Additional Questions Flashcards
The selective barrier surrounding a living cell that enables the cell to concentrate nutrients, retain products, and excrete waste.
Plasma Membrane
A protein that catalyzes a specific chemical reaction.
Enzyme
The copying of one strand of DNA into a complementary RNA sequence.
Transcription
Process by which the sequence of nucleotides in an mRNA molecule directs the incorporation of amino acids into protein.
Translation
Region of DNA that controls a discrete hereditary characteristic of an organism, usually corresponding to a single protein (or set of alternative protein variants) or to a structural, catalytic, or regulatory RNA.
Gene
RNA molecule that specifies the amino acid sequence of a protein.
messenger RNA (mRNA)
The building blocks of proteins
Amino Acid
The total genetic information of a cell or organism as embodied in its complete DNA sequence.
Genome
T/F: Genes and their encoded proteins are co-linear; that is, the order of amino acids in proteins is the same as the order of the codons in the RNA and DNA.
True. Even in eukaryotes, where the coding regions of a gene are often interrupted by noncoding segments, the order of codons in the DNA is still the same as the order of amino acids in the protein.
T/F: DNA and RNA use the same four-letter alphabet.
False. The nucleotide subunits of RNA and DNA differ in two key ways.
First, the backbone in RNA uses the sugar ribose instead of deoxyribose,
which is used in DNA. Second, RNA uses the base uracil in place of the
base thymine, which is used in DNA. Three of the four bases—A, C, and
G—are the same in RNA and DNA.
In the 1940s, Erwin Chargaff made the remarkable observation that in samples of DNA from a wide range of organisms the mole percent of G [G/(A+T+C+G)] was equal to the mole percent of C, and the mole percents of A and T were equal. This was an essential clue to the structure of DNA. Nevertheless, Chargaff’s “rules” were not universal. For example, in DNA from the virus ΦX174, which has a single-stranded genome, the mole percents are A = 24, C = 22, G = 23, and T = 31. What is the structural
basis for Chargaff’s rules, and how is it that DNA from ΦX174 doesn’t
obey the rules?
In double-stranded DNA, which forms the genomes in all cellular life, G pairs with C, and A pairs with T. It is this requirement for base-pairing that necessitates that the number of Gs will equal the number of Cs, and that the numbers of As and Ts will be the same. In bulk samples of DNA, this translates into equivalent mole percents of G and C and of A and T. The virus ΦX174 does not obey the “rules” because its genome is single-stranded DNA. In the absence of a requirement for systematic base-pairing, there is no constraint on the relative amounts of G and C or of A
and T
Which of the following correctly describe the coding relationships (tem-
plate — product) for replication, transcription, and translation?
A. DNA— DNA
B. DNA — RNA
C. DNA — protein
D. RNA — DNA
E. RNA — RNA
F. RNA — protein
G. Protein — DNA
H. Protein — RNA
I. Protein — protein
A. During replication, parental DNA serves as a template for synthesis of
new DNA.
B. During transcription, DNA serves as a template for synthesis of RNA
D. RNA — DNA, called reverse transcription
E. RNA — RNA, called RNA replication, occur in the life cycles of RNA
viruses such as HIV and poliovirus.
F. During translation, RNA (mRNA) serves as the template for synthesis of
protein.
A small packet of genetic material that has evolved as a parasite on the
reproductive and biosynthetic machinery of host cells.
Virus
Organism selected for intensive study as a representative of a large group of species.
Model organism
One of the two divisions of prokaryotes, typically found in hostile environments such as hot springs or concentrated brine.
Archaea
The general term for genes that are related by descent.
Homolog
Living organism composed of one or more cells with a distinct nucleus and cytoplasm.
Eukaryote
Major category of living cells distinguished by the absence of a nucleus.
Prokaryote
T/F: Eukaryotic cells contain either mitochondria or chloroplasts, but not
both.
False. Plant cells contain both mitochondria and chloroplasts.
T/F: Most of the DNA sequences in a bacterial genome code for proteins,
whereas most of the DNA sequences in the human genome do not.
True. Bacterial genomes seem to be pared down to the essentials: most
of the DNA sequences encode proteins, a few encode functional RNAs,
a small amount of DNA is devoted to regulating gene expression, and there are very few extraneous, nonfunctional sequences. By contrast, only about 1.5% of the DNA sequences in the human genome is thought to code for proteins. Even allowing for large amounts of regulatory DNA, much of the human genome is composed of DNA with no apparent function.
The only horizontal gene transfer that has occurred in animals is from the mitochondrial genome to the nuclear genome.
False. In addition to transfers from the mitochondrial genome, there are many examples of transfers of viral genomes; for example, some 1% of the mouse genome arose from copies of a sequence that originated as the genome of the mouse mammary tumor virus. What is rare is the transfer of genes from other species.
The totality of the genetic information carried in the DNA of a cell or an
organism.
Genome
The three-dimensional structure of DNA, in which two DNA chains held
together by hydrogen bonds between the bases are coiled around one
another.
Double Helix
Information-containing element that controls a discrete hereditary characteristic.
Gene
Describes the relative orientation of the two strands in a DNA helix; the
polarity of one strand is oriented in the opposite direction to that of the
other.
Antiparallel
Two nucleotides in an RNA or DNA molecule that are held together by hydrogen bonds.
Base pair
T/F: Human cells do not contain any circular DNA molecules.
False. The human genome consists only of linear molecules, but human cells also contain thousands of mitochondrial DNA molecules, which are circular. They constitute about 1% of the DNA in a human cell.
The start of the coding region for the human f-globin gene reads
5’-ATGGTGCAC-3’. What is the sequence of the complementary strand
for this segment of DNA?
5’-GTGCACCAT-3’.
DNA isolated from the bacterial virus M13 contains 25% A, 33% T, 22% C, and 20% G. Do these results strike you as peculiar? Why or why not? How might you explain these values?
In all samples of double-stranded DNA, the numbers of As and Ts (hence their percentages) are equal since they always pair with each other. The
same is true for G and C. Results such as this one stood out as odd in the
days before the structure of DNA was known. Now itis clear that, while all
cellular DNA is double stranded, certain viruses contain single-stranded
DNA. The genomic DNA of the M13 virus, for example, is single stranded.
Human DNA contains 20% C on a molar basis. What are the mole percents of A, G, and T?
Because C always pairs with G in duplex DNA, their mole percents must be equal. Thus, the mole percent of G, like C, is 20%. The mole percents of A and T account for the remaining 60%. Since A and T always pair, each of their mole percents is equal to half this value: 30%
Full set of chromosomes of a cell arranged with respect to size, shape,
and number.
karyotype
Constricted region of a mitotic chromosome that holds sister chromatids
together.
centromere
Any one of a group of small abundant proteins, rich in arginine and
lysine, that form the primary level of chromatin organization.
histone
Structure composed of a very long DNA molecule and associated proteins that carries part (or all) of the hereditary information of an organism
chromosome
The orderly sequence of events by which a cell duplicates its contents
and divides into two.
cell cycle
Complex of DNA, histones, and non-histone proteins found in the nucleus of a eukaryotic cell.
chromatin
One of the two copies of a particular chromosome in a diploid cell, each copy being derived from a different parent.
homologous chromosome (homolog)
Beadlike structure in eukaryotic chromatin, composed of a short length of DNA wrapped around a core of histone proteins.
Nucleosome
T/F: Human females have 23 different chromosomes, whereas human males
have 24.
True. The human karyotype comprises 22 autosomes and the two sex chromosomes, X and Y. Females have 22 autosomes and two X chromosomes for a total of 23 different chromosomes. Males also have 22 autosomes, but have an X and a Y chromosome for a total of 24 different chromosomes.
T/F: In the living cell, chromatin usually adopts the extended “beads-on-a-
string” form.
False. In living cells, nucleosomes are packed upon one another to generate regular arrays in which the DNA is more highly condensed, usually in the form of a 30-nm fiber. The beads-on-a string form of chromatin is usually observed only after the 30-nm fiber has been experimentally treated to unpack it
T/F: The four core histones are relatively small proteins with a very high proportion of positively charged amino acids; the positive charge helps the histones bind tightly to DNA, regardless of its nucleotide sequence.
True. All the core histones are rich in lysine and arginine, which have basic—positively charged—side chains that can neutralize the negatively charged DNA backbone.
T/F: Nucleosomes bind DNA so tightly that they cannot move from the positions where they are first assembled.
False. By using the energy of ATP hydrolysis, chromatin remodeling complexes can catalyze the movement of nucleosomes along DNA, or even dissociate a nucleosome completely from the DNA.
In the 1950s, the techniques for isolating DNA from cells all yielded molecules of about 10,000 to 20,000 base pairs. We now know that the DNA molecules in all cells are very much longer. Why do you suppose such short pieces were originally isolated?
The DNA molecules in chromosomes are long and exceedingly thin, and therefore very fragile. The techniques in use in the 1950s, which were gentle enough for the isolation of proteins, were much too harsh for DNA. For example, the shearing force exerted by pipetting DNA–sucking it through a small aperture–was sufficient to break it into the observed small pieces. It was a major technical achievement to demonstrate that chromosomes contain a single long DNA molecule
Consider the following statement: a human cell contains 46 molecules of DNA in its nucleus. DO you agree with it? Why or why not?
The number of molecules of DNA ina human cell depends on the type of cell and its stage in the cell cycle. For the vast majority of somatic cells, there are 46 molecules of DNA (chromosomes) per cell prior to replication. After replication, but before completion of cell division, there are 92 molecules of DNA per cell.
List the three specialized DNA sequences and their functions that act to ensure that the number and morphology of chromosomes are constant from one generation of a cell to the next.
Replication origins are the specialized sequences that control the beginning of DNA replication, the process that allows chromosomes to be duplicated. Centromeres are the specialized sequences that permit one copy each of the duplicated chromosomes to be pulled into each daughter cell at cell division. Telomeres are the specialized sequences at the ends of chromosomes that allow ends to be efficiently replicated and also prevent chromosome ends from being recognized as breaks in need of repair.
Less condensed region of an interphase chromosome that stains diffusely.
Euchromatin
Form of transmission of information from cell to cell, or from parent to
progeny, that is not encoded in DNA.
Epigenetic inheritance
Difference in gene expression that depends on the location of the gene on the chromosome.
Position effect
T/F: Deacetylation of histone tails allows nucleosomes to pack together into tighter arrays, which usually reduces gene expression
True. Deacetylation increases the positive charge on the histone tails by unmasking the positive charges on lysines. The increased charge tends
to stabilize chromatin structure, perhaps by allowing the tails to interact
more strongly with the DNA.
T/F: Histone variants are often inserted into already formed chromatin
True. The variant histones are inserted into nucleosomes via a histone-
exchange process catalyzed by ATP-dependent chromatin remodeling complexes.
Why is a chromosome with two centromeres (a dicentric chromosome)
said to be unstable? Wouldn’t a back-up centromere be a good thing for a chromosome, giving it two chances to form a kinetochore and attach to microtubules during mitosis? Wouldn’t that help to ensure that the chromosome didn’t get left behind at mitosis—sort of like using a belt and
braces to keep your pants up?
A dicentric chromosome, containing two centromeres (kinetochores), is inherently unstable. Normally, microtubules from opposite spindle poles attach to a single kinetochore to separate chromatids during mitosis. With two kinetochores, there’s a 50% chance of proper attachment, allowing normal division. However, in the other 50% of cases, microtubules attach to kinetochores on different chromatids, causing each to be pulled in opposite directions, resulting in chromosome breaks and instability.
Giant chromosome in which the DNA has undergone repeated replication without separation into new chromosomes.
Polytene chromosome
Paired chromosomes in meiosis in immature amphibian eggs, in which
the chromatin forms large stiff loops extending out from the linear axis of
the chromosome.
Lampbrush chromosome
Highly condensed, duplicated chromosome with the two new chromosomes still held together at the centromere as sister chromatids.
Mitotic chromosome
T/F: At the final level of condensation each chromatid of a mitotic chromosome is organized into loops of chromatin that emanate from a central axis.
True. It is thought the interphase chromosomes of all eukaryotes are
arranged in loops similar to those observed for lampbrush chromosomes.
A copy of a functional gene that has become irreversibly inactivated by
multiple mutations.
pseudogene
Long blocks of DNA sequence that differ in the number of times they are
present in the genomes of different individuals in a population.
copy number variation (CNV)
Evolutionary process that eliminates individuals carrying mutations that
interfere with important genetic functions.
Purifying selection
Variation between individuals at a certain nucleotide position in the
genome.
Single-nucleotide polymorphism (SNP)
T/F: In a comparison between the DNAs of related organisms such as humans and mice, identifying the conserved DNA sequences facilitates the search for functionally important regions.
True. Humans and mice diverged from a common ancestor long enough ago for roughly two out of three nucleotides to have been changed by random mutation. The regions that have been conserved are those with important functions, where mutations with deleterious effects were eliminated by natural selection. Other regions have not been conserved
T/F: Many human genes so closely resemble their homologs in yeast that the
protein-coding portion of the human gene will substitute for the function
of the yeast gene in yeast cells.
True. Although this statement is not true for all human genes, it is true for many. Even more remarkably, in some cases, the human gene can indeed substitute for the corresponding gene in yeast.
T/F: Gene duplication and divergence is thought to have played a critical role
in the evolution of increased biological complexity.
True. Duplication of chromosomal segments, which may include one or more genes, allows one of the two genes to diverge over time to acquire different but related functions. The process of gene duplication and divergence is thought to have played a major role in the evolution of biological complexity.
Cell type in a diploid organism that carries only one set of chromosomes
and is specialized for sexual reproduction. A sperm or an egg.
Germ cell
A randomly produced, heritable change in the nucleotide sequence of a
chromosome.
Mutation
Any cell of a plant or animal other than a germ cell or germ-line precursor.
Somatic cell
T/F: Both germ-cell DNA stability and somatic-cell DNA stability are essential
for the survival of the species.
True. DNA stability is crucial for survival; excessive mutations in somatic cells can lead to fatal conditions like cancer, while instability in reproductive cells could accumulate harmful mutations, threatening the species’ survival.
The following statement sounds patently false. “The different cells in your body rarely have genomes with the identical nucleotide sequence.” Provide an argument for why it might be true.
Each time the genome is copied for cell division, mutations may occur. In humans, the mutation rate is about one nucleotide change per 10¹⁰ nucleotides per replication. Given the 6.4 × 10⁹ nucleotides in a diploid cell, an average of 0.64 random mutations occur per genome copy. As a result, daughter cells often differ from each other and the parent cell, with additional mutations accumulating over subsequent replications.
Short length of RNA synthesized on the lagging strand during DNA replication and subsequently removed.
RNA primer
Enzyme that joins two adjacent DNA strands together.
DNA ligase
DNA repair process that replaces incorrect nucleotides inserted during DNA replication.
Strand-directed mismatch repair
Enzyme that opens the DNA helix by separating the single strands.
DNA helicase
A protein complex that encircles the DNA double helix and binds to DNA
polymerase, keeping it firmly bound to the DNA while it is moving.
Sliding clamp
Enzyme that binds to DNA and reversibly breaks a phosphodiester bond
in one or both strands, allowing the DNA to rotate at that point.
DNA topoisomerase
Y-shaped region of a replicating DNA molecule at which the two daughter strands are formed
Replication fork
The newly made strand of DNA found at a replication fork that is made in discontinuous segments, which are later joined covalently
Lagging strand
T/F: When read in the same direction (5’-to-3’), the sequence of nucleotides in a newly synthesized DNA strand is the same as in the parental strand used as the template for its synthesis.
False. The nucleotide sequence of a newly synthesized strand is not identical to the parental template strand but is instead complementary to its 3’-to-5’ sequence.
T/F: Each time the genome is replicated, half the newly synthesized DNA is
stitched together from Okazaki fragments.
True. At each replication fork, the leading strand is synthesized continuously, while the lagging strand is synthesized in Okazaki fragments. Since half of the DNA at each fork is assembled from these fragments, half of the genome is synthesized this way.
T/F: E. coli, where the replication fork travels at 500 nucleotide pairs per
second, the DNA ahead of the fork—in the absence of topoisomerase—would have to rotate at nearly 3000 revolutions per minute.
True. If the replication fork moves at 500 nucleotide pairs per second, the DNA ahead must rotate at 48 revolutions per second (or 2880 per minute). To prevent strain on the chromosome, DNA topoisomerase introduces transient nicks ahead of the fork, allowing controlled rotation within a short single-stranded DNA segment.
T/F: Topoisomerase I does not require ATP to break and rejoin DNA strands
because the energy of the phosphodiester bond is stored transiently in a
phosphotyrosine linkage in the enzyme’s active site.
True. When topoisomerase I cleaves DNA, it stores the energy of the backbone phosphodiester bond in a phosphotyrosine bond with the enzyme, which is then used to re-form the phosphodiester bond in DNA.
How would you expect the loss of the 3’-to-5’ proofreading exonuclease
activity of DNA polymerase in E. coli to affect the fidelity of DNA synthesis? How would its loss affect the rate of DNA synthesis? Explain your reasoning.
Loss of the proofreading exonuclease activity of DNA polymerase would compromise the fidelity of DNA synthesis, reducing accuracy by a factor of about 100 in E. coli. This loss could also impact the rate of DNA synthesis, as misincorporated nucleotides, normally corrected by proofreading, might cause the polymerase to pause or stall. However, since misincorporation occurs at a very low frequency (about 1 in 10⁶, detecting a rate change in practice could be challenging.
Discuss the following statement: “Primase is a sloppy enzyme that makes
many mistakes. Eventually, the RNA primers it makes are replaced with DNA made by a polymerase with higher fidelity. This is wasteful. It would be more energy-efficient if a DNA polymerase made an accurate copy in the first place.”
Although the process may seem wasteful, it effectively overcomes the challenge of proofreading during primer formation. Short oligonucleotides bind weakly to single-stranded DNA, making it difficult to distinguish correct from incorrect bases. Primase prioritizes getting a primer in place without concern for accuracy, as these sequences are later removed and replaced by DNA polymerase. Unlike primase, DNA polymerase extends an existing strand, allowing it to firmly hold new nucleotides and accurately assess base pairing. As it fills the gap, it can proofread from the start, making the initial inefficiency a necessary trade-off for accuracy.
SSB proteins bind to single-strand DNA at the replication fork and prevent the formation of short hairpin helices that would otherwise impede
DNA synthesis. What sorts of sequences in single-strand DNA might be able to form a hairpin helix?
Single-strand binding (SSB) proteins prevent the formation of short hairpin helices in single-strand DNA at the replication fork. Hairpin helices can form in single-strand DNA if it contains sequences that are complementary to each other. These complementary regions can fold back and base-pair, creating a stem-loop or hairpin structure.
DNA repair enzymes preferentially repair mismatched bases on the
newly synthesized DNA strand, using the old DNA strand as a template. If mismatches were instead repaired without regard for which strand served as template, would mismatch repair reduce replication errors? Would such a mismatch repair system result in fewer mutations, more mutations, or the same number of mutations as there would have been
without any repair at all? Explain your answers.
Mismatch repair corrects errors in the newly synthesized strand using the parental strand as a reference. If the repair process instead used the erroneous new strand as a template, the mistake would become permanent, erasing the correct sequence. Without the ability to distinguish between strands, repair enzymes would correct replication errors only 50% of the time, introducing as many mutations as if mismatch repair did not exist. In the absence of repair, mismatches persist until replication, leading to a 50% chance of mutant or nonmutant progeny—an outcome similar to indiscriminate repair.
At the completion of replication of the circular genome of the animal
virus SV40, the two daughter circles are interlocked like links in a chain. How do you suppose such interlinked molecules might then separate?
The enzyme topoisomerase II is responsible for unlinking SV40 daughter duplexes. Topoisomerase II introduces a transient double-strand break into one circle, and then guides the second duplex through the first before it reseals the break.
Period during a eukaryotic cell cycle in which DNA is synthesized.
S phase
Large multimeric protein structure that is bound to the DNA at origins of
replication in eukaryotic chromosomes throughout the cell cycle.
Origin recognition complex (ORC)
Special DNA sequence on a bacterial or viral chromosome at which DNA
replication begins.
Replication origin
Enzyme that elongates telomeres, the repetitive nucleotide sequences
found at the ends of eukaryotic chromosomes.
Telomerase
T/F: In a replication bubble, the same parental DNA strand serves as the template strand for leading-strand synthesis in one replication fork and as
the template for lagging-strand synthesis in the other fork.
True
T/F: When bidirectional replication forks from adjacent origins meet, a leading strand always runs into a lagging strand.
True. For a single template strand with its 5’ end on the left and 3’ end on the right, synthesis to the left is continuous (leading strand), while synthesis to the right is discontinuous (lagging strand). When replication forks from adjacent origins collide, a rightward-moving lagging strand will always meet a leftward-moving leading strand.
T/F: If an origin of replication is deleted from a eukaryotic chromosome, the DNA on either side will ultimately be lost, as well, because it cannot be
replicated.
False. If one origin is deleted, the adjacent DNA will still be replicated from a neighboring origin. While replication of this region may be delayed, it will still take place.
Which one of the following statements about the newly synthesized
strand of a human chromosome is correct?
A. It was synthesized from a single origin solely by continuous DNA synthesis.
B. It was synthesized from a single origin solely by discontinuous DNA synthesis.
C. It was synthesized from a single origin by a mixture of continuous and discontinuous DNA synthesis.
D. It was synthesized from multiple origins solely by continuous DNA synthesis.
E. It was synthesized from multiple origins solely by discontinuous DNA
synthesis.
F. It was synthesized from multiple origins by a mixture of continuous and
discontinuous DNA synthesis.
G. It was synthesized from multiple origins by either continuous or discontinuous DNA synthesis, depending on which specific daughter chromosome is being examined.
F. Each newly synthesized strand in a daughter duplex was synthesized by a mixture of continuous and discontinuous DNA synthesis from multiple origins. Consider a single replication origin. The fork moving in one direction synthesizes a daughter strand continuously as part of leading-strand synthesis. The fork moving in the opposite direction synthesizes a portion of the same daughter strand discontinuously as part of lagging-strand synthesis.
A means for repairing double-strand DNA breaks that links two ends with
little regard for sequence homology.
Nonhomologous end joining
Collective term for the enzymatic processes that correct deleterious changes affecting the continuity or sequence of a DNA molecule.
DNA repair
T/F: DNA repair mechanisms all depend on the existence of two copies of the genetic information, one in each of the two homologous chromosomes.
False. Single-strand damage repair, such as base excision or nucleotide excision repair, relies on the two strands of the DNA double helix. However, precise repair of double-strand breaks requires information from a second DNA duplex, such as a sister chromatid or homolog.
T/F: Spontaneous depurination and the removal of a deaminated C by uracil DNA glycosylase leave identical substrates, which are recognized by AP
endonuclease.
True. Both spontaneous depurination and removal of deaminated C by uracil DNA glycosylase leave a sugar that is missing its base, which is the substrate recognized by AP endonuclease.
T/F: Only the initial steps in DNA repair are catalyzed by enzyme that are unique to the repair process; the later steps are typically catalyzed by enzymes that play more general roles in DNA metabolism.
True. The initial steps of DNA repair, such as damage recognition and incision, are repair-specific, while later steps involve enzymes like helicases, DNA polymerases, and ligases, which are commonly used in DNA metabolism.
Discuss the following statement: “The
DNA repair enzymes
that correct damage introduced by deamination and depurination must preferentially recognize such defects on newly synthesized DNA strands.”
The statement is incorrect. DNA defects from deamination and depurination occur spontaneously, not as replication errors, making them equally likely on either strand. If repair enzymes targeted only newly synthesized strands, half of these defects would remain uncorrected. Furthermore, there is no inherent link between these repair processes and replication, as the altered bases are distinct from normal ones and can be recognized in any sequence context. In contrast, replication errors involve normal bases that are mispaired, requiring identification specifically on the newly synthesized strand.
If you compare the frequency of the sixteen possible dinucleotide
sequences in the E. coli and human genomes, there are no striking differences except for one dinucleotide, 5’-CG-3’. The frequency of CG dinucleotides in the human genome is significantly lower than in E. coli and
significantly lower than expected by chance. Why do you suppose that CG dinucleotides are underrepresented in the human genome?
In vertebrate cells, cytosine in the sequence 5’-CG-3’ is selectively methylated. Spontaneous deamination of methyl-C produces thymine, which is recognized and repaired by a specific DNA glycosylase that removes the mismatched T in TG sequences. However, this repair mechanism is not fully effective, making methylated C nucleotides common mutation sites. Over time, the higher mutation rate of CG dinucleotides has led to their loss, explaining their underrepresentation in the human genome.
With age, somatic cells are thought to accumulate genomic “scars” as a
result of the inaccurate repair of double-strand breaks by nonhomologous end joining (NHEJ). Estimates based on the frequency of breaks in
primary human fibroblasts suggest that by age 70, each human somatic
cell may carry some 2000 NHEJ-induced mutations due to inaccurate repair. If these mutations were distributed randomly around the genome, how many protein-coding genes would you expect to be affected? Would you expect cell function to be compromised? Why or why not? (Assume that 2% of the genome—1.5% protein-coding and 0.5% regulatory—is crucial information.)
If inaccurately repaired breaks were randomly distributed in the genome, about 2% would disrupt crucial coding or regulatory regions, affecting around 40 genes per cell. Since gene expression varies by cell type, some mutations would have no impact. Additionally, because the human genome is diploid, the functional allele would often compensate for mutations. While in most cases, 50% of normal protein production is sufficient for cell function, some genes require full expression, meaning certain mutations could still impair cellular function.
