Unit 3

Question 1

3-1. Using principles of hypothesis testing and experimental design, outline and interpret data that demonstrated that DNA is the molecule of heredity. How did experiments on transformed pneumonia-causing bacteria and radioactively-labeled bacterial viruses (Hershey & Chase) inform scientists that the genetic material was indeed DNA and not protein?

Answer 1

Experiments demonstrating that DNA is the molecule of heredity were key in establishing its role in genetic transmission. Griffith’s 1928 transformation experiment showed that a “transforming principle” from heat-killed pneumonia-causing bacteria could make harmless bacteria virulent, suggesting a genetic material transfer. Avery, MacLeod, and McCarty later identified that DNA, not protein, was the transforming principle. This was confirmed by Hershey and Chase in 1952, who used radioactively-labeled bacteriophages to show that only DNA (labeled with phosphorus) entered bacterial cells during infection, while proteins (labeled with sulfur) remained outside. These experiments together provided strong evidence that DNA, and not protein, is the genetic material responsible for inheritance.

Question 2

3-2. Describe the double helix model of DNA, reviewing the evidence that supports this model. What were some specific types of data used by Watson & Crick to determine the structure of DNA?

Answer 2

The structure of DNA can be described as two long strands of nucleotides coiled around each other, forming a twisted ladder-like shape. The evidence that supports the Double Helix Model is X-ray diffraction, and Chargaff’s Rules. Watson and Crick used both X-ray Diffraction Data and Chargaff’s Data combined.

Question 3

3-3. Describe the structure of DNA nucleotide components (phosphate, sugar, A T C G bases), their arrangement in DNA including complementary base pairing, and the types of bonds (hydrogen or covalent) between various components. What are 3’ and 5’ ends of DNA? What is meant by the “antiparallel” structure of DNA?

Answer 3

A DNA nucleotide consists of a Phosphate group (phosphorus atom bonded to four oxygen atoms) A sugar (deoxyribose - 5 carbon with a lock of an oxygen atom at the 2’ position), and nitrogenous bases (Adenine, Thymine, Cytosine, Guanine). Nucleotides are linked together. Each nucleotides phosphate group attaches to the sugar of the next nucleotide, forming a sugar-phosphate backbone. Adenine binds with Thymine through two hydrogen bonds, cytosine pairs with Guanine through three hydrogen bonds. The 5’ end of DNA has a phosphate group attached to the fifth carbon of the sugar, while the 3’ end has a hydroxyl group attached to the third carbon of the sugar. One strand runs in the 5’ to 3’ direction, while the other runs in the opposite direction, from 3’ to 5’.

Question 4

Tell what is meant by “semi-conservative replication”. How did Meselson & Stahl use bacteria and heavy N to determine that DNA replication is semi-conservative, not conservative?

Answer 4

Semi conservative means that each new DNA molecule formed during replication has an original strand of DNA and a newly synthesized strand They added heavy bacteria to a medium with normal nitrogen and allowed them to replicate. From there, they extracted the new DNA and observed that the resulting DNA had an intermediate density - indicating one strand was heavy and one strand was light.

Question 5

3-5. Describe the steps of DNA replication beginning at the “origin of replication” site, including the role of these major enzymes (helicase, primase, DNA polymerase, ligase). Include the role of primers for starting, the use of triphosphate nucleosides as building blocks (& energy), and addition of new nucleotides to the 3’ end of the growing strand.

Answer 5

Helicase unwinds the helix. Primase makes the short single stranded primer. DNA polymerase adds nucleotides to extend the primer. DNA polymerase adds the correct complementary base to the free 3’ OH group of a chain and adds the 5’ phosphate of the new nucleotide to the 3’ OH group on the sugar of the last nucleotide. Triphosphate nucleosides are used for energy when the DNA polymerase cleaves off the outer two phosphate groups - allowing for energy that fuels the condensation reaction of the OH and phosphate groups.

Question 6

3-6. Why does DNA’s anti-parallel arrangement require a difference between replication processes at the leading & lagging strands? What are Okazaki fragments? Why is DNA ligase needed only in replication on the lagging strand?

Answer 6

The antiparallel arrangement of DNA means that the two strands of the double helix run in opposite directions: one runs in the 5’ to 3’ direction, while the other runs in the 3’ to 5’ direction. DNA polymerase, the enzyme responsible for synthesizing new DNA, can only add nucleotides in the 5’ to 3’ direction. Since the leading strand runs in the 3’ to 5’ direction relative to the replication fork, DNA polymerase can consciously synthesize a new strand in the 5’ to 3’ direction as the replication fork opens. The lagging strand runs in the 5’ to 3’ direction relative to the replication fork, opposite to the direction in which DNA polymerase synthesizes DNA. As a result, replication on this strand is discontinuous, The enzyme must work in short stretches moving backwards as the replication fork opens further, leading to the formation of small, discontinuous segments called Okazaki fragments. DNA ligase is the enzyme that joins these fragments together by sealing the breaks (nicks) between them.

Question 7

3-7. Name two ways in which RNA differs chemically from DNA.

Answer 7

RNA contains ribose, while DNA contains deoxyribose, and RNA uses the base uracil (U) in place of thymine (T), which is found in DNA.

Question 8

3-8. RNA polymerase always begins transcription of a gene at a DNA sequence called what? It will stop transcription at a place on the DNA known as what? Compare/contrast each pair of terms: (a) action of DNA polymerase vs RNA polymerase action, (b) primer vs promoter; (c) replication vs transcription

Answer 8

RNA polymerase always begins transcription of a gene at a DNA sequence called the promoter. RNA polymerase will stop transcription of a gene at a DNA sequence called the terminator.
DNA polymerase synthesizes a new DNA strand during DNA replication, using a DNA template. It requires a primer to start synthesis. RNA polymerase synthesizes RNA from a DNA template during transcription without needing a primer. It only transcribes certain gene sequences rather than the entire genome. A Primer is a short sequence of RNA or DNA that provides a starting point for DNA synthesis during replication. A promoter is a DNA sequence that signals the start of transcription, where RNA binds to initiate RNA synthesis. Replication is the process of copying an entire DNA molecule to create two identical DNA strands, transcription is the process of synthesizing an RNA molecule from a DNA template, producing mRNA that carries genetic instructions from DNA to be translated into proteins.

Question 9

3-9. Explain what is meant by a “triplet code”. Examine the mRNA codon chart and explain what is meant by the term “redundant code”. From a mRNA sequence, how do you use the codon chart to translate the mRNA into a protein? NEW VOCAB

Answer 9

The triplet code refers to the fact that each set of three nucleotides in mRNA, called a codon, codes for one specific amino acid in a protein. The redundant code means that multiple codons can code for the same amino acid.

Question 10

3-10. Tell how a point mutation like a base substitution or insertion/deletion can affect the resulting protein. What’s an example of a frameshift mutation? Distinguish silent mutation, missense mutation, and nonsense mutations

Answer 10

A point mutation involves a change in a single nucleotide in the DNA sequence. Base Substitution: A single nucleotide is replaced by another. This may result in:
Silent mutation: No change in the protein sequence due to redundancy in the genetic code (e.g., both GAA and GAG code for glutamic acid).
Missense mutation: A change in one amino acid in the protein (e.g., sickle cell disease is caused by a missense mutation where glutamic acid is replaced by valine in hemoglobin).
Nonsense mutation: The substitution introduces a premature stop codon, leading to a truncated and usually nonfunctional protein.
Insertion/Deletion: A nucleotide is added or removed. If the insertion/deletion is not in a multiple of three nucleotides, it causes a frameshift mutation.

Question 11

3-11. In what three ways do eukaryotic cells modify the messenger RNA (pre-mRNA) after transcription? Tell the main function of the 3’ poly-A tail and the 5’ cap

Answer 11

Addition of a 5’ Cap: Protects the mRNA from degradation, aids in the export of mRNA from the nucleus to the cytoplasm, and Facilitates binding of the mRNA to the ribosome for translation.
Addition of a 3’ Poly-A Tail: Protects mRNA from degradation, and helps in the regulation of translation and facilitates ribosome attachment.
RNA splicing: Allows for alternative splicing, where different combinations of exons are joined, leading to the production of multiple proteins from a single gene.

Question 12

3-12. What are introns and exons? What complex of proteins & snRNA’s splices introns out of the mRNA? Does this splicing occur in the nucleus or in the cytosol? Is mRNA splicing a form of mutation?

Answer 12

Introns: Non-coding sequences of DNA and RNA that are transcribed into pre-mRNA but are removed during RNA splicing. Exons: Coding sequences of DNA and RNA that are transcribed into pre-mRNA and retained in the mature mRNA. They are translated into the amino acid sequence of a protein. The spliceosome is the complex of proteins and small nuclear RNAs (snRNAs) responsible for splicing introns out of pre-mRNA. The spliceosome consists of small nuclear ribonucleoproteins (snRNPs, pronounced “snurps”). RNA splicing occurs in the nucleus of eukaryotic cells before the mRNA is exported to the cytosol for translation. NOT A MUTATION.

Question 13

3-13. Distinguish the functions of each of the following types of RNA: rRNA, tRNA, mRNA, snRNA. Remember that each is made in the nucleus by transcription from the DNA.

Answer 13

rRNA: Forms the core structural and functional components of ribosomes, the molecular machines that synthesize proteins. Role in Translation: Provides a scaffold for ribosome assembly. Catalyzes the formation of peptide bonds between amino acids (ribosome acts as a ribozyme).
tRNA: m Transfers specific amino acids to the ribosome during protein synthesis based on the mRNA codons. Key Features: Contains an anticodon that pairs with a complementary mRNA codon. Carries the corresponding amino acid attached to its 3’ end. Ensures the correct amino acids are incorporated into the growing polypeptide chain.
nRNA: Carries the genetic instructions from DNA to the ribosome for protein synthesis.Key Features: Contains codons, which are sequences of three nucleotides that specify an amino acid. Role in Translation: Serves as a template for the assembly of a polypeptide chain in a ribosome.
sRNA: Plays a critical role in the splicing of pre-mRNA. Key Features: Combines with proteins to form small nuclear ribonucleoproteins (snRNPs) within the spliceosome. Role in RNA Processing: Catalyzes the removal of introns and the joining of exons in pre-mRNA during splicing.

Question 14

3-14. What is the general function of a tRNA? How does a tRNA become bonded to its specific amino acid? Does this occur at the mRNA or before the tRNA arrives at the mRNA? How do the mRNA codon and the tRNA anticodon interact with each other?

Answer 14

tRNA’s general function is to carry specific amino acids to the ribosome during protein synthesis, ensuring the correct sequence of amino acids in the protein. Each tRNA becomes bonded to its specific amino acid through an enzyme called aminoacyl-tRNA synthetase before it interacts with mRNA. This process, known as aminoacylation, occurs in the cytoplasm before the tRNA is brought to the ribosome. During translation, the tRNA’s anticodon, a three-nucleotide sequence, pairs with the complementary mRNA codon, ensuring the correct amino acid is added to the growing polypeptide chain.

Question 15

3-15. What is translation? How does translation begin? Tell what happens in translation elongation to synthesize the protein. What happens during translation when the mRNA stop codon appears on the ribosome? How could the same mRNA be translated multiple times?

Answer 15

Translation is the process by which the genetic code in mRNA is converted into a sequence of amino acids, ultimately synthesizing a protein. It begins when the small ribosomal subunit binds to the 5’ end of the mRNA and the initiator tRNA carrying methionine binds to the start codon (AUG) on the mRNA. This sets the stage for elongation, where the ribosome moves along the mRNA, translating each codon into its corresponding amino acid. During elongation, incoming charged tRNAs with complementary anticodons pair with the mRNA codons, and peptide bonds form between amino acids, elongating the polypeptide chain. When a stop codon (UAA, UAG, UGA) appears in the ribosome’s A site, it triggers the release of the newly synthesized protein and the disassembly of the ribosomal complex. The same mRNA can be translated multiple times because after translation termination, the mRNA can be re-captured by another ribosome and reinitiated, allowing multiple rounds of protein synthesis from the same transcript.

Question 16

3-16. Bacteria can have simultaneous transcription and translation of the same mRNA molecule. Why can’t that occur in eukaryotes?

Answer 16

In bacteria, simultaneous transcription and translation of the same mRNA molecule can occur because both processes happen in the cytoplasm and are not spatially separated. As the bacterial RNA polymerase transcribes the mRNA, ribosomes can immediately attach to the mRNA and begin translating it into protein. In contrast, in eukaryotes, transcription occurs in the nucleus, where the mRNA is synthesized and processed (e.g., capping, splicing, and polyadenylation) before it is transported to the cytoplasm for translation.

Question 17

3-17. A bacterial operon consists of what? Are human genes organized in operons? Bacterial operons with 1 regulatory region and 3 protein-coding genes will have how many promoters? Explain how bacteria can make one mRNA molecule that contains the code for several proteins.

Answer 17

A bacterial operon consists of a regulatory region (including a promoter and an operator) and one or more protein-coding genes that are transcribed together as a single mRNA molecule. Human genes are not organized in operons; each gene typically has its own promoter and is transcribed individually. A bacterial operon with one regulatory region and three protein-coding genes will have one promoter, as the genes are transcribed together as a single mRNA. This allows bacteria to produce multiple proteins from a single mRNA transcript through a process called polycistronic transcription, where the mRNA is translated into several proteins that are often involved in a related metabolic pathway or cellular process, ensuring coordinated gene expression.

Question 18

3-18. Genes in a certain bacterial operon are generally turned off, except when a certain food source is present when they turn on. Is this a repressible operon or an inducible operon? Name a specific example of this type.

Answer 18

This is an inducible operon, where genes are normally turned off but can be turned on in response to the presence of a specific molecule, often a food source. In an inducible operon, the presence of the inducer molecule (such as a nutrient) causes the repressor to be inactivated, allowing gene expression. A specific example of an inducible operon is the lac operon in E. coli, which is turned on when lactose is present. In the absence of lactose, a repressor protein binds to the operator region, blocking transcription. When lactose is available, it binds to the repressor, causing it to release from the operator and allowing transcription of the genes involved in lactose metabolism.

Question 19

3-19. The trp operon genes code for enzymes in a pathway that makes the amino acid tryptophan from a precursor molecule. What are two different ways (one enzymatic and one genetic) a bacterial cell can turn off making tryptophan? Why is the trp operon a good example of end-product feedback inhibition for both enzyme & gene activity?

Answer 19

A bacterial cell can turn off tryptophan production through two mechanisms:

Enzymatic Regulation (Feedback Inhibition): When tryptophan accumulates to high levels, it directly inhibits the activity of the first enzyme in the tryptophan biosynthesis pathway. This is an example of feedback inhibition, where the end product (tryptophan) binds to the enzyme and inhibits its activity, preventing further production of tryptophan from the precursor.

Genetic Regulation (Repression of the trp Operon): The trp operon is regulated by a repressor protein that is normally inactive. When tryptophan levels are high, tryptophan itself binds to the repressor, activating it. The active repressor then binds to the operator region of the trp operon, blocking RNA polymerase from transcribing the operon’s genes, thus preventing the synthesis of the enzymes needed to make tryptophan.

The trp operon is a good example of end-product feedback inhibition because both the enzyme activity (through feedback inhibition) and gene activity (through repression of transcription) are regulated by the end product, tryptophan, ensuring that the cell only produces tryptophan when it is needed, thus conserving resources.

Question 20

3-20. What is the structure of a eukaryotic chromosome - how is DNA arranged with histone proteins?

Answer 20

The structure of a eukaryotic chromosome involves DNA wrapped around histone proteins to form a highly organized and compact structure. DNA is coiled around histone proteins to form nucleosomes, which resemble “beads on a string.” Each nucleosome consists of a segment of DNA wrapped around an octamer of histone proteins (two copies of histones H2A, H2B, H3, and H4). This arrangement helps to package the long DNA molecules into a more compact form, enabling them to fit within the nucleus. The nucleosomes are further coiled and folded to form higher-order structures, eventually resulting in the formation of a chromosome. In addition to histones, other proteins, like histone H1, help to stabilize the structure. This compacting of DNA allows it to be efficiently stored, replicated, and transmitted during cell division.

Question 21

3-21. In each pair of items, which DNA is more likely to be transcribed and why?
(a) DNA in tightly-packed region of chromatin. or DNA in loose, unwound chromatin.
(b) DNA with nucleotide bases that are methylated. or DNA that is un-methylated;
(c) DNA that is wrapped closely to histone or DNA that is not;
(d) DNA that is associated with acetylated histone or DNA with non-acetylated histone.

Answer 21

(a) DNA in loose, unwound chromatin is more likely to be transcribed because loose chromatin (also known as euchromatin) is more accessible to the transcriptional machinery, while tightly-packed chromatin (heterochromatin) is more condensed and generally transcriptionally inactive.

(b) DNA that is un-methylated is more likely to be transcribed because DNA methylation, particularly in promoter regions, typically represses gene expression by preventing transcription factors from binding and activating transcription.

(c) DNA that is not wrapped closely to histones is more likely to be transcribed because DNA that is loosely associated with histones (e.g., euchromatin) is more accessible to transcriptional machinery than DNA that is tightly wrapped around histones, which is more likely to be in a repressed state.

(d) DNA associated with acetylated histones is more likely to be transcribed because acetylation of histones reduces their positive charge, decreasing their affinity for the negatively charged DNA and thereby loosening the chromatin structure, making it more accessible for transcription. Non-acetylated histones are associated with more condensed chromatin and generally repress transcription.

Question 22

3-22. In eukaryotic cells, RNA polymerase cannot begin transcription at the promoter until what proteins are there bound to the DNA in the region of the promoter?

Answer 22

In eukaryotic cells, RNA polymerase cannot begin transcription at the promoter until certain transcription factors are bound to the DNA in the region of the promoter. Specifically, general transcription factors (GTFs) must first assemble at the promoter to form the transcription initiation complex. One of the key GTFs is TFIID, which includes the protein TBP (TATA-binding protein) that binds to the TATA box (a common sequence found in many promoters). Other factors, like TFIIB, TFIIE, TFIIH, and TFIIA, then bind to the complex, helping to recruit RNA polymerase II to the promoter. Once RNA polymerase II is properly positioned, it can begin transcription by unwinding the DNA and synthesizing the RNA strand.

Question 23

3-23. Eukaryotic genes that will be transcribed a great deal have enhancer DNA regions. What is the role of the enhancer? Explain how transcription factors and activator proteins interact with promoter and enhancer regions of the DNA to initiate the process of transcription (major control of eukaryote gene expression)

Answer 23

Enhancers are regulatory DNA sequences found far from the core promoter of eukaryotic genes that play a key role in increasing the transcription of those genes. Enhancers contain binding sites for transcription factors and activator proteins, which are essential for the initiation of transcription. These proteins bind to the enhancer region, and through DNA looping, they interact with the promoter region of the gene, where the RNA polymerase and general transcription factors are located.

Transcription factors bind to both the promoter region (near the gene) and the enhancer region (which can be located upstream, downstream, or even within introns of the gene). Activator proteins, which are a type of transcription factor, bind to enhancers and help increase the efficiency of transcription. The binding of these activators to the enhancer region allows the recruitment of coactivators and the formation of the transcription initiation complex, including RNA polymerase. This complex then facilitates the unwinding of the DNA and the synthesis of RNA.

Overall, the interaction between transcription factors, activator proteins, and the promoter and enhancer regions is a crucial mechanism in the major control of gene expression in eukaryotes. It enables precise regulation of gene activity, allowing cells to control which genes are transcribed based on environmental cues, developmental stages, and cellular needs.

Question 24

3-24. If (almost) all cells in a person’s body have exactly the same genes, then how do various cells differentiate (i.e., become specialized) into different cell types in different body tissues? Briefly describe some specific mechanisms that determine eukaryotic cell specialization, including tissue-specific gene expression.

Answer 24

Although almost all cells in a person’s body have the same genes, they differentiate into specialized cell types through differential gene expression, where certain genes are turned on or off depending on the cell type and its function. The key mechanisms for cell specialization include:

Transcription Factors and Regulatory Proteins: Specific transcription factors are activated in different cell types and bind to promoter and enhancer regions of genes, promoting or repressing gene expression. These factors help drive the expression of genes specific to particular cell functions, such as muscle, nerve, or skin cells.

Epigenetic Modifications: Changes in chromatin structure through DNA methylation and histone modification can silence or activate genes without altering the DNA sequence. For example, in a liver cell, genes responsible for metabolizing toxins may be actively transcribed, while they are silenced in a muscle cell.

RNA Processing: Alternative splicing allows a single gene to produce different mRNA variants, leading to the production of different proteins in different tissues. This adds another layer of regulation for tissue-specific protein production.

Non-coding RNAs: Small RNAs, such as microRNAs, can regulate gene expression post-transcriptionally by binding to mRNA molecules and preventing their translation or promoting their degradation, contributing to cell-specific gene expression patterns.

Signaling Pathways: External signals, such as hormones or growth factors, can activate or repress transcription factors, further influencing the genes that are expressed in a particular cell type. For example, signaling pathways like the Notch or Wnt pathways help determine whether a cell becomes part of the nervous system or another tissue.

These mechanisms work together to ensure that each cell type expresses the appropriate set of genes required for its specialized function in the body.

Question 25

3-25. Tell how responses to extracellular signals can possibly lead to different responses in a cell and turning on/off specific genes.

Answer 25

Responses to extracellular signals can lead to different cellular outcomes and the activation or repression of specific genes through complex signaling pathways. These responses are mediated by cell surface receptors, intracellular signaling molecules, and transcription factors, which ultimately regulate gene expression. Here’s how this process works:

Receptor Activation: Extracellular signals, such as hormones, growth factors, or neurotransmitters, bind to specific cell surface receptors (e.g., G-protein-coupled receptors or receptor tyrosine kinases) on the plasma membrane. This binding activates the receptor and triggers a cascade of intracellular signaling events.

Intracellular Signaling: The activated receptor often triggers secondary messengers, such as cAMP, calcium ions, or inositol phosphates, which propagate the signal inside the cell. These molecules activate a series of intracellular protein kinases or other signaling proteins that amplify the signal.

Activation of Transcription Factors: The signaling cascade leads to the activation or inhibition of transcription factors, which are proteins that bind to specific DNA regions (promoters, enhancers) to either activate or repress gene transcription. For example, activated signaling pathways can activate transcription factors like AP-1, NF-κB, or STATs, which enter the nucleus and interact with DNA to turn genes on or off.

Gene Expression Changes: The changes in transcription factor activity influence the expression of specific genes, leading to cellular responses such as growth, differentiation, metabolism, or apoptosis. For instance, the binding of a growth factor might activate genes involved in cell division, while stress signals might activate genes involved in cell repair or apoptosis.

Cellular Context and Crosstalk: The same extracellular signal can have different effects depending on the cell type and the presence of other signals. For example, the same hormone might stimulate cell division in one cell type but induce differentiation in another, due to differences in the signaling pathways or transcription factors present in each cell.

In summary, extracellular signals lead to different cellular responses by activating signaling pathways that regulate the activity of transcription factors, which in turn control the expression of specific genes, tailoring the cellular response to the external stimulus.

Question 26

3-26. A nucleic acid probe is a short length of single-stranded nucleic acid color-labeled with a fluorescent tag. How can probes be used to measure gene expression (mRNA production) and locate where, within a cell or in an embryo, a certain messenger RNA is located?

Answer 26

Nucleic acid probes can be used to measure gene expression and locate mRNA by hybridizing with complementary mRNA in a sample. In in situ hybridization, the probe, labeled with a fluorescent tag, binds to its target mRNA within tissues or cells, allowing visualization of where the mRNA is expressed. In Northern blotting, probes detect specific mRNA levels in a sample, revealing gene expression by measuring the amount of bound probe. These techniques help identify both the presence and location of specific mRNAs in cells or embryos.

Question 27

3-27. Name an example of how a small, non-coding RNA can affect and even block the expression of specific genes.

Answer 27

An example of how a small, non-coding RNA can block gene expression is through RNA interference (RNAi), where small RNA molecules like microRNAs (miRNAs) or small interfering RNAs (siRNAs) bind to complementary mRNA molecules. This binding either blocks translation or induces degradation of the mRNA, preventing the gene from being expressed. For instance, a miRNA can bind to the 3’ untranslated region (UTR) of a target mRNA, leading to its degradation or inhibition of translation, thereby regulating gene expression post-transcriptionally.

Question 28

3-28. Distinguish/relate these terms: DNA molecule, double helix, gene, chromosome, sister chromatid, pair of homologous chromosomes, human genome (46 chromosomes in 23 homologous pairs).

Answer 28

DNA molecule: A long, single strand of nucleotides that contains genetic information. It is the basic unit of genetic material.
Double helix: The structure of DNA, consisting of two strands of nucleotides twisted around each other, forming a spiral shape. This is the form in which DNA is most commonly found.
Gene: A segment of DNA that encodes the instructions to make a specific protein or RNA molecule, determining traits and functions in an organism.
Chromosome: A long, organized structure made of DNA and proteins (histones) that carries genetic information. Humans have 46 chromosomes in total, organized into 23 pairs.
Sister chromatid: Two identical copies of a chromosome that are connected by a centromere, formed after DNA replication during cell division. Each chromatid contains one copy of the DNA molecule.
Pair of homologous chromosomes: A set of one chromosome from the mother and one from the father, each containing the same genes but potentially different alleles (versions of a gene).
Human genome: The complete set of genetic material in humans, consisting of 46 chromosomes (23 pairs), which includes all the genes and non-coding sequences that make up an individual’s genetic blueprint.

Question 29

3-29. Describe the eukaryotic cell cycle, and tell what happens in each phase, G1, S, G2, M. During what parts of a cell cycle (G1, S, G2, M) would a “chromosome” consist of 2 identical chromatids?

Answer 29

The eukaryotic cell cycle consists of interphase (G1, S, G2) and mitosis (M). Here’s what happens in each phase:

G1 (Gap 1): This is the first phase of interphase where the cell grows, performs normal functions, and prepares for DNA replication. The cell checks for any DNA damage before progressing to the next phase.

S (Synthesis): During this phase, DNA is replicated, meaning each chromosome is duplicated, resulting in two identical sister chromatids joined at the centromere. This ensures that each daughter cell will receive an identical set of chromosomes after division.

G2 (Gap 2): In this phase, the cell continues to grow and prepares for mitosis. The cell checks for DNA errors that may have occurred during replication and makes any necessary repairs.

M (Mitosis): This is the phase where the cell divides into two daughter cells. It includes several stages: prophase, metaphase, anaphase, and telophase, followed by cytokinesis. During mitosis, the sister chromatids are separated into two distinct cells.

A chromosome consists of two identical chromatids during the S, G2, and M phases, as the DNA has been replicated in S phase and the chromatids remain connected until they are separated during anaphase of mitosis.

Question 30

3-30. What happens in mitosis? State changes in chromosome form and movements. Relate chromosome movements to cell changes, spindle fibers (microtubules), and cytokinesis.

Answer 30

Mitosis is the process where a eukaryotic cell divides to form two genetically identical daughter cells. It involves several distinct stages:

Prophase: Chromosomes condense and become visible as distinct structures. Each chromosome consists of two sister chromatids joined at the centromere. The nuclear membrane begins to break down, and spindle fibers (microtubules) start to form from the centrosomes, extending towards the center of the cell.

Metaphase: The chromosomes align at the cell’s equator (metaphase plate). The spindle fibers attach to the centromere of each chromosome via kinetochore proteins, ensuring that each sister chromatid is connected to opposite poles of the cell.

Anaphase: The sister chromatids are pulled apart as the centromeres are split. The spindle fibers shorten, moving the chromatids toward opposite poles of the cell. This ensures that each daughter cell will receive an identical set of chromosomes.

Telophase: The chromatids (now individual chromosomes) reach the poles of the cell. A new nuclear membrane forms around each set of chromosomes, and the chromosomes begin to de-condense back into chromatin.

Cytokinesis: This is the final step where the cytoplasm is divided between the two daughter cells. In animal cells, a contractile ring of actin filaments forms and pinches the cell membrane in two. In plant cells, a cell plate forms to separate the two daughter cells.

The movements of chromosomes during mitosis are driven by the spindle fibers, which pull and push the chromosomes into their correct positions, ensuring that each daughter cell receives an identical set of chromosomes. Cytokinesis then physically divides the cell into two.

Question 31

3-31. A checkpoint in the cell cycle is a control point where “stop” and “go-ahead” signals can regulate the cycle. Name some “signals” that could affect whether the cell progresses through the cycle or not.

Answer 31

Several signals regulate whether a cell progresses through the cell cycle at specific checkpoints:

Internal Signals:

Cyclins and Cyclin-Dependent Kinases (CDKs): These proteins control progression through the cell cycle by activating or inactivating key regulatory proteins. The levels of cyclins fluctuate during the cycle and activate CDKs, which then trigger transitions between phases (e.g., G1 to S, G2 to M).
DNA Damage Response: If DNA is damaged, proteins like p53 can halt the cycle at checkpoints (e.g., G1 checkpoint) to allow for repair. If damage is irreparable, p53 can initiate apoptosis (programmed cell death).
Chromosome Attachment to Spindle: During mitosis, the spindle checkpoint ensures that all chromosomes are properly attached to the spindle fibers before the cell progresses from metaphase to anaphase. Improper attachment can halt the cycle to prevent errors in chromosome separation.
External Signals:

Growth Factors: These are signaling molecules that bind to cell surface receptors, promoting progression through the G1 phase. They activate signaling pathways that stimulate cyclin production, allowing the cell to move into S phase.
Nutrient Availability: Adequate levels of nutrients and energy are required for the cell to progress through the cycle. Insufficient nutrients may cause the cell to pause in G1 or enter a non-dividing state (G0 phase).
Cell Density: Contact inhibition is a process where crowded cells stop dividing. Cells in dense environments may receive signals to halt division, particularly in G1.
These stop and go-ahead signals ensure proper regulation of the cell cycle, allowing the cell to divide only when it is ready and in favorable conditions, and preventing errors that could lead to uncontrolled cell division (e.g., cancer).

Question 32

3-32. The cell cycle is regulated by specific control proteins called cyclins, which must be phosphorylated in order to become active. Recall how cells respond in signaling pathways: (a) distinguish these: kinase, phosphatase, cAMP, (b) give an example of signal amplification in cells.

Answer 32

(a) Kinase is an enzyme that adds phosphate groups to proteins (phosphorylation), activating or deactivating them. Phosphatase is an enzyme that removes phosphate groups from proteins (dephosphorylation), typically inactivating them. cAMP (cyclic AMP) is a secondary messenger in signaling pathways that amplifies signals inside cells, often by activating protein kinases. (b) An example of signal amplification is the G-protein coupled receptor (GPCR) signaling pathway, where the binding of a single signaling molecule (e.g., a hormone) to a receptor can activate many G-proteins, which in turn activate numerous enzymes like adenylyl cyclase, leading to the production of many cAMP molecules and amplification of the cellular response.

Question 33

3-33. What is apoptosis? Why is it called ‘programmed cell death’? Give an example of its role in normal development.

Answer 33

Apoptosis is a form of programmed cell death, a controlled and orderly process where a cell self-destructs in response to internal or external signals. It is “programmed” because the cell follows an intrinsic genetic program that triggers its death, ensuring that it occurs without damaging surrounding tissues. Apoptosis plays a critical role in normal development, such as during digit formation in embryonic development. In the development of human hands and feet, apoptosis removes the cells between developing fingers and toes, allowing them to separate and form distinct digits.

Question 34

3-34. Cancer cells are characterized by uncontrolled cell growth. Cancer results from an accumulation of mutations (many steps) within a cell line. One mutation permanently switches ON a G-protein (so this relay is always activated by GTP). Another mutation causes tyrosine kinase signal receptors to auto-phosphorylate even when there is no signal bound to the receptor. Tell how each of these could contribute to cancer.

Answer 34

Both mutations contribute to cancer by causing uncontrolled cell signaling that drives cell growth and division even in the absence of normal regulatory signals:

G-protein mutation: A mutation that permanently switches ON a G-protein, causing it to remain activated by GTP, leads to continuous activation of downstream signaling pathways. This results in persistent cell growth and division signals, bypassing normal regulatory mechanisms that would usually stop the cell cycle when appropriate. This type of mutation can push the cell into uncontrolled proliferation, a hallmark of cancer.

Tyrosine kinase receptor mutation: If a mutation causes tyrosine kinase receptors to auto-phosphorylate without a signal, the receptor is continuously active, leading to constant activation of signaling pathways involved in cell growth, survival, and division. This can cause abnormal cell proliferation and evade normal cellular controls, contributing to tumor formation and cancer progression.

In both cases, these mutations override the normal signaling pathways that regulate cell division, allowing the cells to grow uncontrollably, a defining characteristic of cancer.

Question 35

3-35. p53 is a tumor-suppressor gene. How do mutations in the p53 contribute to cancer?

Answer 35

Mutations in the p53 gene, which encodes a tumor-suppressor protein, contribute to cancer by disrupting the cell’s ability to regulate the cell cycle and respond to DNA damage. Normally, p53 acts as a “guardian of the genome” by detecting DNA damage and either initiating DNA repair, cell cycle arrest, or apoptosis (programmed cell death) if the damage is irreparable. When p53 is mutated, these critical responses are impaired. This allows damaged cells to continue dividing, accumulating more mutations, and potentially becoming cancerous. As a result, cells with mutated p53 can bypass normal checkpoints and evade apoptosis, leading to uncontrolled growth and tumor formation.

Question 36

3-36. Distinguish diploid (2n) from haploid (n) chromosome number, and tell how they fit together in a sexual reproduction life cycle that includes both meiosis and fertilization.

Answer 36

In a diploid (2n) organism, cells contain two complete sets of chromosomes, one from each parent, while in a haploid (n) organism, cells have only one set of chromosomes. In the context of sexual reproduction:

Meiosis reduces the chromosome number from diploid (2n) to haploid (n). During meiosis, a diploid germ cell (like a sperm or egg) undergoes two rounds of division to produce four haploid gametes, each with one set of chromosomes.

Fertilization occurs when two haploid gametes (sperm and egg) merge, restoring the diploid chromosome number in the zygote (2n). This results in a new organism with a complete set of chromosomes, half from each parent.

Thus, in a sexual reproduction cycle, meiosis generates haploid gametes, and fertilization restores the diploid number, ensuring genetic diversity and the proper chromosome count in the offspring.

Question 37

3-37. Describe the movement of homologous chromosomes and twin chromatids during Mitosis, Meiosis I, and Meiosis II. Compare Meiosis with Mitosis.

Answer 37

Mitosis:
Chromosome movement: Mitosis results in two identical daughter cells, maintaining the diploid chromosome number (2n).
Prophase: Chromosomes condense and become visible as sister chromatids connected by a centromere.
Metaphase: Chromosomes align at the cell equator, and spindle fibers attach to the centromeres of each sister chromatid.
Anaphase: The sister chromatids are pulled apart toward opposite poles of the cell.
Telophase: Chromatids de-condense, and two nuclear membranes form around the separated chromatids.
Cytokinesis: The cytoplasm divides, resulting in two genetically identical diploid daughter cells.
Meiosis I (Reduction division):
Chromosome movement: Meiosis I reduces the chromosome number by half, from diploid (2n) to haploid (n).
Prophase I: Homologous chromosomes pair up (synapsis) and undergo crossing over, exchanging genetic material.
Metaphase I: Homologous chromosome pairs line up at the cell equator, with spindle fibers attaching to the centromeres.
Anaphase I: Homologous chromosomes are separated and move to opposite poles. Sister chromatids remain attached.
Telophase I: Chromosomes reach poles, and the cell divides into two haploid daughter cells.
Cytokinesis: The cytoplasm divides, resulting in two haploid cells, each with one chromosome from each homologous pair.
Meiosis II (Equational division):
Chromosome movement: Meiosis II resembles mitosis but occurs in haploid cells, and its goal is to separate sister chromatids.
Prophase II: Chromosomes condense, and new spindle fibers form in both haploid cells.
Metaphase II: Chromosomes align at the equator of each haploid cell.
Anaphase II: Sister chromatids are pulled apart to opposite poles.
Telophase II: Chromatids reach the poles, and nuclear membranes reform.
Cytokinesis: The cytoplasm divides, producing four non-identical haploid gametes.
Comparison of Meiosis and Mitosis:
Mitosis results in two genetically identical diploid cells, maintaining the chromosome number.
Meiosis reduces the chromosome number by half, resulting in four genetically diverse haploid cells (gametes), which is crucial for sexual reproduction.
In summary, Mitosis ensures growth and tissue repair, while Meiosis facilitates genetic diversity through the reduction of chromosome number and the shuffling of genetic material.

Question 38

3-38. Given the 2n number for a particular cell, tell the number of cells and the number of chromosomes in each as a result of (a) mitosis and (b) meiosis.

Answer 38

Let’s assume the 2n number (diploid number) for a particular cell is 6 (so the cell has 6 chromosomes in total).

(a) Mitosis:
Starting cell: 1 cell with a diploid (2n) number of 6 chromosomes.
Result: Mitosis produces 2 genetically identical cells, each with the same diploid number (2n = 6).
Number of cells: 2
Number of chromosomes in each cell: 6
(b) Meiosis:
Starting cell: 1 cell with a diploid (2n) number of 6 chromosomes.
Result: Meiosis reduces the chromosome number by half, producing 4 genetically diverse haploid cells (n = 3), each with half the chromosome number of the original cell.
Number of cells: 4
Number of chromosomes in each cell: 3
In summary:

Mitosis: 1 cell → 2 cells, each with 6 chromosomes.
Meiosis: 1 cell → 4 cells, each with 3 chromosomes.

Question 39

3-39. Why is meiosis called “reduction division”?

Answer 39

Meiosis is called “reduction division” because it reduces the chromosome number by half. During meiosis, a diploid cell (2n) undergoes two rounds of division (Meiosis I and Meiosis II), resulting in four haploid cells (n), each containing half the number of chromosomes as the original cell. This reduction is crucial for sexual reproduction, ensuring that when gametes (sperm and egg) fuse during fertilization, the resulting zygote has the correct diploid chromosome number.

Question 40

3-40. Name something that happens to the two members of a homologous pair in meiosis that does not happen to them during mitosis.

Answer 40

In meiosis, the two members of a homologous pair undergo synapsis during Prophase I, where they physically pair up and align closely. This is followed by crossing over, where homologous chromosomes exchange genetic material, increasing genetic diversity. This process does not occur in mitosis, where homologous chromosomes do not pair or exchange genetic material.

Question 41

3-41. A cell with 40 chromosomes (20 homologous pairs) undergoes meiosis: what are the products? (give the number of cells and number of chromosomes in each)

Answer 41

If a cell with 40 chromosomes (20 homologous pairs) undergoes meiosis, the result is 4 haploid cells, each with 20 chromosomes (half of the original number). Specifically, during meiosis:

Meiosis I reduces the chromosome number by half, separating the homologous chromosome pairs, resulting in 2 cells, each with 20 chromosomes (haploid).
Meiosis II further separates the sister chromatids of each chromosome, resulting in 4 haploid cells, each with 20 chromosomes.
Thus, the products are 4 haploid cells, each containing 20 chromosomes.

Question 42

3-42. Distinguish and relate these terms: chromosome, DNA double-helix molecule, gene locus (location), allele (specific form of a gene).

Answer 42

A chromosome is a structure composed of a long strand of DNA coiled around proteins called histones. The DNA double-helix molecule refers to the spiral shape of the DNA molecule, made of two strands of nucleotides twisted around each other. A gene locus is the specific physical location of a gene on a chromosome. An allele is a variant form of a gene that can exist at a particular locus, with different alleles contributing to variations in traits. For example, the gene for eye color may have different alleles, such as a brown allele or a blue allele, located at the same locus on homologous chromosomes.

Question 43

3-43. For the following pairs of terms, define and distinguish between them: gene & allele; genotype & phenotype; homozygous & heterozygous; dominant & recessive alleles

Answer 43

Gene vs. Allele: A gene is a segment of DNA that codes for a specific trait or function, while an allele is a specific version or variant of a gene. For example, a gene might code for eye color, and the alleles could be for brown or blue eyes.

Genotype vs. Phenotype: Genotype refers to the genetic makeup of an organism, specifically the alleles it carries for a given trait (e.g., BB or Bb for eye color). Phenotype is the physical expression or observable traits resulting from the genotype (e.g., brown eyes or blue eyes).

Homozygous vs. Heterozygous: An organism is homozygous for a gene if it has two identical alleles for that gene (e.g., BB or bb). It is heterozygous if it has two different alleles for a gene (e.g., Bb).

Dominant vs. Recessive Alleles: A dominant allele is one that expresses its trait even when only one copy is present (e.g., a single brown eye allele, B, will result in brown eyes). A recessive allele only expresses its trait when two copies are present (e.g., blue eyes occur only with the genotype bb).

Question 44

3-44. Distinguish Mendel’s principle of segregation from the principle of independent assortment in terms of alleles moving into gametes. State the events of meiosis that explain each of those.

Answer 44

Mendel’s Principle of Segregation states that for any given gene, the two alleles (one from each parent) separate during the formation of gametes, so each gamete receives only one allele. This principle is explained by Meiosis I, specifically during Anaphase I, when homologous chromosomes (each containing one allele for the gene) are separated into different gametes.

Mendel’s Principle of Independent Assortment states that the alleles for different genes are inherited independently of each other, meaning the distribution of one pair of alleles does not affect the distribution of another. This is explained by the random orientation of homologous chromosome pairs during Metaphase I of meiosis. The way one pair of chromosomes lines up and separates does not influence how other pairs line up, allowing for the independent assortment of alleles for different traits.

In summary, the Principle of Segregation is explained by the separation of homologous chromosomes in Anaphase I, while the Principle of Independent Assortment is explained by the random arrangement of chromosomes in Metaphase I.

Question 45

3-45. When given the parental genotypes for one trait, e.g.—Aa mating with AA—set up a Punnett Square for the cross, showing the possible gametes and offspring genotypes. (a) From the Punnett Square, predict the ratio of offspring genotypes. (b) When provided dominance information for those alleles (simple=complete vs incomplete), predict the ratio of offspring phenotypes.

Answer 45

Given Parental Genotypes:
Aa x AA

(a) Punnett Square for Genotype Prediction:
We start by setting up the Punnett Square to show the possible gametes and offspring genotypes.

Parent 1 (Aa) can produce two types of gametes: A and a.
Parent 2 (AA) can produce only one type of gamete: A.
The Punnett Square looks like this:

A (from AA) A (from AA)
A (from Aa) AA AA
a (from Aa) Aa Aa
Offspring Genotypes:
50% AA (homozygous dominant)
50% Aa (heterozygous)
So, the genotype ratio is:

1 AA : 1 Aa
(b) Phenotype Prediction with Complete Dominance:
If the alleles exhibit complete dominance, where A is dominant over a, then both AA and Aa will show the dominant phenotype. Since A is dominant, both genotypes will produce the dominant phenotype.

100% of the offspring will have the dominant phenotype.
Phenotype Ratio:

100% Dominant Phenotype