Mastering Genetics Problem Set #2

False The degeneracy of the genetic code means that an amino acid may be coded for by more than one codon. However, a single codon can only ever specify one amino acid.

An addition and a deletion mutation A single addition or deletion would change the reading frame of the protein, but if another mutation occurred to cancel the effects of the first mutation, only those amino acids between the mutations would change.

"A polycistronic mRNA may be transcribed if the gene products are used in the same pathway or needed at the same time" is false. This statement is false. Polycistronic mRNAs are produced only in prokaryotes. In eukaryotes, a single gene is transcribed at a time.

are three bases in mRNA that code for an amino acid

the code is triplet

AUG; methionine

there can be more than one codon for a particular amino acid

Introns are removed from mRNA before translation.

61 There are 61 codons that code for amino acids and 3 stop codons that do not code for an amino acid.

transcription Transcription is initiated when the cell signals for the expression of a particular gene and involves the synthesis of RNA from a DNA template

mRNAs are made from genes

The addition of a 7-mG cap at the 5' end of the transcript and the addition of a poly-A sequence at the 3' end of the message.

5'-capping, 3'-poly(A) tail addition, splicing

False

RNA that is removed during RNA processing

UGAGCC 1. The RNA polymerase reads the sequence of DNA bases from only one of the two strands of DNA: the template strand. 2. The RNA polymerase reads the code from the template strand in the 3' to 5' direction and thus produces the mRNA strand in the 5' to 3' direction. 3. In RNA, the base uracil (U) replaces the DNA base thymine (T). Thus the base-pairing rules in transcription are A→U, T→A, C→G, and G→C, where the first base is the coding base in the template strand of the DNA and the second base is the base that is added to the growing mRNA strand.

the specific sequence of bases along the DNA strands In eukaryotes, binding of RNA polymerase II to DNA involves several other proteins known as transcription factors. Many of these transcription factors bind to the DNA in the promoter region (shown below in green), located at the 3' end of the sequence on the template strand. Although some transcription factors bind to both strands of the DNA, others bind specifically to only one of the strands. Transcription factors do not bind randomly to the DNA. Information about where each transcription factor binds originates in the base sequence to which each transcription factor binds. The positioning of the transcription factors in the promoter region determines how the RNA polymerase II binds to the DNA and in which direction transcription will occur.

Noncoding sequences called introns are spliced out by molecular complexes called spliceosomes. A cap consisting of a modified guanine nucleotide is added to the 5' end of the pre-mRNA. A poly-A tail (50-250 adenine nucleotides) is added to the 3' end of the pre-mRNA.

Transcription, 5' cap addition, addition of poly-A tail, exon splicing, passage through nuclear membrane

Exon splicing Correct. Introns must be removed from eukaryotic pre-mRNA; prokaryotic mRNA does not contain introns.

It provides a site for ribosome binding in the cytoplasm. To protect the transcript from degradation. Correct. The 5' cap is essential for recognition of the mRNA by ribosomes in the cytoplasm.

Exon/intron boundaries are typically characterized by a 5' GU splice junction and a 3' AG splice junction. Correct. These splice junctions are recognized by the spliceosome so that accurate removal of introns is possible.

small RNAs associated with protein complexes in the nucleus Correct. snRNPs recognize the 5' and 3' splice junctions and the branch point sequence, excise the intron, and splice together the exons.

A longer than usual final transcript Correct. Such a mutation could block intron removal, resulting in a longer than usual transcript.

rRNA Ribosomal RNA and ribosomes form the site of protein translation. Transfer RNAs work to bring amino acids to the ribosome. After a tRNA contributes its amino acid to the growing polypeptide chain, it must be "recharged" with a new amino acid. This is done independently of rRNA.

3′ end The 3′ end of a tRNA molecule contains the amino acid binding site.

It attaches a specific amino acid to a tRNA molecule. Aminoacyl tRNA synthetase catalyzes the charging reaction that links a specific amino acid to a tRNA molecule.

True Each aminoacyl tRNA synthetase enzyme recognizes only one amino acid, but each enzyme can often recognize several tRNAs because there is usually more than one codon for each amino acid.

is attached to the tRNA occupying the A site

It is catalyzed by peptidyl transferase.

aminoacyl-tRNA synthetase

Initiation

tRNA

Termination

translation

a formyl group is attached to the initiating methionine This modification is not present on methionine residues added during elongation.

complementary base pairing between mRNA and DNA Transcription, not translation, is dependent on this association.

The small ribosomal subunit binds to an mRNA sequence near the 5' end of the transcript

nucleolus

cytoplasm

cytoplasmic ribosomes

nucleus

the ribosomes on the rough ER

True

A mutation in the operator sequence Correct. Such a mutation could prevent binding of the repressor, allowing expression under all conditions.

It ensures that a cell dedicates resources to the production of enzymes involved in lactose metabolism only when lactose is available in the environment. Correct. The cell expends energy to produce the proteins necessary for lactose metabolism only when lactose is present.

True Correct. When the repressor binds to the operator, RNA polymerase cannot transcribe the structural genes.

True Lactose indirectly induces or stimulates the transcription of genes involved in its metabolism.

The lac Z, Y, and A genes would not be expressed. If lactose could not bind to the repressor, the repressor would stay bound to the operator and repress the transcription of the lac Z, Y, and A genes.

It decreases the levels of cAMP in the cell, repressing transcription from the lac operon. Glucose decreases the levels of cAMP in the cell, preventing formation of the CAP-cAMP complexes necessary for the stimulation of transcription from the lac operon.

True

operon

genes of an operon

promoter

regulatory gene

operator

repressor

inducer

Operon is not transcribed. The trp operon is regulated through negative control only. When tryptophan is present, the operon genes are not transcribed. When lactose is absent, the repressor protein is active, and transcription is turned off.

Operon is transcribed, but no sped up through positive control. When lactose is present, the repressor protein is inactivated, and transcription is turned on.

Operon is transcribed quickly through positive control. When glucose is absent, another regulatory protein (CAP) binds to the promoter of the lac operon, increasing the rate of transcription if lactose is present.

Methylation of histone tails in chromatin can promote condensation of the chromatin. Acetylation of histone tails is a reversible process. DNA is not transcribed when chromatin is packaged tightly in a condensed form. Acetylation of histone tails in chromatin allows access to DNA for transcription. Some forms of chromatin modification can be passed on to future generations of cells. One of the mechanisms by which eukaryotes regulate gene expression is through modifications to chromatin structure. When chromatin is condensed, DNA is not accessible for transcription. Acetylation of histone tails reduces the attraction between neighboring nucleosomes, causing chromatin to assume a looser structure and allowing access to the DNA for transcription. If the histone tails undergo deacetylation, chromatin can recondense, once again making DNA inaccessible for transcription. Recent evidence suggests that methylation of histone tails can promote either the condensation or the decondensation of chromatin, depending on where the methyl groups are located on the histones. Thus, methylation can either inactivate or activate transcription, and demethylation can reverse the effect of methylation. Changes in chromatin structure may be passed on to future generations of cells in a type of inheritance called epigenetic inheritance.

Both the fantasin gene and the imaginin gene will be transcribed at high levels when activators specific for control elements A, B, C, D, and E are present in the cell. The fantasin gene will be transcribed at a high level when activators specific for control elements A, B, and C are present in the cell. Control elements C, D, and E are distal control elements for the imaginin gene.

RNA polymerase is a holoenzyme composed of a five-subunit core enzyme and a sigma (σ) subunit. Different types of σ subunits aid in the recognition of different forms of bacterial promoters. The bacterial promoter is located immediately upstream of the starting point of transcription (identified as the +1 nucleotide of the gene). The promoter includes two short sequences, the -10 and -35 consensus sequences, which are recognized by the σ subunit.

The RNA polymerase holoenzyme first binds loosely to the promoter sequence and then binds tightly to it to form the closed promoter complex. An open promoter complex is formed once approximately 18 bp of DNA around the -10 consensus sequence are unwound. The holoenzyme then initiates RNA synthesis at the +1 nucleotide of the template strand.

The RNA-coding region is the portion of the gene that is transcribed into RNA. RNA polymerase synthesizes RNA in the 5′ → 3′ direction as it moves along the template strand of DNA. The nucleotide sequence of the RNA transcript is complementary to that of the template strand and the same as that of the coding (nontemplate) strand, except that the transcript contains U instead of T.

Most bacterial genes have a pair of inverted repeats and a polyadenine sequence located downstream of the RNA-coding region. Transcription of the inverted repeats produces an RNA transcript that folds into a stem-loop structure. Transcription of the polyadenine sequence produces a poly-U sequence in the RNA transcript, which facilitates release of the transcript from the DNA.

1. TFIID binds to the TATA box. 2. TFIIA, TFIIB, TFIIF, and RNA pol II bind. 3. TFIIE and TFIIH bind. 4. Synthesis of the pre-mRNA begins at the +1 nucleotide. 5. A 5' cap is added to the pre-mRNA. 6. Spliceosome complexes carry out intron splicing. 7. A poly-A tail is added to the pre-mRNA.

False

genetic disorder at the cellular level The location of genetic mutations and how they occur vary greatly. However, all cancers result from a genetic disorder at the cellular level

G1 Cells can exit the cell cycle and enter G0 or be committed to initiate DNA synthesis late in G1.

False The longest stage of interphase is S; cells typically spend about 7 hours in this stage. The shortest stage of interphase is G2 (3 hours), although the shortest stage of the entire cell cycle is mitosis (1 hour).

Precise replication of DNA A cell checks for precise replication of DNA at the G2/M checkpoint.

Cyclin-dependent kinase

False

The ability to form secondary tumors at another site Cells from malignant tumors can break off from the main tumor, travel through the blood or lymph, and establish secondary tumors at another site.

Programmed cell death

It is a natural process involved in a protective mechanism against cancer formation.

Oncogenes are genes that induce or maintain uncontrolled cellular proliferation associated with cancer. They are mutant forms of proto-oncogenes, which normally function to regulate cell division.

Gene amplification, repositioning of regulatory sequences, point mutations, translocations.

An addition and a deletion mutation