quiz 4 deck Flashcards

1
Q

List key characteristics of genetic material

A

Must contain complex information
must replicate faithfully
must encode the phenotype
must have the capacity to vary

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2
Q

Describe how several key experiments demonstrated that DNA was the genetic material.

A

Avery Macleod and mccarty’s experiment revealed the transforming principle to be DNA ; In a very simple experiment, Oswald Avery’s group showed that DNA was the “transforming principle.” When isolated from one strain of bacteria, DNA was able to transform another strain and confer characteristics onto that second strain. DNA was carrying hereditary information

Griffith experiment demonstrated transformation in bacteria; the transforming bacteria ; In this experiment, Griffith mixed the living non-virulent bacteria with a heat inactivated virulent form. He subsequently infected mice with this mixture and much to his surprise, the mice developed pneumonia and died. Furthermore, he was able to isolate colonies of the virulent strain from these mice.

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3
Q

Outline the primary and secondary structures of DNA.

A

The primary structure of dna is held together with phosphodiester bonds while the secondary structure is held with hydrogen bonds

Primary structure of DNA is the sequence of bases in a strand (for example: TAAAGGCCATTTTGGCGTTTGTC?.) Secondary structure of DNA is the interactions between bases that allow for the formation of more complex structures. DNA’s secondary structure tends to be a double helix.

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4
Q

Describe how nucleotides are joined together in a polynucleotide strand.

A

The nucleotides join together by means of a phosphodiester bond (Fig. 4.7). The phosphodiester bond links the phosphate group of one base to –OH group on the 3-carbon of the sugar of another base.

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5
Q

Explain the antiparallel and complementary characteristics of DNA

A

DNA is made up of two strands. Each strand has a backbone made up of alternating sugars and phosphate groups. The two strands are linked by complementary nitrogenous bases. The strands are oriented in opposite directions, making the structure “antiparallel”.

Antiparallel:
The antiparallel characteristic of DNA refers to the arrangement of the two strands of the DNA molecule in opposite directions. In other words, one strand runs in the 5’ to 3’ direction while the other strand runs in the 3’ to 5’ direction. This antiparallel arrangement is essential for the complementary base pairing between the nitrogenous bases of the two strands.

Complementary:
The complementary characteristic of DNA refers to the specific pairing of the nitrogenous bases on the two strands of the DNA molecule. Adenine (A) always pairs with thymine (T), and cytosine (C) always pairs with guanine (G). This base pairing is specific and complementary, meaning that the sequence of one strand determines the sequence of the other strand. This complementary base pairing allows DNA to be replicated accurately during cell division and also provides the basis for protein synthesis.

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6
Q

Describe the different secondary structures of DNA and what factors produce these differences

A

The double helix structure is the most stable and common form of DNA, where two complementary strands of DNA are wrapped around each other in a helical shape. The stability of the double helix structure is due to the hydrogen bonding between complementary nitrogenous base pairs (A-T and C-G) and the hydrophobic interactions between stacked base pairs.

The single-stranded hairpin loop structure occurs when a single strand of DNA folds back on itself, forming a hairpin-like structure. This structure is stabilized by hydrogen bonding between complementary nucleotides on the same strand. Hairpin loops are often involved in regulating gene expression and can also form during DNA replication and repair processes.

Factors that influence the secondary structure of DNA include the sequence of nucleotides, the presence of chemical modifications (such as methylation), and environmental factors such as temperature and pH. These factors can affect the stability of the hydrogen bonds and the hydrophobic interactions between base pairs, and can also influence the formation and stability of hairpin loops.

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7
Q

List some of the genetic implications of DNA structure

A

Replication: The complementary base pairing of the two strands of DNA allows for accurate replication of genetic information during cell division.

Mutations: Changes in the DNA structure can lead to mutations, which can result in altered genetic information and potentially lead to genetic disorders.

Gene expression: The secondary structure of DNA can affect the accessibility of genes to transcription factors, which can influence gene expression and ultimately determine cell fate and function.

Genetic variation: The sequence and structure of DNA can vary between individuals, contributing to genetic variation and diversity within a population.

Genetic engineering: Knowledge of the structure of DNA has allowed for the development of genetic engineering techniques, such as gene editing and gene therapy, which have the potential to treat genetic disorders and enhance human health.

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8
Q

Describe special structures that occur within DNA and RNA

A

DNA double helix: The double helix structure is the most common and stable form of DNA, in which two complementary strands of DNA are wrapped around each other in a helical shape.

RNA secondary structures: RNA molecules can form a variety of secondary structures, including hairpin loops, stem-loop structures, and pseudoknots. These structures are stabilized by complementary base pairing within the RNA molecule.

Telomeres: Telomeres are repetitive DNA sequences located at the ends of chromosomes that protect the genetic information from degradation during cell division.

Centromeres: Centromeres are regions of DNA that play a critical role in cell division, as they are responsible for the proper separation of chromosomes during mitosis and meiosis.

Ribosomes: Ribosomes are complex structures made up of RNA and protein that facilitate the translation of genetic information from mRNA into proteins.

Histones: Histones are proteins that package DNA into chromatin, helping to regulate gene expression by controlling the accessibility of genes to transcription factors.

Introns and exons: In eukaryotic DNA, genes are often composed of both coding regions (exons) and non-coding regions (introns). Introns are spliced out of the RNA transcript before it is translated into a protein, while exons are joined together to form the final mRNA molecule

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9
Q

Explain how large amounts of DNA are packed into a cell

A

Positive supercoiling occurs when the DNA helix is over-twisted, causing it to become more tightly coiled than the relaxed state. This form of supercoiling is commonly found in prokaryotic DNA, where it helps to compact the genome into a smaller space and allows for more efficient DNA replication and transcription. Positive supercoiling can also help to stabilize DNA by reducing the strain caused by torsional forces during transcription and DNA replication.

Negative supercoiling, on the other hand, occurs when the DNA helix is under-twisted, causing it to become more loosely coiled than the relaxed state. This form of supercoiling is commonly found in eukaryotic DNA, where it helps to open up the DNA double helix structure and facilitate DNA replication and transcription. Negative supercoiling also helps to relieve the torsional stress that accumulates during DNA replication and transcription

he first level of organization involves the winding of DNA around histone proteins to form nucleosomes. Nucleosomes are composed of a histone octamer (eight histone proteins) around which approximately 147 base pairs of DNA are wrapped. The DNA-histone complex is then coiled into a fiber-like structure called chromatin.

The second level of organization involves the folding of chromatin into loop domains, which are stabilized by proteins called scaffold proteins. These loops are further compacted into larger domains, which are arranged in a non-random manner to allow for specific gene expression patterns.

The final level of organization involves the formation of chromosomes during cell division. Chromosomes are composed of tightly compacted chromatin fibers that are further condensed into a highly compact structure visible under a microscope.

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10
Q

Describe how supercoiling is a consequence of over-rotating or under-rotating a DNA helix

A

Supercoiling is a consequence of over-rotating or under-rotating the DNA helix. Over-rotation of the DNA helix leads to positive supercoiling, which causes the DNA to become more tightly coiled than the relaxed state. This can occur during cellular processes such as DNA replication and transcription. Under-rotation of the DNA helix leads to negative supercoiling, which causes the DNA to become more loosely coiled than the relaxed state. Negative supercoiling is necessary for various cellular processes such as DNA replication and transcription. Supercoiling can have both positive and negative effects on DNA structure and function, depending on the specific context and extent of the supercoiling

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11
Q

List the different types of chromatin found in eukaryotic chromosomes.

A

Euchromatin: This is the less condensed, more transcriptionally active form of chromatin. Euchromatin is often found in regions of the genome that are actively transcribed, such as protein-coding genes. Euchromatin is composed of loosely packed nucleosomes and is more accessible to DNA-binding proteins such as transcription factors.

Heterochromatin: This is the highly condensed, less transcriptionally active form of chromatin. Heterochromatin is often found in regions of the genome that are not actively transcribed, such as centromeres and telomeres. Heterochromatin is composed of tightly packed nucleosomes and is less accessible to DNA-binding proteins.

There are also two subtypes of heterochromatin:

a) Constitutive heterochromatin: This is a type of heterochromatin that is always present and highly condensed in all cells of an organism. It typically contains highly repetitive DNA sequences that do not code for genes.

b) Facultative heterochromatin: This is a type of heterochromatin that can switch between the condensed and decondensed states depending on cellular needs. For example, during development, some genes may be silenced by being packaged into facultative heterochromatin until they are needed for specific functions.

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12
Q

Describe the nucleosome structure

A

A nucleosome is the structural unit of DNA packaging in eukaryotes. A nucleosome is basically DNA segments surrounded by histone protein octamers resembling a thread coiled around a spool. A nucleosome is the fundamental unit of chromatin.

The nucleosome is the fundamental subunit of chromatin. Each nucleosome is composed of a little less than two turns of DNA wrapped around a set of eight proteins called histones, which are known as a histone octamer.

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13
Q

Outline the higher order chromatin structure found within a chromosome.

A

30-nm fiber: The 10-nm fiber of nucleosomes can be further compacted into a 30-nm fiber through a process of chromatin fiber condensation. The 30-nm fiber is thought to involve further interactions between the histone H1 and linker DNA, as well as interactions between adjacent nucleosomes.

Chromosome territories: The 30-nm fiber is then organized into a higher order structure where each chromosome occupies a specific region of the nucleus called the chromosome territory. Within the chromosome territory, the chromatin is further organized into loops of varying sizes, which are thought to be anchored to a scaffold composed of non-histone proteins. The loop domains are thought to be involved in regulating gene expression and replication timing.

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14
Q

Discuss how chromatin structure changes over time

A

During development: Chromatin structure changes dramatically during embryonic development, as cells differentiate into different tissues and cell types. The chromatin of stem cells is generally more open and accessible, allowing for greater flexibility in gene expression. As cells differentiate, the chromatin structure becomes more compact and specialized, resulting in specific gene expression patterns.

In response to environmental cues: Chromatin structure can also change in response to environmental cues, such as exposure to stress or changes in nutrient availability. For example, some genes that are normally repressed may become activated in response to stress, requiring changes in chromatin structure to allow for increased gene expression.

During DNA replication: Chromatin structure also changes during DNA replication, as the nucleosomes must be disassembled and reassembled on the newly synthesized DNA strands. This process is tightly regulated to ensure proper replication and maintenance of chromatin structure.

During cell division: Chromatin structure changes during cell division, as the chromatin must be condensed and packaged into chromosomes to ensure proper segregation to daughter cells.

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15
Q

Describe the special features of centromeres and telomeres

A

Centromeres: Centromeres are specialized regions of chromosomes that are essential for proper segregation during cell division. They are characterized by a specific DNA sequence that serves as a binding site for proteins that attach to spindle fibers, which pull the chromosomes apart during cell division. The centromere also helps to organize the chromosome into a compact and functional structure, and defects in centromere structure or function can lead to chromosomal instability and genetic disorders.

Telomeres: Telomeres are repetitive DNA sequences found at the ends of linear chromosomes. They serve to protect the chromosome ends from degradation and fusion with other chromosomes. Telomeres shorten with each round of cell division, eventually leading to cellular senescence or apoptosis. Telomerase, an enzyme that adds telomere repeats to chromosome ends, is active in some cells, such as stem cells and cancer cells, but is generally inactive in most somatic cells.

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16
Q

Describe different types of sequences found in eukaryotic chromosomes.
* Describe the characteristics of unique sequence DNA, moderately repetitive DNA, and highly repetitive DNA

A

Moderately repetitive DNA: These are short sequences that are repeated 10-1000 times in the genome. They are generally dispersed throughout the genome. Highly repetitive DNA: These are very short DNA sequences(<100bp) which are organised as long tandem repeats.

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17
Q

Outline the endosymbiotic theory (chloro and mito vs DNA)

A

The endosymbiotic theory proposes that eukaryotic cells evolved from a symbiotic relationship between two different types of prokaryotic cells: a host cell and an engulfed cell. The engulfed cell eventually became a permanent component of the host cell, forming mitochondria or chloroplasts, and providing energy through aerobic respiration or photosynthesis. This theory is supported by various lines of evidence, including the similarities between mitochondria and bacteria, the presence of circular DNA in mitochondria and chloroplasts, and the observation that both organelles divide independently of the host cell.

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18
Q

Give characteristics of the mitochondrial genome

A

five basic functions of mitochondrial genes:
respiration and oxidative phosphorylation
translation
transcription
rna processing
import of proteins into mitochondira

viable cells, no variation

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19
Q

Describe patterns of evolution in mitochondrial DNA.

A

Variation in the size of mitochondrial genomes between species is due to differences in noncoding DNA Sequences,

genes for most proteins and enzymes are found in mitochondria and are encoded by nuclear DNA

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20
Q

Explain how Meselson and Stahl demonstrated that DNA replication in E. coli occurs in a semiconservative
fashion

A

The experiment done by Meselson and Stahl demonstrated that DNA replicated semi-conservatively, meaning that each strand in a DNA molecule serves as a template for synthesis of a new, complementary strand

The Meselson Stahl Experiment. A centrifuge was used to separate DNA molecules labeled with isotopes of different densities. This experiment revealed a pattern that supports the semiconservative model of DNA replication.

Messelson and Stahl’s experiment supported the semi-conservative mode of replication. The DNA was first replicated in 14N medium which produced a band of 14N and 15N hybrid DNA. This eliminated the conservative mode of replication.

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21
Q

Predict the results that Meselson and Stahl would have obtained if replication occurred by conservative or
dispersive replication.

A

If the dispersive model of DNA replication had been correct Meselson and stahl would have observed that DNA extracted from bacterial cells following a second round of DNA replication in 14 and would have been only of hybrid density and somewhat lighter than after one round of replication

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22
Q

Outline semiconservative replication by theta and linear replication

A

in semiconservative replication by theta, the double-stranded DNA molecule unwinds at a specific origin of replication, forming a “theta” structure with a replication fork at each end. The DNA strands separate, and each serves as a template for a new complementary strand to be synthesized, resulting in two identical daughter DNA molecules. (DNA IS CIRCULAR)

In linear replication, the double-stranded DNA molecule has linear ends, and replication occurs bidirectionally from multiple origins of replication along the length of the DNA molecule. The leading strand is synthesized continuously, while the lagging strand is synthesized in short Okazaki fragments that are later joined together. The result is also two identical daughter DNA molecules.

bidirectional: two replication forks present at the end of the replication bubble, the forks process outward from the bubble in both directions until they meet
Unidirectional: only one replication fork present

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23
Q

Explain how 5’to 3’ replication occurs on the two antiparallel strands of DNA.

A

During DNA replication, the double-stranded DNA molecule is unwound, and each strand serves as a template for the synthesis of a new complementary strand. DNA polymerases can only add nucleotides to the 3’ end of a growing strand, so replication occurs in the 5’ to 3’ direction. Because the two strands of DNA are antiparallel, one strand is oriented in the 5’ to 3’ direction, while the other is oriented in the 3’ to 5’ direction. The strand that runs in the 5’ to 3’ direction is called the leading strand and is synthesized continuously by DNA polymerase III. The strand that runs in the 3’ to 5’ direction is called the lagging strand and is synthesized discontinuously in short fragments called Okazaki fragments. These fragments are later joined together by DNA ligase to form a continuous strand in the 5’ to 3’ direction.

24
Q

Summarize the different enzymes and proteins involved in bacterial replication and their functions

A

Initiator protein: binds to oigin and seperates strands of DNA to initiate replication
DNA helicase: unwinds DNA at replication fork
Single stranded binding proteins: attach to a single stranded DNA and prevent secondary structures from forming
DNA gyrase: moves ahead of the replication fork, making and resealing breaks in double helical DNA to release the torque that builds up as a result of unwinding at the replication fork

DNA primase: synthesised a short RNA primer to provide a 3’-OH group for the attachment of NDA nucleotides
DNA polymerase 3: elongates a new nucleotide stand from the 3’
DNA polymerase 1: removes primers and replaces them with DNA
DNA ligase: joins okazaki fragments by sealing breaks in the sugar-phosphate backbone of newly synthesized DNA

25
Q

Describe the mechanisms that ensure the fidelity of DNA replication

A

proofreading . DNA polymearase 1 does 3’t o 5’ exonuclease activity to remove the incorrectly paired nucleotide, if the 3’OH group not in right position to accept the next nucelotide, polymerization is stalled and this nucleotide is removed

Mismatch pair: correct errors after replication is complete; incorrect pairing causes deformity in secondary structure which is recognized and fixed by enzymes, they use methylation to know which nucleotide is incorrect

26
Q

Explain how replication is initiated once and only once at thousands of eukaryotic origins.

A

Origin recognition complex: initiator

-binds to defined origins to initiate replication
-ORC directly recruits and loads helicase onto the origin of the dsDNA

Replication LIcensing FActors
-proteins that ensures that replication only takes place one at each origin
-required at the origin before replication can be initiated
-removed after DNA has been replicated
-prevented from reinitiating replication until mitosis complete

27
Q

Explain some differences in DNA polymerases of bacteria and eukaryotes

A

Bacterial DNA polymerases have fewer subunits compared to eukaryotic DNA polymerases.

Eukaryotic DNA polymerases are more complex and have more specialized functions than bacterial DNA polymerases.

Bacterial DNA polymerases have higher error rates compared to eukaryotic DNA polymerases.

Eukaryotic DNA polymerases are more processive and can replicate longer stretches of DNA without dissociating from the template strand.

Eukaryotic DNA polymerases are more likely to have proofreading mechanisms, such as 3’-5’ exonuclease activity, to correct errors during replication.

Eukaryotic DNA polymerases are more sensitive to inhibition by certain chemotherapeutic agents than bacterial DNA polymerases.

Some eukaryotic DNA polymerases, such as DNA polymerase gamma, are specialized for replication of mitochondrial DNA, while no bacterial DNA polymerase has this function.

28
Q

Describe how nucleosomes are reassembled on newly synthesized DNA molecules

A

EUkaryotes wrapped around histoproteins forming a nucelosome

creation of nucleosome requires: -disrution of original nucleosomes on parental DNA
-redistributuion of preexisting histones on the new DNA
-the addition of newly synthesized histones to complete the formation of new nucleosomes

newly assembled octamers have old and new histones
nucleosome reassembly is facilitated by histone chaperons

29
Q

Explain how the ends of eukaryotic chromosomes are replicated.

A

Eukaryotic chromosomes have linear ends, called telomeres, which shorten with each cell division due to the end replication problem, where DNA polymerases cannot fully replicate the ends of linear DNA molecules. To overcome this, eukaryotic cells utilize telomerase, an enzyme that contains an RNA template that can extend the 3’ end of the chromosome, allowing for complete replication of the telomere.

30
Q

Explain the end of the replication problem.

A

The end replication problem refers to the challenge of fully replicating the ends of linear DNA molecules. DNA polymerases can only add nucleotides in a 5’ to 3’ direction, which means that the lagging strand of DNA being replicated cannot be fully replicated at the 5’ end. This results in the loss of a few nucleotides with each round of replication, leading to the shortening of the chromosome over time. To address this issue, organisms have evolved specialized mechanisms to maintain telomeres, which are regions of repetitive DNA at the ends of chromosomes. Telomerase, an enzyme that adds DNA to the ends of chromosomes, is one such mechanism that can counteract the shortening of telomeres and extend the lifespan of cells

telomerase serves as a primer for nucleotides to be added bc there is not enough space

31
Q

Explain how telomerase extends telomeres.

A

Telomerase is an enzyme that contains an RNA template sequence that is complementary to the repetitive DNA sequence found at the end of telomeres.

Telomerase binds to the 3’ end of the telomere overhang and uses its RNA template to synthesize new DNA onto the end of the chromosome, extending the telomere.

The enzyme then moves along the chromosome, continuing to add DNA repeats until the entire overhang has been replicated.

This process of telomere extension by telomerase allows cells to maintain the length of their telomeres and continue to divide without suffering from the effects of telomere shortening
.
While telomerase is active in embryonic cells and some adult stem cells, it is usually inactive in most adult somatic cells, leading to telomere shortening and eventual senescence or apoptosis

. However, some cancer cells can reactivate telomerase to maintain telomere length and become immortalized.
G rich 3’ overhang is extended by telomerase telomerase has an rna component with 12-22 nucleotides complementary to the g rich sequence, the c rich strand is synthesized and the small gap at end is not important since telomerase extends the chromosome at each replication

32
Q

Explain how the structure of RNA allows it to participate in a variety of cellular functions.

A

RNA is a versatile molecule that participates in a variety of cellular functions due to its unique structure. Unlike DNA, RNA is single-stranded, which allows it to fold into complex secondary and tertiary structures, enabling it to interact with other molecules in the cell. RNA also contains four different nucleotide bases, like DNA, but it includes uracil instead of thymine, allowing for unique base pairing interactions with DNA and RNA. RNA can act as a messenger between DNA and protein, as in the case of messenger RNA (mRNA), which carries genetic information from the DNA in the nucleus to ribosomes in the cytoplasm, where it is translated into protein. RNA can also act as a catalyst, as in the case of ribozymes, which are RNA molecules with enzymatic activity that can cleave other RNA molecules. Additionally, RNA plays important roles in gene regulation, splicing, and catalyzing chemical reactions, among other functions.

33
Q

Summarize the differences in structure of RNA and DNA.

A

DNA:

Double-stranded helix
Deoxyribose sugar
Bases: Adenine (A), Thymine (T), Cytosine (C), Guanine (G)
RNA:

Single-stranded
Ribose sugar
Bases: Adenine (A), Uracil (U), Cytosine (C), Guanine (G)
Additionally, RNA can have several different types, including messenger RNA (mRNA), transfer RNA (tRNA), and ribosomal RNA (rRNA), each with unique functions in the process of protein synthesis.

34
Q

List the different classes of RNA and their functions.

A

Messenger RNA (mRNA):

Transcribed from DNA and carries the genetic information from the DNA to the ribosomes, where it serves as a template for protein synthesis.
Transfer RNA (tRNA):

Carries specific amino acids to the ribosome during protein synthesis, and helps to translate the genetic information carried by mRNA into the sequence of amino acids that make up a protein.
Ribosomal RNA (rRNA):

Along with proteins, makes up the ribosome, which is responsible for the synthesis of proteins.
Small nuclear RNA (snRNA):

Involved in the processing of pre-mRNA, including splicing out introns and joining together exons to produce mature mRNA.
Small nucleolar RNA (snoRNA):

Involved in the processing of rRNA, including modification of nucleotides and cleavage of rRNA precursors.
MicroRNA (miRNA):

Involved in the regulation of gene expression by targeting specific mRNAs for degradation or by inhibiting their translation.
Long non-coding RNA (lncRNA):

Involved in diverse functions including gene regulation, chromatin remodeling, and X chromosome inactivation

35
Q

Identify the parts of a typical transcription unit

A

DNa template needs a transcription unit

Promoter: DNA sequence that the transcription apparatus binds to initiate transcription

RNA coding region: sequence of DNA nucleotide that is copied into an rna molecule

Terminator: sequence of nucleotides that signals where transciption is to end.

36
Q

Give the substrate for transcription and how it is used to create a polyribonucleotide chain

A

Transcription requires a substrate

the initiation of rna synthesis does not require a primer, because RNA has no OH group

New nucleotides are added to the 3’ end of the RNA molecule

DNA unwinds at the fron of the transcription bubble

37
Q

Compare RNA polymerases of bacteria and eukaryotes

A

RNA Polymerases in Bacteria:
-elongation stage carried out by rna polymerase core enzyme, involves the unwinding of only short regions of the DNA by rna polymerase, generates positive supercoils ahead of transcription buuble, and its rna polymerase has frequent pauses

One type of RNA polymerase (RNAP) is responsible for transcribing all types of RNA, including mRNA, tRNA, and rRNA.
Only four subunits in the RNAP enzyme.
No dedicated initiation factors required for transcription initiation.
RNAP recognizes specific DNA sequences called promoters to initiate transcription.
The sigma factor helps RNAP recognize promoters and initiates transcription.
Termination of transcription occurs through the rho-dependent or rho-independent mechanisms.
RNA Polymerases in Eukaryotes:

Three types of RNA polymerases (Pol I, Pol II, and Pol III) responsible for transcribing different types of RNA.
More complex structure with many subunits, including general transcription factors (GTFs) required for initiation.
Transcription initiation requires multiple GTFs and a complex set of regulatory proteins.
Eukaryotic RNA polymerases recognize promoters with specific sequence elements (e.g., TATA box) and regulatory sequences.
Termination of transcription occurs through a polyadenylation signal or by a mechanism involving the Rat1 and Xrn2 exonucleases.

38
Q

Describe the three major stages of transcription (bacterial)

A

Initiation:
RNA polymerase (RNAP) binds to a specific DNA sequence called the promoter region.
RNAP unwinds the DNA double helix, exposing the template strand for transcription.
Sigma factor, a protein that helps RNAP recognize promoter sequences, dissociates from RNAP after transcription initiation.

Elongation:
RNAP synthesizes a complementary RNA strand using the template DNA strand.
RNAP moves along the DNA template strand in the 3’ to 5’ direction and adds nucleotides to the 3’ end of the growing RNA chain in the 5’ to 3’ direction.
The RNA chain grows until the RNAP reaches the termination site.

Termination:
Termination of transcription occurs when RNAP reaches a specific DNA sequence called the termination site, which signals RNAP to release the RNA transcript.
In some cases, termination is aided by the Rho factor, a protein that destabilizes the interaction between RNAP and the DNA template.
The newly synthesized RNA molecule is released from the DNA template, and the DNA double helix re-forms.

39
Q

Describe consensus sequences found in a typical bacterial promoter.

A

Bacterial transcription: initiation, a part of promoter recognition the consensus sequence: is the sequences that comprise most commonlyencountered nucleotides at a specific site, sequences btwn consensus sequences vary greatly suggestion not important for promoter recognition, upstream is - x axis (-10 consensus sequence and -35)

40
Q

Explain the differences in rho-dependent and rho-independent termination in prokaryotes

A

RHO dependent terminantors: DNA sequence of terminator causes RNA polymerase to pause, RUT site: DNA sequence upstream of the terminator that serves as binding site for the RHO facto, the rho factor follows the rna polymerase and unwinds the DNA- RNA hybrid in the transition bubble, ending transcription

RHA independent terminators: contain inverted repeats that bind to eacho other when transcribed firing a hairpin, string of adenine nucleotides following the inverted repeats resulting ina string of uracils, RNA polymerase pauses at the uracils allowing for the formation of the hairpin, destabilizing the DNA-RNA pairing

https://youtu.be/ot075b_LLHg

41
Q

Summarize the general principles of bacteria transcription.

A

Bacterial transcription is the process by which the information encoded in the DNA sequence of a bacterial genome is used to generate RNA molecules.
The process is initiated when RNA polymerase, a multisubunit enzyme, recognizes and binds to a specific DNA sequence called the promoter region.
Once bound to the promoter, RNA polymerase unwinds the DNA double helix and begins synthesizing a complementary RNA molecule in the 5’ to 3’ direction.
The RNA polymerase moves along the DNA template, adding nucleotides to the growing RNA chain and synthesizing a transcript that is complementary to the template strand of the DNA.
Transcription continues until the RNA polymerase reaches a termination signal, which causes the RNA transcript to be released from the DNA template and the RNA polymerase to dissociate from the DNA.
The resulting RNA transcript can undergo further processing, such as splicing, editing, or modification, before it is translated into a protein or performs other functions within the cell.

42
Q

Describe eukaryotic promoters and how they differ from bacterial promoters.

A

Eukaryotic promoters:

Eukaryotic promoters are DNA sequences located upstream of the transcription start site that are recognized by RNA polymerase II and other transcription factors.
Eukaryotic promoters are typically more complex and larger than bacterial promoters, ranging from hundreds to thousands of base pairs in length.
Eukaryotic promoters often contain several distinct sequence elements, including the TATA box, initiator element, and downstream promoter element, which are recognized by specific transcription factors.
Eukaryotic promoters also often contain enhancers and silencers, which are DNA sequences that can positively or negatively regulate transcription from a distance.
Eukaryotic promoters are often regulated by multiple layers of control, including epigenetic modifications, chromatin remodeling, and transcription factor binding.

Bacterial promoters:
Bacterial promoters are DNA sequences located upstream of the transcription start site that are recognized by RNA polymerase and specific sigma factors.
Bacterial promoters are typically shorter and less complex than eukaryotic promoters, usually consisting of a -35 and -10 consensus sequence recognized by sigma factors.
Bacterial promoters are often regulated by specific regulatory proteins or small molecules that bind to the promoter region and modulate transcription.
Bacterial promoters can also be subject to feedback inhibition or attenuation, which allows cells to rapidly respond to changes in environmental conditions.
Bacterial promoters are generally simpler and less regulated than eukaryotic promoters, although there are some exceptions in bacterial systems that use complex regulatory networks to control gene expression.

43
Q

Describe the process of transcription in eukaryotes

A

Transcription proceeds through three main stages: initiation, elongation, and termination.

During elongation, RNA polymerase II adds nucleotides to the growing RNA chain and synthesizes a transcript that is complementary to the template strand of the DNA.

After the RNA polymerase II transcribes a short sequence of nucleotides, it pauses to allow the RNA transcript to be processed by capping, splicing, and polyadenylation factors before resuming elongation.

The completion of transcription occurs when RNA polymerase II reaches a specific termination signal, which causes the RNA transcript to be released from the DNA template and the RNA polymerase II to dissociate from the DNA.

The resulting RNA transcript is called pre-mRNA, and it undergoes further processing, including splicing, 5’ capping, and polyadenylation, to produce mature mRNA that is exported from the nucleus to the cytoplasm for translation into protein.

44
Q

Explain the nature of colinearity between genes and proteins.

A

Colinearity refers to the correspondence between the linear order of nucleotides in a gene and the linear order of amino acids in the corresponding protein.
In other words, the sequence of nucleotides in a gene determines the sequence of amino acids in the protein it encodes.
This principle is known as the genetic code, and it specifies the relationship between codons (groups of three nucleotides) in DNA or mRNA and the amino acids they encode.
The genetic code is nearly universal, meaning that it is shared by most organisms and is largely conserved across evolutionary time.
The colinearity between genes and proteins means that changes in the DNA sequence of a gene can lead to changes in the amino acid sequence of the protein it encodes.
These changes can have significant effects on the structure and function of the protein, and they can impact the phenotype of an organism.
Conversely, changes in the amino acid sequence of a protein can provide insights into the evolutionary history and relationships of different organisms, as well as the functional properties of the protein

45
Q

Define an intron and explain where introns are found

A

Introns: noncoding sequence between exons, both exons and introns are transcribed into RNA, introns are later removed by rna processing, introns tend to be longer than exons

46
Q

Summarize the different regions of a typical mRNA molecule

A

5’ untranslated region (UTR): This region is located at the 5’ end of the mRNA molecule, and contains regulatory elements that affect translation initiation and mRNA stability.

Coding sequence (CDS): This region contains the nucleotide sequence that encodes the amino acid sequence of the protein product. It is bounded by the start codon (usually AUG) at the 5’ end, and the stop codon (UAA, UAG, or UGA) at the 3’ end.
3’ untranslated region (UTR): This region is located at the 3’ end of the mRNA molecule, and contains regulatory elements that affect mRNA stability, localization, and translation termination.

Poly(A) tail: This is a string of adenine nucleotides that is added to the 3’ end of the mRNA molecule after transcription. It protects the mRNA from degradation and enhances translation efficiency.

Cap structure: This is a modified guanine nucleotide that is added to the 5’ end of the mRNA molecule after transcription. It plays a role in translation initiation and mRNA stability.

Splice sites: These are specific sequences at the boundaries between exons and introns that are recognized by the splicing machinery during pre-mRNA processing. They determine which exons are included in the final mRNA molecule.

Regulatory elements: These are short sequences of nucleotides within the UTRs or CDS that interact with proteins, RNAs, or small molecules to modulate mRNA stability, translation efficiency, or splicing

47
Q

Describe the 5’cap and its function.

A

-a guanine base with an added methyl group added to end
- additional methyl groups added to the next two adjacent nucleotides
-cap binding proteins recognize and attach to the cap
-ribosome binds to proteins and moves toward the start codon
-protects mRNA from degradation by 5’ exonucleases

48
Q

Describe the poly(A) tail, how it is added, and its function

A

-50-250 adenine added to the 3’ end of the mRNA
–protection against 3’ exonuclease degradation so no essential info is lost to transcribe a protein
-facilitates attachment of mRNA to the ribosome
-aids in stability

49
Q

Explain the types of processing that occur in pre-mRNA

A

Addition of 5’ cap: facilitates binding of ribosome to end of mRNA, increases mRNA

3’ cleavage and addition of poly A tail: increases stability of mRNA, facilitates binding of ribosome to mRNA
RNA splicing: removes noncoding introns from premrna, facilitates export of mrna to cytoplasm, allows for multiple proteins to be produced through alternative splicing
RNA editing: alters nucleotide sequence of mRNA

50
Q

Explain what RNA editing is

A

Protein coding sequence of mrna is altered after transcription
- translated amino acid sequence differs from, that encoded by the gene

  • guideRNAs: serve as a template for alterations made
51
Q

Summarize the structure of tRNAs.

A

-Modified RNA nucleotide bases
-CLoverleaf structure (an acceptor arm: contains 5’ and 3’ end of trna, this is where the amino acid attaches to the trna

-anticodon that pairs with eh corresponding codon on mRNA

52
Q

Describe how tRNAs are processed

A

Transfer RNAs (tRNAs) are transcribed from DNA in the nucleus of eukaryotic cells or in the cytoplasm of prokaryotic cells. Once transcribed, tRNAs undergo several processing steps to generate mature, functional tRNA molecules. In eukaryotes, these processing steps include the removal of extra nucleotides from the 5’ and 3’ ends of the tRNA precursor, the cleavage of the precursor at a specific site to generate the mature tRNA molecule, the addition of a CCA sequence to the 3’ end of the tRNA, and the post-transcriptional modification of specific nucleotides within the tRNA molecule. These modifications include the addition of methyl groups, thiolation of sulfur atoms, and isomerization of base pairs, among others, and are critical for the stability, structure, and function of the tRNA molecule.

53
Q

Outline the process of tRNA charging

A

Binding of tRNA to their appropriate amino acids,

Aminoacyl-tRNA syntheses: attaches an amino acid to a tRNA,

TRNA charging is the chemical reaction in which an aminoacyl-tRNA synthetase attaches an amino acid to its corresponding tRNA

54
Q

Explain how translation is initiated.

A

Five necessary components:
mRNA, Small and large subunit of ribosome, initiation factor (3 proteins), initiator tRNA, GTP

3 main steps: -Mrna binds to small subunit of ribosome, initiator tRNA binds to mRNA (codon: anticodon), large ribosomal subunit joins initiation complex

55
Q

Describe the process of elongation, by which amino acids are added to the growing polypeptide chain according
to the codons contained in the mRNA

A

After initiation and formation of the initiation complex, the ribosome moves along the mRNA in the 5’ to 3’ direction to identify the next codon.
A new aminoacyl-tRNA, carrying an amino acid specified by the codon, binds to the A site of the ribosome.

The ribosome catalyzes the formation of a peptide bond between the amino acid on the A site tRNA and the growing polypeptide chain on the P site tRNA, forming a dipeptide.

Translocation occurs, which involves the movement of the ribosome down the mRNA by one codon, shifting the A site tRNA to the P site and the P site tRNA to the E site, where it is released.

The A site is now open for the next aminoacyl-tRNA to bind, and the process of elongation repeats until the ribosome encounters a stop codon.

56
Q

Describe the process of termination that occurs when a ribosome encounters a stop codon.

A

When a ribosome reaches a stop codon (UAA, UAG, or UGA) in the mRNA, a release factor protein binds to the A site of the ribosome.

The release factor triggers hydrolysis of the ester bond between the polypeptide chain and the tRNA in the P site, releasing the completed protein.

The release factor also causes the ribosome to dissociate into its subunits, freeing the mRNA and allowing it to be translated again.

57
Q

Describe the processes of mRNA surveillance.

A

Nonsense-mediated mRNA decay (NMD):
Targets mRNAs with premature termination codons (PTCs) located upstream of the normal stop codon.
The presence of a PTC triggers the binding of Upf proteins to the mRNA.
Upf proteins recruit factors that promote degradation of the mRNA by ribonucleases.

Transfer messenger RNA (tmRNA) surveillance:
Recognizes stalled ribosomes on mRNAs lacking a stop codon.
Trans-translation occurs where tmRNA acts as both a tRNA and mRNA.
tmRNA directs the stalled ribosome to translate a peptide tag onto the nascent polypeptide and directs the ribosome to the 3’ end of the tmRNA for degradation
.
Nonstop mRNA decay:
Targets mRNAs that lack a stop codon due to a mutation or transcription error.
A ribosome encounters the 3’ end of the mRNA and continues translating until it reaches a ribosome rescue factor.
The rescue factor recruits ribonucleases to degrade the mRNA.

No-go decay (NGD):
Targets mRNAs with stall-inducing sequences (SISs) that cause ribosomes to pause or stall during translation.
Paused ribosomes are recognized by NGD factors that recruit ribonucleases to degrade the mRNA.
NGD is important for preventing the accumulation of stalled ribosomes and abnormal protein production