MIC1008 Tutorial 15/11/2023

What are exons?

• Set of genes that are adjacent to one another in the genome and are coordinately controlled

• Genetic information coding for an amino acid sequence that will form a functional polypeptide/protein

• Intervening non-coding segments of DNA

• Site where repressor proteins bind

Answer explanation

Exons contain the actual genetic information coding for protein. In contrast, introns are intervening non-coding segments of DNA. During the splicing process of pre-mRNA modification, introns are removed from the sequence. Alternative splicing patterns allow multiple exon sequences to be created from a single gene, resulting in multiple proteins.

An operon is a set of genes that are adjacent to one another in the genome and are coordinately controlled. The silencer region is a site of a gene where repressor proteins bind to regulate gene expression.

What is the role of promoter region in the regulation of gene expression?

• Alter gene expression by binding transcription factors

• Bind negative regulators to inhibit gene expression

• Recruit RNA polymerase and transcription factors to DNA

• Allow for alternative splicing and recombination of genetic components

Answer explanation

The promoter region is the site of a gene where RNA polymerase and other transcription factors bind to DNA, upstream from the gene locus. A mutation in this region commonly results in a decrease in the amount of gene transcribed.

An enhancer region is a stretch of DNA that alters gene expression by binding transcription factors, while a silencer region is a site on the gene where repressor proteins bind. Introns are intervening non-coding segments of DNA that are not expressed in the final protein. Alternative splicing patterns of introns and exons allows for multiple proteins to be generated from a single gene.

During translation, which site in the ribosome allows for tRNA molecules to enter the complex?

• A site

• P site

• E site

• R site

Answer explanation

The ribosomal complex has three sites where tRNA molecules can be oriented during the process of translation: the A site, the P site, and the E site. During polypeptide elongation, a tRNA with an attached amino acid will enter at the A site. It will then move to the P site, now holding the growing polypeptide chain. All tRNAs no longer holding an amino acid will exit the ribosome at the E site.

*May help to remember the following:

A = Admit

E = Exit/Escape

Which of the following most accurately describes the chronological order of ribosome biogenesis in eukaryotes?

• Ribosomal proteins are translated in the cytoplasm. At the same time, rRNA genes are being rapidly transcribed in the nucleolus. rRNA is transported to the cytoplasm where the rRNA and ribosomal proteins form the 60S and 40S ribosomal subunits.

• Ribosomal proteins are translated in the cytoplasm and transported to the nucleolus. At the same time, rRNA genes are being rapidly transcribed in the cytoplasm. rRNA and ribosomal proteins form the 60S and 40S subunits in the cytoplasm.

• Ribosomal proteins are translated in the cytoplasm and transported to the nucleolus. At the same time, rRNA genes are being rapidly transcribed in the nucleolus. rRNA and ribosomal proteins form the 50S and 30S subunits in the nucleolus and are then transported to the cytoplasm for functional ribosome assembly.

• Ribosomal proteins are translated in the cytoplasm and transported to the nucleolus. At the same time, rRNA genes are being rapidly transcribed in the nucleolus. rRNA and ribosomal proteins form the 60S and 40S subunits in the nucleolus and are then transported to the cytoplasm for functional ribosome assembly.

Answer explanation

Ribosomal proteins are translated in the cytoplasm and rRNA genes are transcribed in the nucleolus. Following protein translation, these proteins enter the nucleus through nuclear pores and localize to the nucleolus. Here, transcribed rRNA associates with the ribosomal proteins to form the 60S and 40S eukaryotic ribosomal subunits. Prokaryotes have 50S and 30S subunits. The ribosomal subunits then translocate to the cytoplasm where they join together to form fully functional ribosomes

Which of the following is a recognized mode of alternative splicing?

• Exon skipping

• Alternative acceptor site

• Alternative donor site

• All of the above

Answer explanation

The correct answer is all of the first three answers (all of the above). Exon skipping is the most common mode in mammals and occurs when an exon is spliced out of the primary transcript. Mutually exclusive exon splicing occurs when one of two exons is retained, but not both. Alternative donor site occurs when an alternative 5' splice junction is used which will change the 3' end of the upstream exon. Alternative acceptor site occurs when there is an alternative 3' split junction and the 5' end of the downstream exon is changed.

If the mRNA sequence is AGA UUU CGA GCC, what are the anticodon of corresponding tRNAs?

Explain how a single nucleotide change can have different effects on protein function.

Answer explanation

Nucleotide changes in the third position of codons may not change the amino acid and would have no effect on the protein (code degeneration). Other nucleotide changes that change important amino acids or create or delete start or stop codons would have severe effects on the amino acid sequence of the protein.

It is reasonable to conclude from the pedigree that

• the trait shown is autosomal recessive

• II-4 and II-5 have the same genotype

• the trait cannot be X linked recessive because II-7 is unaffected

• if II-2 had a partner with ß thalassaemia the chance that their first child would be affected with thalassaemia is 1/2

The chance that the next child of II-8 and II-9 is a male affected with ß thalassaemia is

• 1/4

• 3/4

• 1/8

• 2/3

Answer explanation

Probability of thalassemia x Probability of male

=1/4 x 1/2

=1/8

The chance that II-1 is heterozygous for ß thalassaemia is

• 1/4

• 3/4

• 1/8

• 2/3

Answer explanation

3 potential events for non-affected individuals:

TT Tt Tt

Heterozygous = 2/3