BIOLOGY 1A03

Midterm 2 Review Session

Presented by the McMaster Biology Society!

DISCLAIMER

"This review session is intended to benefit the students of 1A03 and is not compulsory. Though the information discussed in this review session has been reviewed by the course instructors, it is still the student's responsibility to confirm the accuracy of this information using the course lecture notes, tutorials, and text."

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5

T2M3: Translation

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01

tRNA

ACTIVATION:

• Each aminoacyl tRNA synthetase is specific to the type of RNA and corresponding amino acid
• Causes attachment of amino acid to tRNA using ATP
• tRNA becomes charged

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Biological Function:

Enables the translation of the info in the mRNA genetic message to a polypeptide

Anticodon is written in 3’ → 5’ direction and aligns with mRNA in 5’ → 3’.

Translation Initiation

Assembling an Initiation Complex

Assembling an Initiation Complex

• A site: Aminoacyl site
• P site: Peptidyl site
• E site: Exit site

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Each incoming tRNA is delivered in association with a GTP- bound elongation factor.

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Practice Question

During the process of translation, which of the following describes what occurs once the ribosomal translation complex is completely assembled?

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A) The initiation factors join onto the P site of the ribosome.

B) The initiation factors convert the consensus sequences into termination STOP codons.

C) The initiation factors bind directly to the E site of the ribosome complex.

D) The initiation factors are then released.

E) The initiation factors link onto elongation factors of the exons.

Practice Question

During the process of translation, which of the following describes what occurs once the ribosomal translation complex is completely assembled?

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A) The initiation factors join onto the P site of the ribosome.

B) The initiation factors convert the consensus sequences into termination STOP codons.

C) The initiation factors bind directly to the E site of the ribosome complex.

D) The initiation factors are then released.

E) The initiation factors link onto elongation factors of the exons.

One Gene, One Enzyme Hypothesis

Each gene contains the information to make each enzyme

Hypothesis

One Gene, One Enzyme Hypothesis

One Gene, One Enzyme Hypothesis

One gene, one enzyme hypothesis:

• one gene codes for one enzyme

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One gene, one polypeptide hypothesis:

• Srb & Horowitz further proved that genes not only code for enzymes but also proteins

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One gene, one… hypothesis:

• present day scientists found that post-translational modifications can change the protein into different forms

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Practice Question

Consider the biochemical pathway of the synthesis of arginine. If similar protocols were implemented to that of researchers Beadle & Tatum and Srb & Horowitz, which used Neurospora crassa, what would be observed for the arg 2 mutant that is placed solely on minimal medium?

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A) Growth would occur on minimal medium.

B) Growth would be observed at a very slow rate and then growth would increase rapidly at an indefinite rate.

C) Growth would occur on the minimal medium depending upon the quantity of arg 2 mutants used for the experiment.

D) No growth would be observed on the minimal medium alone.

Practice Question

Consider the biochemical pathway of the synthesis of arginine. If similar protocols were implemented to that of researchers Beadle & Tatum and Srb & Horowitz, which used Neurospora crassa, what would be observed for the arg 2 mutant that is placed solely on minimal medium?

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A) Growth would occur on minimal medium.

B) Growth would be observed at a very slow rate and then growth would increase rapidly at an indefinite rate.

C) Growth would occur on the minimal medium depending upon the quantity of arg 2 mutants used for the experiment.

D) No growth would be observed on the minimal medium alone. If it is a mutant, it cannot grow on a minimal medium!

T2M4: The Complex Proteome

02

Proteome Complexity

Single genes can encode multiple proteins due to：

• RNA processing
• Post-translational modifications

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RNA Processing - Alternate Splicing

Protein Product Ａ

Protein Product Ｃ

Protein ProductＢ

Isoforms－different types of mature mRNA

Post-Translational Modifications

Phosphorylation: reversible covalent attachment of phosphate group to serine, threonine, tyrosine amino acids by kinases

Methylation: reversible covalent addition of a methyl group

Acetylation: add acetyl group to amino acid Residue in a protein.

Practice Question

Although the human genome contains only about 20,000–25,000 protein-coding genes, the human proteome includes over one million distinct proteins. Which of the following best explains how this extensive protein diversity is achieved?

A) Each gene can be replicated multiple times during transcription, increasing the number of proteins produced.

B) Alternative RNA splicing and post-translational modifications allow a single gene to give rise to multiple protein isoforms with different structures and functions.

C) Ribosomes can translate both coding and non-coding regions of DNA, generating diverse protein products.

D) Mutations during DNA replication increase the total number of functional genes that code for unique proteins.

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Practice Question

Although the human genome contains only about 20,000–25,000 protein-coding genes, the human proteome includes over one million distinct proteins. Which of the following best explains how this extensive protein diversity is achieved?

A) Each gene can be replicated multiple times during transcription, increasing the number of proteins produced.

B) Alternative RNA splicing and post-translational modifications allow a single gene to give rise to multiple protein isoforms with different structures and functions.

C) Ribosomes can translate both coding and non-coding regions of DNA, generating diverse protein products.

D) Mutations during DNA replication increase the total number of functional genes that code for unique proteins.

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Detecting a Signal

Glucose Absorption

PTM’s of Insulin

1. PRE-PROINSULIN

2. PROINSULIN

3. 2 Dimer chains form insulin

In the lumen of rER

In the lumen of rER

In Golgi

Insulin Receptors

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２）　

１）

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４）　

５）

Practice Question

Following a meal, blood glucose levels rise and trigger a signaling cascade that results in glucose uptake by target cells. Which of the following correctly outlines the sequence of events involved in this process?

A) Pancreas detects high blood glucose → Insulin binds to receptor → Receptor dimerizes and phosphorylates → Glucose transporters move to cell membrane → Blood glucose decreases

B) Liver detects high blood glucose → Insulin binds to receptor → Receptor releases glucose → Glucose levels decrease

C) Pancreas detects high blood glucose → Insulin binds to receptor → Receptor deactivates → Glucose transporters are inhibited → Blood glucose decreases

D) Muscle cells detect high blood glucose → Pancreas releases glucagon → Receptor dimerizes → Glucose uptake increases

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Practice Question

Following a meal, blood glucose levels rise and trigger a signaling cascade that results in glucose uptake by target cells. Which of the following correctly outlines the sequence of events involved in this process?

A) Pancreas detects high blood glucose → Insulin binds to receptor → Receptor dimerizes and phosphorylates → Glucose transporters move to cell membrane → Blood glucose decreases

B) Liver detects high blood glucose → Insulin binds to receptor → Receptor releases glucose → Glucose levels decrease

C) Pancreas detects high blood glucose → Insulin binds to receptor → Receptor deactivates → Glucose transporters are inhibited → Blood glucose decreases

D) Muscle cells detect high blood glucose → Pancreas releases glucagon → Receptor dimerizes → Glucose uptake increases

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T3M1: Modulating Transcription

03

Gene Categories

Regulated

• Only expressed when needed
• E.g. enzymes

• Constitutively expressed
• E.g. structural proteins, ribosomal proteins

Housekeeping

Levels of Regulation

• Controls the amount of mRNA produced
• Transcription occurs when proteins bind to promoter regions and bind RNA polymerase
• Controlling proteins that bind to promoter and inhibit transcription

• Prevents immediate translation of mRNA
• Ribosome binds to cap (or Shine-Dalgarno sequence in prokaryotes) and initiates translation
• Rate of translation and mRNA stability impacts amount of protein translated

• Translated polypeptide doesn’t fold immediately
• More than 12 exist to regulate the ability to become (in)active
• Allows cells to have a stockpile of a protein that is inactive, faster to activate than make from scratch

Transcriptional

Translational

Post-Translational

The Glucose Effect

• B-galactosidase only made when there is an absence of glucose and a presence of lactose
• Gene that codes for the enzyme is a regulated gene, even if it is vital to survival
• Regulation of B-galactosidase is controlled by negative and positive processes

Practice Question

In an environment where there is E. coli growing, where does expression of the lac operon increase?

Practice Question

ANSWER: c

Practice Question

True or False: Regulation of B-galactosidase is only controlled by negative regulation.

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If true, explain how.

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If false, explain why.

Practice Question

FALSE - Regulation of B-galactosidase is controlled by negative and positive processes

• Negative regulation (decreasing the amount produced): through binding of the repressor to the operator and preventing RNA polymerase from binding and transcribing the gene
• Positive regulation (increasing amount produced): cAMP bind to CRP and bind to cAMP binding site on operon, RNA polymerase can then be recruited by the activator complex

T3M2: Prokaryotic Transcriptional Regulation

04

Resource availability

When glucose is fully depleted:

• Increase in amount of B-galactosidase and Lactose permease

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Lactose permease: transport protein (embeds in cell membrane and allows for import of lactose into bacterial cell)

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Lactase (B-galactosidase): cleaves lactose into glucose and galactose

Operon Model

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• Operon: prokaryotic cluster of genes with similar function that has a operator & promoter
• Operator: the “on/off” switch of an operon

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This model is for transcriptional regulation in prokaryotes

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Negative Regulation

WAIT! What happens when lactose is present?

When lactose levels are low, glucose levels are high

Positive Regulation of the Lac Operon

• Positive regulation happens in presence of lactose, and B-galactosidase and lactose permease are expressed.
• High glucose levels causes inhibition of adenylyl cyclase (which catalyses production of cAMP from ATP).
• CRP-cAMP complex binds to DNA and further activates transcription

Practice Question

Which of the following elements, when non-functional, would lead to a reduction in the total amount of functional lacZ produced by a bacterial cell?

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A) The repressor protein.

B) The lac I gene.

C) The lac Y gene.

D) The CRP-cAMP binding site.

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Practice Question

Which of the following elements, when non-functional, would lead to a reduction in the total amount of functional lacZ produced by a bacterial cell?

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A) The repressor protein. No repressor = more lacZ!

B) The lac I gene. No LacI = no repressor = more lacZ!

C) The lac Y gene. Unrelated to the amount of lacZ produced

D) The CRP-cAMP binding site. No recruitment of RNA pol = less transcription = less lacZ!

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T3M3: Eukaryotic Transcriptional Regulation

05

Chromatin Remodelling

Regulating Eukaryotic Transcription

• Transcriptions factors have α-helical domains that bind specifically to DNA
• Fitting within major grooves as well as forming H-bonds
• When the interaction is strong enough, it recruits other transcription factors and RNA polymerase to activate the gene

Regulating Eukaryotic Transcription

• Specific DNA sequences (aka cis-sequences) required to start transcription
• TATA box and transcriptional start sites form part of the core promoter*
• *Binding site required for RNA polymerase and associated transcription factors.

Practice Question

Which of the following best explains how chromatin remodeling facilitates transcription in eukaryotic cells?

A) DNA is methylated at cytosine bases, tightening the chromatin and promoting transcription factor binding.

B) Histone acetylation reduces the positive charge on histone tails, loosening DNA–histone interactions and allowing transcription factors to access DNA.

C) The TATA box is methylated, which recruits RNA polymerase II directly to the promoter.

D) Transcription factors bind to nucleosomes, compacting chromatin and silencing gene expression.

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Practice Question

Which of the following best explains how chromatin remodeling facilitates transcription in eukaryotic cells?

A) DNA is methylated at cytosine bases, tightening the chromatin and promoting transcription factor binding.

B) Histone acetylation reduces the positive charge on histone tails, loosening DNA–histone interactions and allowing transcription factors to access DNA.

C) The TATA box is methylated, which recruits RNA polymerase II directly to the promoter.

D) Transcription factors bind to nucleosomes, compacting chromatin and silencing gene expression.

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Regulating Eukaryotic Transcription

Transcription Repression by Methylation

DNA methylation (at CpG islands) silences gene expression and stops transcription initiation

UNMETHYLATED

METHYLATED

Developmental Gene Expression

Practice Question

Fetuses have (higher/lower) oxygen affinity because the expression of (__________) globin is on and (___________) globin is off.

Practice Question

Fetuses have (higher/lower) oxygen affinity because the expression of gamma-globin is on and beta-globin is off.

T3M4: Turning off the Signal

06

DNA Microarray Analysis

Post-transcriptional regulation

Transcribed from protein coding genes, forms hairpin loops due to complementary base pairing. Recruits RISC complex to prevent translation. Also called micro RNA.

Exact complements of the mRNA target. Recruits RISC complex to initiate degradation. Also called small interfering RNA.

siRNA

miRNA

Post-translational regulation

Ubiquitin + Proteasome

Practice Question

True or false:

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1. siRNA recruits the RISC complex for degradation initiation while miRNA recruits RISC to prevent translation.

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1. siRNA forms a hairpin loop, miRNA doesn’t.

Practice Question

True or false:

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1. siRNA recruits the RISC complex for degradation initiation while miRNA recruits RISC to prevent translation.

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1. siRNA forms a hairpin loop, miRNA doesn’t.

More Practice!

Practice Question

Suppose the terminal adenine of CCA at the 3’ end of the tRNA was missing. Which of the following would be impacted?

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A) The amino acid would be easily able to attach to the tRNA, since there would be a stable attachment point during tRNA inactivation.

B) The tRNA molecule would be able to undergo post-transcriptional modifications involving the addition of a signal sequence at the 5’ end.

C) The tRNA codon region would be free to form a three-dimensional, J-structured anticodon region.

D) The amino acid would not be able to attach to the tRNA, since there would not be an attachment point during tRNA activation.

Practice Question

Suppose the terminal adenine of CCA at the 3’ end of the tRNA was missing. Which of the following would be impacted?

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A) The amino acid would be easily able to attach to the tRNA, since there would be a stable attachment point during tRNA inactivation.

B) The tRNA molecule would be able to undergo post-transcriptional modifications involving the addition of a signal sequence at the 5’ end.

C) The tRNA codon region would be free to form a three-dimensional, J-structured anticodon region.

D) The amino acid would not be able to attach to the tRNA, since there would not be an attachment point during tRNA activation.

Practice Question

Describe the differences between housekeeping and regulated genes.

Practice Question

ANSWER: Housekeeping genes are constitutively expressed whereas regulated genes are only expressed when they are needed.

Q&A

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