Exam 1 Study Questions (Ch 1, 3, 6, 7)

Question 1

What are the structural differences between deoxyribose and ribose sugars?

What are the structural differences between purine and pyrimidine nitrogenous bases? (Ch 1)

Answer 1

Ribose is in RNA. Deoxyribose is in DNA. Ribose has a hydroxyl (OH) group on the 2’ carbon, while deoxyribose has just a hydrogen (H) at the 2’ carbon.

Purines have a double-ring structure. Adenine (A) and Guanine (G) are purines. Pyrimidines have a single-ring structure. Cytosine (C), Thymine (T), and Uracil (U) are pyrimidines.

Question 2

What is supercoiling?

What is the difference between negatively and positively supercoiled DNA? (Ch 1)

Answer 2

Supercoiling is the coiling of a closed duplex DNA in space so that it crosses over its own axis. It creates tension in the DNA molecule. It occurs only in closed DNA with no free ends.

Positive supercoiling occurs if DNA is twisted in the same direction as the helix. Negative supercoiling occurs if DNA is twisted in the opposite direction as the helix. Negative supercoiling creates tension that can be relieved by denaturing the helix and promotes denaturing of helix.

Question 3

What is a point mutation?

What are the differences between transitions, transversions, indels, and frameshift mutations? (Ch 1)

Answer 3

A point mutation is a mutation that changes a single base pair.

Transition mutation: Replaces a purine with a purine or a pyrimidine with a pyrimidine. A <—> G & C <—> T(U)

Transversion mutation: Replaces a purine with a pyrimidine or vice versa. A <—> T(U) & A <—> C & G <—> T(U) & G <—> C

Indel mutation: Insertions and deletions (can lead to frameshift).

Frameshift mutation: Shifts the reading frame

Question 4

What is the relationship between DNA modification and mutation rate? (Ch 1)

Answer 4

At mutation hotspots, mutation frequency is increased by at least an order of magnitude. Many mutational hotspots result from the presence of modified bases.

Question 5

Be able to define silent mutation, null mutation, loss of function mutation, gain of function mutation, and dominant negative. (Ch 1)

Answer 5

Silent mutation: Neutral substitutions. Substitution in a protein that cause changes in amino acids but does not affect activity. Change in nucleotide but no change in protein coding.

Null mutation: Completely eliminates function of gene.

Loss-of-function mutation: Prevents the normal gene product from being produced or renders it inactive.

Gain-of-function mutation: The altered gene product possesses a new molecular function or a new pattern of gene expression.

Dominant negative mutation: Results in a protein that interferes with the normal function of the wild-type protein.

Question 6

What is a polymorphic locus? (Ch 1)

Answer 6

A polymorphic locus is a locus with multiple functional alleles.

Question 7

What is the reading frame and how is it determined? (Ch 1)

Answer 7

The reading frame is all the ways that a nucleotide sequence can be translated into polypeptide. In DNA, there are 6 possible reading frames: 3 forward and 3 reverse. In mRNA, there are 3 possible reading frames.

Usually only one of the possible reading frames is used. Others are closed by frequent termination signals.

Question 8

What is the relationship between the genetic code, codons, frameshift mutations, and the ORF? (Ch 1)

Answer 8

The genetic code is the relationship between a DNA sequence and the sequence of the corresponding polypeptide. It is read in triplet combinations of ribonucleotides called codons. The codons are nonoverlapping and are read from a fixed starting point.

The open reading frame (ORF) is a sequence of DNA consisting of triplets that can be translated into amino acids starting with an initiation codon and ending with a termination codon.

A frameshift mutation shifts the reading frame.

Question 9

What is the general structure of an mRNA? (Ch 1)

Answer 9

Untranslated 5’ region (5’ UTR, leader)
Coding region
Untranslated 3’ region (3’ UTR, trailer)

Question 10

What general steps are involved in RNA processing in eukaryotes? (Ch 1)

Answer 10

5’ cap
3’ poly A tail
removal of introns through splicing

Question 11

What is the difference between cis- and trans-acting elements? (Ch 1)

Answer 11

Cis-acting elements: Affect only the contiguous stretch of DNA.

Trans-acting elements: Can act on any copy of a gene in the cell.

Question 12

What are homologous genes?

How are the introns and exons of homologous genes related? (Ch 3)

Answer 12

Homologous genes: Genes that share a common ancestor. Should share common features that preceded their evolutionary separation.

When two genes are related, the similarities in sequence between their exons is greater than between their introns. Changes in exon sequence are constrained by selection against mutations that alter the function of the polypeptide (negative selection). Introns change more rapidly than exons.

Question 13

What will happen to the sequences of exons and introns when they are under negative or positive selection? (Ch 3)

Answer 13

Negative selection: Changes in exon sequence are constrained by selection against mutations that alter the function of the polypeptide. Exon sequence is under selective pressure to produce a polypeptide with a specific function. Introns change more rapidly than exons.

Positive selection: Under positive selection an individual with an advantageous mutation in an exon has a greater fitness relative to those without the mutation. Introns may be more highly conserved than exons if the gene is under positive selection.

Question 14

What is the relationship between complexity of a species and the structure of introns and exons? (Ch 3)

Answer 14

Introns are short in unicellular eukaryotes but can be many kb in multicellular eukaryotes. The overall length of a eukaryotic gene is determined largely by the length of its introns.

Question 15

What is alternative splicing and what are the evolutionary advantages and disadvantages associated with this process? (Ch 3)

Answer 15

Alternative splicing: Production of different polypeptides by including or excluding individual exons or choosing between alternative exons.

Question 16

Describe the relationship between interrupted gene structure and the structure of the protein encoded by the gene. (Ch 3)

Answer 16

Interrupted genes: Genes for which the coding sequence is not continuous due to the presence of introns.

Question 17

What is a gene family?

What is a gene superfamily? (Ch 3)

Answer 17

Gene family: A set of genes within a genome that encode related or identical RNA or proteins. Derived by duplication of an ancestral gene followed by accumulation of changes in sequence between the copies.

Superfamily: A set of genes all related by presumed descent from a common ancestor. Now show considerable variation.

Question 18

What are orthologous genes? (Ch 3)

Answer 18

Orthologous genes: Related genes in different species. Special type of homologue. Gene diverges after speciation.

Question 19

What are the differences between repetitive and nonrepetitive DNA?

What types of sequences would you find contained within each?

Where is each type found in the genome? (Ch 6)

Answer 19

Highly repetitive DNA has a very short repeating sequence and no coding function (Satellite DNA). Short repeating sequence units.

Satellite DNA is located in large blocks that can have distinct physical properties.

Satellite DNA is often the major constituent of centromeric heterochromatin.

Question 20

What are pseudogenes? (Ch 6)

Answer 20

Pseudogenes: Inactive but stable components of the genome derived by mutation of an ancestral active gene.

Usually inactive because of mutations that block transcription, translation, or both

Question 21

What is a gene cluster and how are they created via unequal recombination? (Ch 6)

Answer 21

Gene cluster: A group of adjacent genes that are identical or related (rRNA and histone proteins).

Duplication and expansion into clusters often begins with unequal crossover (recombination that occurs between sites that are similar but not perfectly aligned).
– Increases the number of repeats on one chromosome and decreases on the other.

Question 22

What are the similarities and differences between satellite, minisatellite, VNTR, and microsatellite regions? (Ch 6)

Answer 22

Satellite DNA: Highly repetitive DNA has a very short repeating sequence and no coding function (not transcribed).

Minisatellites: Repeating unit length is 10-100 bp. Greater number of repeats than microsatellites. Rate of crossover at minisatellite loci is 10X that of other parts of the genome.

Microsatellites: Repeating unit length is <10 bp

Variable Number of Tandem Repeats (VNTR): Microsatellites that have < 20 repeats and a 2-6 bp repeat

Microsatellite and minisatellite loci can be used for gene mapping and DNA fingerprinting

Question 23

Describe how the result of an unequal crossing over event between genes (or exons) is different than the result when unequal crossing over occurs within a gene (or exon). (Ch 6)

Answer 23

Unequal crossing over within a gene cluster, but between genes, results in:
– Increase in the number of repeats on one chromosome
– Decrease in the number of repeats on the other.

If crossing over occurs within a gene, the result depends on the similarity of the genes:
– If the genes are nearly identical, there is little change in the sequence of either gene
– If the genes are not identical, then the recombinant genes will be different from either of the original genes.

Question 24

What effect does the presence of interrupted genes have on the rate of unequal crossing over?

Why does it have this effect? (Ch 6)

Answer 24

Interrupted genes are an obstacle to unequal crossing over. Exons of adjacent genes are similar. Introns have diverged and are not similar enough for pairing.

Restriction of pairing to the exons limits the continuous length of DNA that can cross over. Lowers the chances of unequal crossing over.

Question 25

Describe the types and genomic structure of rRNA genes in prokaryotes and eukaryotes. (Ch 6)

Answer 25

rRNA is the primary product of transcriptional events in a cell (80-90% of cellular RNA is rRNA).

Number and location of major rRNA genes varies greatly in different domains of life:
Prokaryotes
– 1-10 copies
– Dispersed throughout the genome
Eukaryotes
– 100+ copies
– Clustered in specific locations
– Organized as tandem repeats
– rDNA

Question 26

What are the processing steps required to convert the 45S rRNA precursor to 28S, 18S, and 5S rRNA? (Ch 6)

Answer 26

Large rRNA precursor
– 45S rRNA tandem repeats Processed into 18S, 5.8S, and 28S rRNA
– snoRNA “guide RNAs” required for processing

1. 45S precursor rRNA
2. Chemical modification: 2’-O-methylated nucleotide and pseudouridine monophosphate
3. Cleavage and trimming of ends
4. Small subunit rRNA (18S) and large subunit rRNA (5.85S, 28S, 5S)

Question 27

How does the structure of the nucleolus change throughout the cell cycle and how is this related to the organization of rDNA containing chromosomes? (Ch 6)

Answer 27

Nucleolus: Region of nucleus where rRNA synthesis, processing, and ribosomal subunit synthesis occurs.

During interphase, the 5 chromosomes that carry the 48S large rRNA genes loop their DNA into the single nucleolus.

The nucleolus disappears during M phase.

Each 48S rRNA-containing chromosome forms a small nucleolus during telophase as the chromosomes disperse. These small nucleoli merge to form a large nucleolus during G1.

Question 28

How is variability incorporated into minisatellite loci versus how it is incorporated into microsatellite loci? (Ch 6)

Answer 28

Minisatellite variability is produced by unequal crossing over.

Microsatellite and VNTR variability is produced by intrastrand mispairing and slippage during replication.

Question 29

How can viruses combine their capsids with their nucleic acid genomes? (Ch 7)

Answer 29

Capsid: The external protein coat of a virus
particle.

The length of DNA that can be incorporated into a virus is limited by the structure of the capsid. Nucleic acid within the capsid is extremely condensed.

Two primary methods of constructing capsid with nucleic acids:
– Assemble the capsid around the nucleic acid
– Construct the capsid and load nucleic acid into the empty structure

Question 30

Describe the structure of the bacterial chromosome including specific protein components. (Ch 7)

Answer 30

Natural closed DNA is negatively
supercoiled.

The nucleoid is composed of individual domains, each of which can retain their own supercoiled status.

Bottlebrush Nucleoid Structure: Supercoiled loops emanating from a central core
– Topologically isolated loops are approximately 10kb in length

Nucleoid Associated Proteins (NAPs): Regulators of nucleoid structure and gene expression

Protein HU (NAPs): Histone-like structure. Dimer that plays a role in DNA flexibility. Often associated with regions of distorted DNA.

Protein H-NS (NAPs): Histone-like structure. Preference for AT-rich regions. Interacts with other expression- modulating proteins. Affects large-scale gene silencing and microdomain boundaries

Question 31

What are the nuclear matrix and metaphase scaffold? (Ch 7)

Answer 31

Metaphase DNA is arranged as 60 kb loops attached to a central proteinaceous scaffold.

Metaphase scaffold: A proteinaceous structure in the shape of a sister chromatid pair.

During interphase, the metaphase scaffold expands to fill the entire nucleus and is called the nuclear matrix.

Question 32

What are MARs and what is their function with respect to eukaryotic chromosome structure? (Ch 7)

Answer 32

Interphase DNA is attached to the nuclear matrix at specific DNA sequences called Matrix Attachment Regions (MARs).

The MARs are A-T rich but do not have a specific consensus binding sequence.

MARs often contain: Cis-acting transcription regulatory sites, 5’ intron sites, and Topoisomerase II recognition sites.

MARs can also bind to metaphase scaffold.

Proteins regulate association of MARs with matrix to regulate transcription.

Question 33

Describe classically defined euchromatin and heterochromatin. (Ch 7)

Answer 33

Euchromatin: Less tightly packed than mitotic chromosomes.

Heterochromatin: Remains densely packed at chromosomal density throughout interphase.

Question 34

Describe the modern classifications of chromatin into open active and closed inactive. How do these regions correspond to the classical definitions? (Ch 7)

Answer 34

Open Active Chromatin: Euchromatin. Actively transcribed genes. Minority of genome.

Closed Inactive Chromatin: Euchromatin and heterochromatin. Euchromatic portion is considered quiescent. Heterochromatic portion is subdivided into different subtypes.

Question 35

Describe the similarities and differences between facultative and constitutive heterochromatin. (Ch 7)

Answer 35

Facultative Heterochromatin: Regulated chromatin. Developmentally repressed genes. Barr bodies. (Only turned on in specific situations, almost always off).

Constitutive Heterochromatin: “Permanently” condensed. Usually replicates later in the S phase. Repeat rich. Reduced recombination frequency. Reduced gene density with low levels of transcription. (Not transcribed, packed tight all the time).

Question 36

How is the DNA associated with different chromosomes organized within the interphase nucleus into chromosome territories?

What are transcription factories and how are they related to gene expression? (Ch 7)

Answer 36

Chromosome Territories: Distinct regions of the interphase nucleus associated with chromatin from specific chromosomes.

Chromosomes are not entangled within territories. They do interact at periphery of territories. Homologues are separated in the nucleus.

Heterochromatin found at periphery of nucleus. Gene dense regions found towards center.

Active genes often found at territorial borders, clustered together in interchromosomal spaces enriched in
transcriptional machinery called “transcription factories”.

Transcriptionally active regions of chromatin are actively extended towards the center of the nucleus and transcription factories.

Question 37

What is G-banding?

How are G-bands and interbands related to the structure of chromosomes? (Ch 7)

Answer 37

G-banding: Metaphase spread digested with trypsin and stained with Giemsa.

Chromosomes have the appearance of a series of striations called G-bands.

G bands are lower in GC content than interbands. Genes tend to be enriched in the interbands.

In humans, bands are approximately 107 bp in size.

Question 38

Describe the structure of the centromere.

Via what structures are the centromeres attached to the spindle fibers? (Ch 7)

Answer 38

Centromere: A constricted region of a chromosome that includes the
site of spindle attachment.

Eukaryotic chromosomes are held on the mitotic spindle by the attachment of microtubules to the kinetochore that forms in its centromeric region.

Microtubule Organizing Center: A region from which microtubules emanate.

Cohesins: Proteins that hold sister chromatids together. Gradually degraded during anaphase to allow segregation to proceed.

Characterized by:
– Centromere-specific histone H3 variant CENP-A
– Post-translationally modified H2A and H3
– Often contain satellite DNA- rich heterochromatin

Kinetochore: Protein structure at centromere that facilitates microtubule attachment

Question 39

What are the functions of telomeres? (Ch 7)

Answer 39

Telomeres are required for:

1. Protection of linear chromosomal ends:
   Double stranded DNA breaks are typically targeted for degradation and repair.
   Double stranded break repair.
2. Extension of chromosomal ends:
   Ends of chromosomes are unable to naturally be extended.
3. Paring of homologous chromosomes and recombination.

Question 40

Describe the molecular structure of the telomere.

What is the general structure of a t-loop and a G-quadruplex and where in a telomere will you find these structures? (Ch 7)

Answer 40

Typically consists of a simple repeat with the sequence GGGTTA.

A single stranded extension of the GT-rich strand is created after synthesis of the telomere
– Created from partial degradation of the CA-rich strand
– Can form a G-quadruplex (lots of G’s, facilitated by metal ion)

Stability of telomeres due to t-loop and shelterin
– 5-10 kb in size
– 3′ repeating unit of the G+T-rich strand forms a loop by displacing its homolog in an upstream region of the telomere
– Catalyzed by TRF2 protein

Question 41

How is telomerase binding and number of telomere repeats added to the telomere controlled in yeast versus how it is controlled in humans?

Which specific proteins are involved in the process and what are their functions? (Ch 7)

Answer 41

Yeast
Rif 1 and 2
Rap 1
Cdc13 - telomerase recruitment
Human
TRF 1 and 2
Tin 2
Tpp 1
Pot 1- telomerase inhibition
Rap 1

Question 42

How can yeast cells that have lost their telomeres escape senescence and how is this related to telomerase function? (Ch 7)

Answer 42

Escape from senescence in yeast can occur if:

1. Telomerase is reactivated
2. Chromosomes are circularized
3. Unequal crossing over to restore telomeres