paradox

Answer 30

This apparent paradox (called the C-value paradox) can be explained by the fact that not all nuclear DNA encodes genes – much of the DNA in larger genomes is non-gene coding. In fact, in many organisms, genes are separated from each other by long stretches of DNA that do not code for genes or any other genetic information. Much of this “non-gene” DNA consists of transposable elements of various types, which are an interesting class of self-replicating DNA elements discussed in more detail in a subsequent chapter. Other non-gene DNA includes short, highly repetitive sequences of various types.

Question 31

characteristics of model organisms

Answer 31

(1) they are small, (2) fast growing with a short generation time, (3) produce lots of progeny from matings that can be easily controlled, (4) have small genomes (small Cvalue), and (5) are diploid (i.e. chromosomes are present in pairs)

Question 32

The most commonly used model organism are:

Answer 32

The prokaryote bacterium, Escherichia coli, is the simplest genetic model organism and is often used to clone DNA sequences from other model species. Yeast (Saccharomyces cerevisiae) is a good general model for the basic functions of eukaryotic cells. The roundworm, Caenorhabditis elegans is a useful model for the development of multicellular organisms, in part because it is transparent throughout its life cycle, and its cells undergo a well-characterized series of divisions to produce the adult body. The fruit fly (Drosophila melanogaster) has been studied longer, and probably in more detail, than any of the other genetic model organisms still in use, and is a useful model for studying development as well as physiology and even behavior. The mouse (Mus musculus) is the model organism most closely related to humans, however there are some practical difficulties working with mice, such as cost, slow reproductive time, and ethical considerations. The zebrafish (Danio rerio) has more recently been developed by researchers as a genetic model for vertebrates.Unlike mice, zebrafish embryos develop quickly and externally to their mothers, and are transparent, making it easier to study the development of internal structures and organs. Finally, a small weed, Arabidopsis thaliana, is the most widely studied plant genetic model organism. This provides knowledge that can be applied to other plant species, such as wheat, rice, and corn.

Question 33

How would the results of the cross in humans have been different if heredity worked through blending inheritance rather than particulate inheritance?

Answer 33

f heredity worked through blending inheritance rather than particulate inheritance, the results of a genetic cross in humans would be significantly different: No Clear Dominance or Recessiveness: In particulate inheritance (Mendelian genetics), traits are passed as discrete units (genes), leading to dominant and recessive traits. If blending inheritance were true, there would be no dominant or recessive traits—offspring would always exhibit an intermediate phenotype. Loss of Genetic Variation Over Generations: In blending inheritance, traits would continuously mix and dilute over generations, making populations more homogenous. Eventually, extreme traits (e.g., very tall or very short individuals) would disappear, as each generation would move toward an average. No Reappearance of Parental Traits: In particulate inheritance, recessive traits can reappear in later generations if two carriers pass on the recessive allele. Under blending inheritance, once traits mix, they cannot separate out in future generations. For example, if a tall person and a short person had a child of medium height, future generations would never have very tall or very short individuals again. Continuous Variation in All Traits: Instead of distinct categories (e.g., blood types A, B, AB, O), all traits would exist on a spectrum. Instead of distinct eye colors (e.g., blue, brown), there would be a continuous range of shades between parental colors. Lack of Mendelian Ratios: In crosses like those Gregor Mendel observed (e.g., a 3:1 ratio for dominant and recessive traits), such ratios would not exist. Instead, all offspring would show blended, intermediate traits rather than distinct dominant and recessive categories. In reality, particulate inheritance explains why genetic variation persists over generations, allowing for the reappearance of traits and the maintenance of genetic diversity. This is why Mendelian genetics (particulate inheritance) correctly describes heredity, while blending inheritance has been disproven.

Question 34

Imagine that astronauts provide you with living samples of multicellular organisms discovered on another planet. These organisms reproduce with a short generation time, but nothing else is known about their genetics. a) How could you define laws of heredity for these organisms? b) How could you determine what molecules within these organisms contained genetic information? c) Would the mechanisms of genetic inheritance likely be similar for all organisms from this planet? d) Would the mechanisms of genetic inheritance likely be similar to organisms from earth?

Answer 34

If astronauts brought back living multicellular organisms from another planet with a short generation time but unknown genetics, we would take several steps to define their heredity laws and determine how their genetic material works. (a) How could you define laws of heredity for these organisms? To define laws of heredity, you would follow a method similar to Gregor Mendel’s experiments: Controlled Crossbreeding: Cross different individuals with distinct traits (e.g., color, size, shape) and observe how these traits are passed to offspring. Statistical Analysis of Offspring Traits: Track the ratios of traits in each generation to determine if inheritance follows a particulate (Mendelian-like) or blending pattern. Test for Dominance & Recessiveness: Identify whether traits follow a dominant-recessive pattern or if inheritance is polygenic (continuous variation). Track Generations Over Time: Observe inheritance patterns across multiple generations to see if certain traits disappear, reappear, or follow predictable ratios. Look for Environmental Influence: Expose organisms to different environmental conditions to determine if heredity is influenced by external factors or if traits are strictly inherited. (b) How could you determine what molecules within these organisms contain genetic information? To find the genetic material, we would follow a strategy similar to how DNA was discovered as Earth's hereditary molecule: Cellular Extraction & Biochemical Analysis: Break open cells and isolate major biological molecules (proteins, lipids, carbohydrates, nucleic acids). Determine which molecule varies across individuals but remains constant across an individual's cells. Transformation Experiments: Similar to Griffith’s experiment, introduce potential genetic material from one organism into another and see if traits are inherited. If a molecule (like DNA on Earth) carries genetic information, transferring it to another organism should result in inherited changes. Use of Enzymes to Break Down Specific Molecules: Treat cells with enzymes that degrade proteins, lipids, or nucleic acids to see which one prevents inheritance. If a nucleic acid-like molecule is broken down and traits are no longer inherited, that suggests it is the genetic material. Fluorescent Tagging & Microscopy: Label different cellular components (e.g., proteins, nucleic acids) with fluorescent markers to see which molecule is passed from parent to offspring. Sequencing & Molecular Analysis: If possible, analyze the structure of the genetic material and compare it to Earth’s DNA/RNA. (c) Would the mechanisms of genetic inheritance likely be similar for all organisms from this planet? Likely yes, but with variations: If all life on the planet shares a common ancestor, they might have similar hereditary mechanisms. However, some species could use different molecules for inheritance, similar to how DNA vs. RNA functions across different viruses and organisms on Earth. There might be alternative genetic systems where information is stored in proteins, lipids, or entirely new biomolecules. (d) Would the mechanisms of genetic inheritance likely be similar to organisms from Earth? Possibly, but not necessarily identical. If the planet's life evolved under similar chemical and physical conditions as Earth, it might use DNA, RNA, or similar nucleic acids. However, it is possible that: They have entirely different genetic molecules (e.g., XNA, a synthetic nucleic acid on Earth). They could replicate differently (e.g., no double-helix, no base pairing like A-T, G-C). They could use different storage and transfer mechanisms (e.g., proteins instead of nucleic acids for information storage). But some form of information storage, replication, and inheritance must exist for them to function as living, evolving organisms. Conclusion: To understand these alien organisms, we would observe inheritance patterns, isolate their genetic material, and compare their biological molecules to those on Earth. While their genetics may share universal principles (e.g., replication, mutation, transmission), they could have entirely unique inheritance mechanisms that challenge our current understanding of life.

Question 35

It is relatively easy to extract DNA and protein from cells; biochemists had been doing this since at least the 1800’s. Why then did Hershey and Chase need to use radioactivity to label DNA and proteins in their experiments?

Answer 35

Hershey and Chase used radioactive labeling in their famous 1952 experiment to conclusively prove that DNA, not protein, is the genetic material. Although DNA and proteins had been extracted from cells since the 1800s, distinguishing which molecule was responsible for heredity required a precise method. Here’s why they needed radioactivity: 1. Both DNA and Proteins Are Present in Cells Every cell contains both DNA and proteins, so simply extracting them wasn’t enough to determine which one carried genetic information. To figure out which molecule was being transferred into bacterial cells by viruses, they needed a way to track each molecule separately. 2. DNA and Proteins Have Different Chemical Compositions DNA contains phosphorus (P) but no sulfur (S). Proteins contain sulfur (S) but little to no phosphorus (P). Hershey and Chase exploited this difference by using radioactive isotopes to label each molecule uniquely: Radioactive phosphorus-32 ( 32 P ) ( 32 P) labeled DNA. Radioactive sulfur-35 ( 35 S ) ( 35 S) labeled proteins. 3. Tracing Which Molecule Entered Bacteria They used bacteriophages (viruses that infect bacteria), which consist of only DNA and a protein coat. By labeling DNA with 32 P 32 P and proteins with 35 S 35 S, they could track which part of the virus was injected into bacteria. Results: 32 P 32 P (DNA) was found inside infected bacteria, proving DNA entered the cell. 35 S 35 S (protein) was found outside, showing the protein coat remained outside. 4. Why Not Just Extract DNA and Proteins? Extracting DNA and protein alone wouldn’t show which one was actually responsible for heredity. Without labeling, there would be no way to tell if the genetic material that got into the bacteria was DNA or protein. Conclusion: Hershey and Chase’s use of radioactive labeling allowed them to trace the movement of DNA and protein separately. This was the key to proving that DNA, not protein, is the molecule of inheritance, revolutionizing our understanding of genetics.

Question 36

Compare Watson and Crick’s discovery with Avery, MacLeod and McCarty’s discovery. a) What did each discover, and what was the impact of these discoveries on biology? b) How did Watson and Crick’s approach generally differ from Avery, MacLeod and McCarty’s? c) Briefly research Rosalind Franklin on the internet. Why is her contribution to the structure of DNA controversial?

Answer 36

What did each discover, and what was the impact of these discoveries on biology? Avery, MacLeod, and McCarty (1944) – DNA as the Genetic Material Discovery: They demonstrated that DNA, not protein, was the molecule responsible for genetic inheritance. How? They conducted experiments on Streptococcus pneumoniae bacteria, showing that purified DNA from a virulent (disease-causing) strain could transform a non-virulent strain into a virulent one. Impact: This was the first strong evidence that DNA, rather than protein, carried genetic information. Their work laid the foundation for later experiments, like the Hershey-Chase experiment (1952) and the Watson-Crick model (1953). Watson and Crick (1953) – Structure of DNA Discovery: They determined that DNA has a double-helix structure, with complementary base pairing (A-T, G-C). How? They built models based on data from X-ray diffraction (Rosalind Franklin's work) and Chargaff’s rules (A = T, G = C). Impact: Explained how DNA replicates and stores genetic information. Provided a structural basis for modern genetics, leading to advances like gene sequencing, genetic engineering, and biotechnology. (b) How did Watson and Crick’s approach generally differ from Avery, MacLeod, and McCarty’s? Aspect Avery, MacLeod, & McCarty Watson & Crick Methodology Experimental biology Theoretical modeling Approach Conducted lab experiments to isolate DNA and prove its role in heredity Used model-building and existing data to deduce DNA's structure Main Contribution Proved DNA is the genetic material Discovered DNA's structure (double helix) Key Evidence Bacterial transformation experiments X-ray diffraction data (Rosalind Franklin) & Chargaff’s rules Avery, MacLeod, & McCarty were experimental scientists, focusing on biochemical techniques to prove DNA's function. Watson & Crick were more theoretical, relying on model-building and existing data rather than experiments. (c) Rosalind Franklin’s Contribution and Controversy Who was Rosalind Franklin? She was a British scientist specializing in X-ray crystallography. Her Photo 51 was a crucial piece of evidence that showed DNA had a helical structure. Why is her contribution controversial? Her data was used without her direct permission: Maurice Wilkins (her colleague) showed Photo 51 to Watson without Franklin’s knowledge. This helped Watson & Crick finalize the double-helix model, but Franklin was not credited appropriately. She was not included in the Nobel Prize: Watson, Crick, and Wilkins won the Nobel Prize in 1962 for the discovery of DNA’s structure. Franklin had died in 1958, and Nobel Prizes are not awarded posthumously. However, her contributions were largely overlooked during her lifetime. Watson’s own remarks about Franklin: In his book The Double Helix, Watson described Franklin in a dismissive and sexist manner, further sparking debate over her treatment in scientific history. Her Legacy Today, Franklin is widely recognized as a key contributor to DNA’s discovery. Many awards, buildings, and institutions now bear her name in honor of her work. Conclusion Avery, MacLeod, & McCarty proved that DNA is the genetic material. Watson & Crick discovered the double-helix structure of DNA. Rosalind Franklin's work was essential to Watson & Crick’s discovery, but her role was undervalued during her lifetime, making her contribution controversial.

Question 37

List the information that Watson and Crick used to deduce the structure of DNA.

Answer 37

Watson and Crick used several key pieces of scientific evidence to deduce the double-helix structure of DNA. These sources of information included experimental findings from other scientists and their own model-building efforts. Key Information Used by Watson and Crick Rosalind Franklin’s X-ray Diffraction Data (Photo 51) Franklin’s X-ray crystallography images (especially Photo 51) showed a characteristic X-shaped pattern, indicating that DNA had a helical structure. The X-ray data also suggested that DNA had a uniform diameter of about 2 nanometers. Chargaff’s Rules Erwin Chargaff found that: The amount of adenine (A) always equals thymine (T). The amount of guanine (G) always equals cytosine (C). This suggested a base-pairing mechanism. Linus Pauling’s Alpha-Helix Model of Proteins Linus Pauling had previously discovered that proteins could form helical structures. Watson and Crick applied this idea to DNA, considering it might also form a helical shape. Density and Structural Measurements DNA’s constant width (from Franklin’s data) meant that purines (A & G, larger bases) must always pair with pyrimidines (T & C, smaller bases). A purine-purine pair would be too wide, and a pyrimidine-pyrimidine pair would be too narrow. Antiparallel Strands The uniform structure suggested that the two strands run in opposite directions (antiparallel). This arrangement was necessary for hydrogen bonding between complementary bases. Model Building Watson and Crick physically constructed 3D models of DNA using metal rods and connectors. By adjusting bond angles and spacing, they confirmed the double-helix model. Final Model of DNA (1953) Double helix with two antiparallel strands. Sugar-phosphate backbone on the outside. Base pairs (A-T, G-C) held together by hydrogen bonds on the inside. Uniform 2-nm width. One complete turn of the helix every 10 base pairs. Their discovery revolutionized genetics and led to major breakthroughs in DNA replication, gene expression, and biotechnology.

Question 38

Refer to Watson and Crick’ a) List the defining characteristics of the structure of a DNA molecule. b) Which of these characteristics are most important to replication? c) Which characteristics are most important to the Central Dogma?

Answer 38

a) Defining Characteristics of the Structure of a DNA Molecule (Watson & Crick Model) Double Helix – DNA consists of two twisted strands forming a helical structure. Antiparallel Strands – The two strands run in opposite directions: one from 5' to 3', the other from 3' to 5'. Sugar-Phosphate Backbone – The deoxyribose sugar and phosphate groups form the outer backbone of the molecule. Complementary Base Pairing – Bases pair Adenine (A) with Thymine (T) and Guanine (G) with Cytosine (C) via hydrogen bonds. Uniform Diameter – The width of the helix is consistent (2 nm) because a purine (A or G) always pairs with a pyrimidine (T or C). Major and Minor Grooves – The helical twist creates grooves that allow proteins (e.g., transcription factors) to interact with DNA. Base Stacking – The nitrogenous bases stack tightly due to hydrophobic interactions, stabilizing the structure. One Full Helical Turn Every 10 Base Pairs – The helix twists every 3.4 nm, containing 10 base pairs per turn. (b) Characteristics Most Important to DNA Replication Complementary Base Pairing – Ensures that each strand serves as a template for accurate copying of the genetic code. Antiparallel Strands – Directs semi-conservative replication, where one strand is synthesized continuously (leading strand) and the other discontinuously (lagging strand). Hydrogen Bonds – These bonds between base pairs are weak enough to break easily, allowing DNA to unzip for replication. Major & Minor Grooves – Provide access points for DNA polymerase and other replication enzymes. (c) Characteristics Most Important to the Central Dogma (DNA → RNA → Protein) Complementary Base Pairing – Essential for transcription, where RNA polymerase reads DNA and synthesizes mRNA. Antiparallel Strands – Dictates the direction of transcription, ensuring mRNA synthesis occurs correctly from the template strand. Major & Minor Grooves – Provide binding sites for transcription factors and RNA polymerase. Base Sequence – The specific sequence of DNA codes for genes, ultimately determining protein structure and function. Conclusion Replication relies on complementary base pairing and strand directionality. The Central Dogma depends on base pairing, strand orientation, and grooves for transcription and gene expression. These structural features make DNA both stable for genetic storage and flexible for gene expression and duplication.

Question 39

Which model organism can be used most efficiently to identify genes related to: i) eye development ii) skeletal development iii) photosynthesis iii) cell division iv) cell differentiation v) cancer

Answer 39

Best Model Organisms for Identifying Genes Related to Specific Biological Processes Different model organisms are best suited for studying different biological processes due to their genetic similarity to humans, ease of manipulation, and established research history. (i) Eye Development → Drosophila melanogaster (Fruit Fly) Why? Drosophila has well-characterized eye development pathways, especially involving genes like Pax6, which is also crucial for human eye development. Fast reproduction, easy genetic manipulation, and well-studied developmental biology. Alternative: Mus musculus (Mouse) – Closer to humans but slower to study. (ii) Skeletal Development → Mus musculus (Mouse) Why? Mice share 90% of genes with humans, making them ideal for studying bone growth, cartilage formation, and skeletal disorders. Genetic manipulation techniques (e.g., knockout mice) allow for targeted gene studies. Alternative: Zebrafish (Danio rerio) – Transparent embryos allow real-time visualization of skeletal formation. (iii) Photosynthesis → Arabidopsis thaliana (Thale Cress) Why? Arabidopsis is the most widely used plant model for studying photosynthesis and genetic regulation in plants. Its small genome, fast growth cycle, and ease of genetic manipulation make it efficient for photosynthesis research. Alternative: Chlamydomonas reinhardtii (Green Algae) – A unicellular photosynthetic organism that is useful for studying chloroplast function. (iv) Cell Division → Saccharomyces cerevisiae (Budding Yeast) Why? S. cerevisiae has highly conserved cell cycle genes, making it ideal for studying mitosis and regulatory proteins like cyclins and CDKs. Yeast divides rapidly, allowing quick experiments on cell cycle progression. Alternative: Schizosaccharomyces pombe (Fission Yeast) – Also widely used for cell cycle studies. (v) Cell Differentiation → Caenorhabditis elegans (Nematode Worm) Why? C. elegans has a fully mapped cell lineage, meaning every single cell division and differentiation event is known. Genes related to developmental biology, apoptosis, and organ formation are well studied in this organism. Alternative: Drosophila melanogaster – Useful for studying differentiation in embryonic and neural development. (vi) Cancer → Mus musculus (Mouse) Why? Mice develop cancers similar to humans, making them ideal for studying oncogenes, tumor suppressor genes, and drug testing. Genetic knockout and transgenic models help identify cancer-related mutations. Alternative: Zebrafish (Danio rerio) – Can develop tumors similar to humans, and their transparency allows direct observation of tumor growth.

Question 40

DNA helices that are rich in G-C base pairs are harder to separate (e.g. by heating) than A-T rich helices. Why?

Answer 40

DNA helices that are rich in G-C base pairs are harder to separate than A-T rich helices because G-C pairs form three hydrogen bonds, while A-T pairs form only two. This extra hydrogen bond increases the stability and melting temperature (Tm) of G-C rich DNA. Key Reasons Why G-C Rich DNA is Harder to Separate: More Hydrogen Bonds G-C pairs have three hydrogen bonds (≡). A-T pairs have only two hydrogen bonds (=). More hydrogen bonds = stronger interactions between DNA strands. Higher Thermal Stability More hydrogen bonds mean more energy (heat) is required to break them. G-C rich regions have a higher melting temperature (Tm) than A-T rich regions. Base Stacking Interactions G-C pairs also have stronger π-π stacking interactions between adjacent base pairs. These hydrophobic interactions further stabilize the DNA helix. Practical Applications PCR (Polymerase Chain Reaction): G-C rich regions require higher denaturation temperatures (~95°C or more). Sometimes, DMSO (dimethyl sulfoxide) or other denaturants are used to help separate the strands. Genome Stability: G-C rich regions are often found in promoters and regulatory regions of genes. These regions may be more resistant to mutations because they are structurally stable. Conclusion The extra hydrogen bond and stronger stacking interactions in G-C pairs make DNA harder to separate than A-T rich DNA, requiring higher temperatures or special conditions for denaturation.

Question 41

T/F all DNA in an organism contains genes.

Answer 41

false, Not all DNA in an organism contains genes.