BIS 002A: WEEK 8,9,10 STUDY GUIDE KEY

The figure below depicts an important reaction in the biosynthesis of a nucleic acid polymer. Like other exergonic cellular biochemical reactions involving an “external” energy source ATP (in a deoxy form dATP) is utilized in the creation of a new strand of DNA during replication. Key differences between the previous ATP-requiring reactions and DNA synthesis are that

A. In DNA synthesis, the dATP is not immediately returned to the cellular pool of ADP or AMP to be regenerated into ATP.

B. The enzyme that catalyzes the reaction can also utilize three other deoxynucleotide triphosphates in addition to dATP

C. The hydrolysis of the phosphates on dATP is NOT exergonic in the synthesis of new DNA because the dAMP is captured on the growing strand

1. In DNA synthesis, the dATP is not immediately returned to the cellular pool of ADP or AMP to be regenerated into ATP.

2. 2. The enzyme that catalyzes the reaction can also utilize three other deoxynucleotide triphosphates in addition to dATP

Reasoning:

• Statement A: In DNA synthesis, dATP is incorporated into the growing DNA strand and is not recycled back into the cellular ATP pool immediately. This is different from typical ATP-utilizing reactions where ATP is hydrolyzed and ADP/AMP is regenerated.

• Statement B: DNA polymerase can use dATP along with three other deoxynucleotide triphosphates (dGTP, dCTP, dTTP) for DNA synthesis.

If a double strand of DNA is analyzed and contains 30 percent G bases, what percent of the strand contains C bases?

30 %

Reasoning:

• Chargaff’s rule states that in a double-stranded DNA molecule, the amount of guanine (G) always equals cytosine (C).

• If 30% of the bases are guanine, then 30% must also be cytosine

The sequence of one strand of DNA is 5'-GCTTTAG-3'. The sequence of the complementary strand would be:

5’-CTAAAGC-3’

Reasoning:

• DNA strands are complementary and antiparallel

Newly synthesized DNA is illustrated in the Figure. The Okazaki fragments have not yet been processed. Which of the Okazaki fragments was synthesized first?

THE ONE ON THE LEFT

Reasoning:

• Okazaki fragments are synthesized on the lagging strand in the opposite direction of the replication fork movement.

• The fragment that appears further away from the replication fork (on the left) must have been synthesized first.

2Use the replication fork figure to answer this question. The replication fork is moving from left to right. The strand labeled________is the template for the _______ and is created from __________. Okazaki fragments will be located on strand _____.

B, LEADING STRAND, 5’ TO 3’, A

Reasoning:

• The leading strand is synthesized continuously in the 5' → 3' direction.

• The template strand for the leading strand must be 3' → 5'.

• Okazaki fragments are synthesized on the lagging strand (B).

Why is an RNA primer considered essential during DNA synthesis by DNA polymerase?

DNA polymerase requires a free 3'-OH group.

Reasoning:

• DNA polymerase can only add nucleotides to a pre-existing 3'-OH group.

• The RNA primer provides this necessary 3'-OH group for DNA polymerase to begin synthesis.

An mRNA has the sequence 5'-AUGAAAUCCUAG-3'. What is the template (non-coding/ complementary) DNA strand for this sequence?

5'-CTAGGATTTCAT-3'

Reasoning:

• The template strand is complementary and antiparallel to the mRNA sequence.

• Given mRNA: 5'-AUGAAAUCCUAG-3',

• The complementary DNA template strand is 3'-TACTTTAGGATC-5',

• When written in 5' → 3' direction, it is 5'-TACTTTAGGATC-3'.

Since only one DNA strand is copied into mRNA to provide the information for protein synthesis why is the redundant information content of DNA so important to life?

Redundant information allows precise copying of information from both strands during replication and for correctly repairing damage if either strand is damaged

Reasoning:

• DNA redundancy means that if one strand is damaged, the complementary strand provides the correct sequence for repair.

• Both strands act as templates during replication, ensuring high-fidelity DNA duplication.

Which of the following sites would you predict to be present in the gene encoding a tRNA

1,3,4

Reasoning:

• Promoter (#1): Necessary for transcription initiation.

• Transcription Initiation Site (#3): The site where transcription starts.

• Transcription Termination Site (#4): The site where transcription stops.

• Ribosome Binding Site (#2), Translation Initiation Site (#5), and Translation Termination Site (#6) are not present in tRNA genes because tRNA is not translated into a protein

In bacteria, the site on the mRNA that binds to the ribosome has a sequence that is complementary to the end of the 16S rRNA present in the small subunit of the ribosome. The sequence of the 16S rRNA is sequence 5’- CCUCCU-3’. Based on this information, which of the following sequences could act as a ribosome binding site?

5’- AGGAGG-3’

Reasoning:

• The ribosome binding site (Shine-Dalgarno sequence) in bacterial mRNA must be complementary to the 3’-CCUCCU-5’ sequence in the 16S rRNA of the ribosome.

• The complementary sequence to CCUCCU is AGGAGG.

• This sequence ensures correct alignment of the mRNA with the ribosome for translation.

In a certain mutant strain of bacteria, the enzyme leucyl-tRNA synthase mistakenly attaches isoleucine to leucyl-tRNA 10% of the time instead of attaching leucine. These bacteria will synthesize

proteins in which isoleucine is inserted at some positions normally occupied by leucine

Reasoning:

• Leucyl-tRNA synthetase normally attaches leucine to tRNA^Leu.

• In this mutant, it mistakenly attaches isoleucine 10% of the time.

• This means some tRNA^Leu molecules will carry isoleucine instead of leucine.

• During translation, ribosomes will insert isoleucine in positions where leucine should have been..

_________ is the process of removing the __________ from pre-mRNA and joining the remaining __________ together to form a ‘mature’ mRNA molecule.

splicing, introns, exons

Reasoning:

• Splicing is the process of removing introns from pre-mRNA and joining together exons to form mature mRNA.

• Introns are non-coding sequences that must be removed.

• Exons are coding sequences that remain in the final transcript

A mutation occurs such that a spliceosome cannot remove one of the introns in the mRNA. What effect will this have?

The resulting mRNA will be shipped from the nucleus to the cytoplasm where it will be translated by a ribosome

Reasoning:

• If a spliceosome fails to remove an intron, the mRNA is still exported to the cytoplasm.

• However, the presence of the intron may lead to incorrect protein synthesis or premature translation termination.

The promoter is

a short sequence of DNA where RNA polymerase first attaches when a gene is to be transcribed

Reasoning:

• The promoter is a DNA sequence that serves as the binding site for RNA polymerase.

• It is not where repressors bind (that’s the operator).

• It initiates transcription by allowing RNA polymerase to start reading the DNA template.

Gene regulatory proteins function by:

enhancing or inhibiting the activity of RNA polymerase at a promoter

Reasoning:

• Gene regulatory proteins control gene expression by binding to DNA near the promoter.

• They either enhance RNA polymerase binding (activators) or block it (repressors).

• This directly affects transcription rates.

Which of the following is true of both positive and negative gene regulation

They involve transcription factors (proteins or RNA) binding to DNA.

Reasoning:

• Both positive and negative regulation involve transcription factors binding to DNA.

• Positive regulation: Activators bind to increase transcription.

• Negative regulation: Repressors bind to decrease transcription.

• This is a fundamental mechanism in gene expression control.

Transcription factors function by

activating or blocking the activity of RNA polymerase

Reasoning:

• Transcription factors work by either:

  • Helping RNA polymerase bind to the promoter (activators).

  • Preventing RNA polymerase from binding (repressors).

• This determines whether a gene is transcribed or not.

Some cellular genes are expressed constitutively and are called housekeeping genes. Examples include the genes encoding the subunits of RNA polymerase, ribosomal rRNAs, ribosomal proteins, tRNA, and enzymes required for basic metabolic and repair pathways. Based on our discussion in lecture, the expression levels of these genes is most likely controlled using

Promoter strength

Reasoning:

• Housekeeping genes are always expressed at consistent levels.

• Their transcription is controlled by the strength of their promoter rather than regulation by activators or repressors.

• Strong promoters increase transcription, while weak promoters reduce it.

An operon consists of two structural genes (A and B) that code for the enzymes Aase and B-ase, respectively. The operon also includes P (promoter) and RE (regulatory element) upstream of the first gene in the operon. When a certain compound (X) is added to the growth medium of E. coli, the separate enzymes A-ase and B-ase are both synthesized at a 50-fold higher rate than in the absence of X. Which one of the following statements could be true of such an operon?

1. Adding X to the growth medium causes a transcription factor bound at its RE to be released, thereby allowing RNA polymerase to initiate transcription at the promoter (P).

2. Adding X to the growth medium causes a transcription factor to bind its RE, thereby allowing RNA polymerase to initiate transcription at the promoter (P).

• Reasoning:

  • The operon is induced by compound X.

  • If a repressor is removed from the regulatory element (A), RNA polymerase can bind and start transcription.

  • If an activator binds to the regulatory element (B), it helps RNA polymerase initiate transcription.

  • Both mechanisms lead to a 50-fold increase in enzyme synthesis.

7The pathway below shows the feedback inhibition caused if product I or product F accumulate. Suppose the cell has too much product I, what would be the first immediate thing(s) you’d likely see once compound I became too abundant?

1. There would be a reduction in production of G

2. There would be increased production of F

Reasoning:

• Feedback inhibition occurs when too much of a product accumulates.

• If product I accumulates:

  • It inhibits the enzyme producing compound G (A).

  • The excess substrate (compound C) is diverted to produce more of product F (B).

The figure to the right diagrams a metabolic pathway that might controls the color of flowers in certain plants: The genotype of a diploid wild-type flower is AA BB CC DD, and its phenotype is light purple. Alleles A, B, C, D each encoded enzymes that catalyze the eponymous reactions in the Figure. a, b, c, and d are loss-of-function alleles. What will the mostly likely phenotype of the AA Bb cc DD double mutant?

ORANGE

Within a group of mutants (genetically related organisms with slightly different genotypes) with the same overt phenotype, DNA sequencing studies mapping the location of the mutations determined that mutations in different organisms were on different chromosomes. This indicates that

This basically says that mutations in different genes can lead to the same phenotype and implies, when making the reasonable assumption that a phenotype is a product of a multistep biochemical pathway. . .

different genes can govern different individual steps in the same biological pathway

REASONING: This could kind of work if the mutations at the different loci actually influence the production of products that have the same catalytic activity but is not as good as B given what we have covered in class (likey good to rewrite this particular answer choice)

In the network below we see the functional relationships network of transcription factors, the genes encoding them, environmental factors and the genes responsible for a complex bacterial phenotype called heterocyst formation (hgl, hep, and nif). Which interpretation of the figure is most reasonable?

1. NITROGEN LIMITATION “TURNS-ON” hgl expression

2. The NtcA transcription factor can act both negatively and positively

REASONING:

1. YES, NITROGEN LEADS TO 2-OG INCREASE WHICH UP-REGULATES NtcA that in turn activates nrrA, which in turn activated hetR, which activates hgl and help

2. Yes, it positively activates nrrA and negatively influence ccbP

Within the human population, 200 different alleles exist for a specific locus. Each person carries at most —— alleles for this locus

2

REASONING-

YES, WE MUST REASONABLY ASSUME THAT WE ARE REFERRING TO ALLELES OF ONE GENE AND THUS ONE GENE AT EACH LOCUS. THEN YOU HAVE ONE ALLELE AT A LOCUS AND SINCE YOU HAVE ONE COPY FROM MOM AND ONE FROM DAD YOU HAVE 2 ALLELES. IF MOM OR DAD WAS A DIFFERENT PERSON, YOU MIGHT HAVE A DIFFERENT SET OF 2 ALLELES.