Genetics Exam #2 Questions

Why did geneticists believe, even before direct experimental evidence was obtained, that the genetic code would turn out to be composed of triplet sequences and be nonoverlapping?

* triplet code is much simpler than a four-letter code
* triplet-- given 20 amino acids, minimal use of the 4 DNA bases would require 3 bases per amino acid, thus providing 64 possible code words
* adding or deleting a multiple of 1 or 2 bases changed the reading frame, but adding or deleting multiples of 3 bases did not

what experimental evidence provided the initial insights into the compositions of codons encoding specific amino acids?

* homopolymer codes tested to see if any amino acids were incorporated into newly synthesized proteins


* use of RNA heteropolymers determined the means of the composition of triplet code words corresponding to amino acids
* the codon CAGUGG can only be read as CAG UUG but not as CAG GUG UGG
* homopolymer and heteropolymer RNAs were introduced into an in vitro system

how were the specific sequences of triplet codons determined experimentally?

* triplet binding assay took advantages of the observation that ribosomes, when presented in vitro with an RNA sequence as short as three ribonucleotides, will bind to it and form a complex similar to what is found in vivo
* both triplet binding assay and repeating copolymers of known sequences

how were the experimentally derived triplet codon assignments verified in studies using bacteriophage MS2?

* the bacteriophage MS2 was a virus with only 3500 ribonucleotides (RNA) and it coded for only 3 genes
* the amino acids of their gene’s products were used to confirm (or refute) the possible amino acid coded by the codons

Review the Chapter Concepts list on p. 283. These all center around how genetic information is stored in DNA and transferred to RNA prior to translation into proteins. Write a short essay that summarizes the key properties of the genetic code and the process by which RNA is transcribed on a DNA template.

* the encoded form that genetic information is stored in is nearly universal
* transcription is the process of genetic code transferring from DNA to RNA, and this translated form exists as triplet codons as a set of three ribonucleotides and each stands for part of RNA’s compromise
* the four ribonucleotides of RNA can be arranged into 64 different three-letter sequences for the 20 amino acids in proteins
* there is one start codon and three stop codons
* the transcription process is slightly different in prokaryotes because the transcription must be processed before transcription

assuming the genetic code is a triplet, what effect would the addition or loss of two nucleotides have on the reading frame? what about the addition or loss of three, six, or nine nucleotides?

* frameshift mutations result from the addition or deletion of one or more nucleotides within a gene and subsequently the mRNA transcribed from it
* the gain or loss of two nucleotides there would be a frameshift mutation
* if the addition or loss was three, six, or nine nucleotides the frame of reading would be reestablished, this means that it only is reestablished in triplets

The mRNA formed from the repeating tetranucleotide UUAC incorporates only 3 amino acids but the use of UAUC incorporates 4 amino acids. Why?

the triplets UUA and CUU, which are produced from the first sequence, code the same amino acid

in studies of the amino acid sequence of wild-type and mutant forms of tryptophan synthetase in E. coli, the following changes have been observed

* gly
* arg
* thr
* ser
* ile
* glu
* val
* ala

determine a set of triplet codons in which only a single-nucleotide change produces each amino acid change

a single nucleotide change may lead to change in the amino acid sequence. the mutation or change may be random. for tryptophan synthase a single mutation in a single codon can lead to conversion of glycine to either glutamine or arginine. three substituent mutations may convert it to threonine, isoleucine, and serine. a mutation in glutamine can convert it to valine or alanine.

predict the amino acid sequence produced during translation by the following start hypothetical mRNA sequences (note that the second sequence was formed from the first by a deletion of only one nucleotide):

* sequence 1: 5’ AUGCCGGAUUAUAGUUGA 3’
* sequence 2: 5’ AUGCCGGAUUAAGUUGA 3’

what type of mutation gave rise to sequence 2?

* sequence #1: AUGCCGGAUUAUAGUUGA
* AUG - CCG - GAU - UAU - AGU - UGA
* met - pro - asp - tyr - ser - stop
* sequence #2: AUGCCGGAUUAAGUUGA
* AUG - CCG - GAU - UAA - GUUGA
* met - pro - asp - stop
* frameshift mutation which removed a nucleotide within a gene occurred, specifically a deletion mutation. the nucleotide deleted is the 12th base (U) from the left in sequence 1, resulting in a new sequence, in which a terminating triplet UAA is introduced. this means the second sequence will be terminated before the first one.

present an overview of various forms of posttranscriptional RNA processing in eukaryotes and provide an example for each

during post-transcriptional process, primary transcript RNA is converted into a mature RNA. this process takes place prior to the translation process. post transcriptional processing of RNA in eukaryotes involves three steps, namely capping, poly-adenylation and splicing


1. capping: in this step, 7-methylguanosine is added to 5’ end of RNA. this helps recognition of mRNA by the ribosome during translation process.
2. poly-adenylation: in this step, polmerase A tail is added to the 3’ end of RNA. around 259 adenines are added to the 3’ end of the RNA after the cleavage
3. splicing: it is the process in which introns, the non-coding regions are removed from the pre-RNA and exons, the coding regions are joined together to form a mature RNA

Isoginkgetin is a cell-permeable chemical isolated from the ginko biloba tree that binds to and inhibits snRNPs.

a.) what types of problems would you anticipate in cells treated with isoginkgetin?

b.) would this be most problematic for E.coli cells, yeast cells, or human cells and why?

* a. isoginkgetin which inhibits snRNPs (small nuclear ribonucleoproteins) by binding to them will affect the splicing of nascent mRNA chain. when splicing nascent mRNA chains there can be two types of self-splicing or spliceosome mediated splicing.
* in self-splicing, the introns self-splice themselves and no protein is involved in this method of splicing
* spliceosome mediated splicing involves several short RNA molecules called small nuclear RNAs (named U1, U2, etc) these short RNA molecules are associated with snRNPS to form the spliceosomal complex carries out intro splicing. when the snRNPs are inhibited, the nascent pre-mRNA containing into does not undergo intro splicing. due to non-splicing, introns are included in the mature mRNA. inclusion of introns into the mature mRNA results in the translation of intronic neucleotides.
* thus, a protein with additional amino acids may be formed which may be long than the normal protein. this protein will be dysfunctional and will affect the concerned functions such as cell metabolism
* b. this inhibition of pre-mRNA splicing by isoginkgetin will be most problematic for human cells. E. coli cells and yeast cells will not be affected since they do not possess introns or very few introns are present in their genome. due to this, intron splicing does not take place or is a rare phenomenon in yeast or human cells. in such a situation, an inhibitor of spliceosome will be ineffective since such a mechanism is absent. human cells are eukaryotic cells. in human cells, a majority of the genome is composed of introns. splicing of this intron is majorly conducted by the spliceosome complex. human cells will face certain problems because isoginkgetin directly inhibits the major splicing mechanism- spliceosome-mediated splicing. as mentioned above, this process involved the snRNPs which are inhibited by isoginkgetin

review the chapter concepts list on p. 312. These all relate to the translation of genetic information stored in mRNA into proteins and how chemical information in proteins imparts function to those molecules. Write a short essay that discusses the role of ribosomes in the process of translation as it relates to these concepts.

* the ribonucleotide sequence of messenger RNA (mRNA) reflects genetic information stored in the DNA of genes and corresponds to the amino acids sequences in proteins encoded by those genes


* the process of translation decodes the information in mRNA, leading to the synthesis of polypeptide chains
* translation involves the interactions of mRNA, tRNA, ribosomes, and a variety of translation factors essential to the initiation, elongation, and termination of the polypeptide chain
* proteins achieve a three-dimensional conformation that arises from the primary amino acid sequences of the polypeptide chains making up each protein
* the function of any protein is closely tied to its three-dimensional structure, which can be disrupted by mutation

contrast the roles of tRNA and mRNA during translation, and list all enzymes that participate in the transcription and translation process.

Transfer RNAs are “adaptor” molecules in that they provide a way for amino acids to interact with sequences of bases in nucleic acids. Amino acids are specifically and individually attached to the 3’ end of tRNAs that possess a three-base sequence (the anticodon) to base pair with three bases of mRNA. Messenger RNA, on the other hand, contains a copy of the triplet codes that are stored in DNA, the sequences of bases in mRNA interact three at a time, with the anticodons of tRNAs

* enzymes involved in transcription include RNA polymerase (e. coli) and RNA polymerase I, II, III (eukaryotes)
* those involved in translation include the following: aminoacyl tRNA synthetases, peptidyl transferase, GTP-dependent release factors

Francis Crick proposed the “adaptor hypothesis” for the function of tRNA. Why did he chose that description?

It was reasoned that there would not be sufficient affinity between amino acids and nucleic acids to account for protein synthesis. For example, acidic amino acids would not be attracted to nucleic acids. With an adaptor molecule, specific hydrogen bonding could occur between nucleic acids and specific covalent bonding could occur between an amino acid and nucleic acid tRNA.

During translation, what molecule bears the codon? What about the anticodon?

the sequence of base triplets in mRNA constitutes the sequence of codons. A three-base portion of the tRNA constitutes the anticodon

The ‘a’ chain of eukaryotic hemoglobin is composed of 141 amino acids. What is the minimum number of nucleotides in an mRNA coding for this polypeptide chain?

141 amino acids = 3 nucleotides per amino acid = at least 423 nucleotides (426 including the termination codon) if assuming that 5’ cap, poly-A-tail, and UTR are omitted

Assuming that each nucleotide in an mRNA is 0.34 nm long, how many triplet codes can simultaneously occupy the space in a ribosome that is 20nm in diameter?

20/0.34 = 58.82 nucleotides or 59 nucleotides

Summarize the steps involved in charging tRNAs with their appropriate amino acids.

before translation can begin, tRNA’s have to bind to their respective amino acids (charging or aminoacylation)


1. the amino acid is activated with ATP and becomes aminoacyladelyic acid
2. a complex is formed with the enzyme aminoacyl synthetase
3. once the pyrophosphate is released from the enzyme complex, tRNA can bind to the complex
4. amino acid is transferred to the tRNA (binds to the adenine residue of the 3’ end)
5. the charged tRNA (aminoacyl tRNA) is now ready to bind to the ribosome and start translation (replication) process

When a codon in an mRNA with the sequence 5’ -UAA- 3’ enters the A site of a ribosome, it is not recognized by a tRNA with a complementary anticodon. Why not? What recognizes it instead?

the stop codon is recognized by RF1

discuss the potential difficulties of designing a diet to alleviate the symptoms of phenylketonuria.

* picking food that is low/free of phenylalanine
* avoiding high protein diets, cautious with all meats, beans and nuts, dairy products
* specific low protein breads and pastas

Individuals with phenylketonuria cannot convert phenylalanine to tyrosine. Why don’t these individuals exhibit a deficiency of tyrosine?

all tyrosine needed is from the diet, most natural proteins contain this amino acids

early detection and adherence to a strict dietary regimen have prevented much of the intellectual disability that used to occur in those with phenylketonuria (PKU). affected individuals now often lead normal lives and families. For various reasons, such individuals tend to adhere less rigorously to their diet as they get older. Predict the effect that mothers with PKU who neglect their diets might have on newborns.

developmental delays and permanent intellectual disability

Infantile cardiomyopathy is a devastating disorder that is fatal during the first year of life due to defects in the function of heart muscles resulting from mitochondrial dysfunction. A study, performed by Götz et al. \[(2011). Am. J. Hum. Genet. 88:635–642), identified two different causative mutations in the gene for mitochondrial alanyl-tRNA synthetase (mtAlaRS). One mutation changes a leucine residue at amino acid position 155 to arginine (p.Leu155Arg). The other mutation changes arginine at position 592 to tryptophan (p.Arg592Trp). The mtAlaRS enzyme has an N-terminal domain (amino acids 36–481) that catalyzes tRNA aminoacylation and an internal editing domain (amino acids 484–782) that catalyzes deacylation in the case that the tRNA is charged with the wrong amino acid.

(a) Consider the position of the disease causing missense mutations in the mtAlaRS gene in the context of the known protein domains of this enzyme. What predictions can you make about how these mutations impair protein synthesis within mitochondria in different ways?

(b) Which mutation would you predict has a more severe impairment of translation in mitochondria, and why?

a.) the missence mutation in the mtAlaRS gene may occur at two positions: the one in which leucine is substituted by arginine at 155th position. leucine is a non-polar/hydrophobic amino acid while arginine is a polar amino acid with basic charge. therefore, the folding pattern of protein related to the sequene of amino acids and this its function will be altered. in mitochondria, tRNA aminoacylation will get affected by this mutation. similarily, at position 592, arginine is substituted by tryptophan, which is an amino acid with an aromatic ring which means the molecular weight is high compared to arginine. thus, it also alters the strucure and folding pattern of protein leading to malfunctioning. in mitochondria, duacylation in the case when tRNA is charged with wrong amino acid will not take place, leading to the incorporation of wrong amino acids in the polypeptide chain

b.) the second mutation (ex: the substitution of arginine by tryptophan is more severe because the result of it would be no proofreading system on the addition of wrong amino acids, and completely different (maybe lethal) protein may get synthesized

In this chapter, we focused on the regulation of gene expression in bacteria. Along the way, we found many opportunities to consider the methods and reasoning by which much of this information was acquired. From the explanations given in the chapter, what answers would you propose to the following fundamental questions?

(a) How do we know that bacteria regulate the expression of certain genes in response to the environment?

when certain activities are supplied to growth media, organisms respond with the production of certain enzymes

In this chapter, we focused on the regulation of gene expression in bacteria. Along the way, we found many opportunities to consider the methods and reasoning by which much of this information was acquired. From the explanations given in the chapter, what answers would you propose to the following fundamental questions?

(b) What evidence established that lactose serves as the inducer of a gene whose product is related to lactose metabolism?

when lactose is added to the medium, E. coli responds with the production of enzymes involved in lactose metabolism. when lactose was removed such enzymes decreased in concentration

In this chapter, we focused on the regulation of gene expression in bacteria. Along the way, we found many opportunities to consider the methods and reasoning by which much of this information was acquired. From the explanations given in the chapter, what answers would you propose to the following fundamental questions?

(c) What led researchers to conclude that a repressor molecule regulates the lac operon?

using a mutant of E. coli that produces relatively large amounts of the repressor researchers have shown that a repressor binds specifically to DNA containing the operator portion of the lac operon

In this chapter, we focused on the regulation of gene expression in bacteria. Along the way, we found many opportunities to consider the methods and reasoning by which much of this information was acquired. From the explanations given in the chapter, what answers would you propose to the following fundamental questions?

(d) How do we know that the lac repressor is a protein?

the lac repressor was shown to be a protein using radioactive labeling of sulfur amino acids

In this chapter, we focused on the regulation of gene expression in bacteria. Along the way, we found many opportunities to consider the methods and reasoning by which much of this information was acquired. From the explanations given in the chapter, what answers would you propose to the following fundamental questions?

(e) How do we know that the trp operon is a repressible control system, in contrast to the lac operon, which is an inducible control system?

contrary to the lac system in which lactose in the medium causes the synthessi of lac-utilizing structural proteins, addition of tryptophan to the medium represses the transcription of proteins for the synthesis of tryptophan

Contrast positive versus negative control of gene expression.

* positive: the regulatory molecule stimulates transcription
* negative: removal of the regulatory molecule from DNA stimulates transcription

Contrast the role of the repressor in an inducible system and in a repressible system

* inducible system: repressor interacts with operator and inhibits transcription, it is inactivated by inducer to permit transcription
* repressible system: repressor is normally inactive until activated by a corepressor, enabling it to bind to operator and inhibit transcription

For the lac genotypes shown in the following table, predict whether the structural genes (Z) are constitutive, permanently repressed, or inducible in the presence of lactose.

* I+O+Z+= Inducible because a repressor protein can interact with the operator to turn off transcription
* I-O+Z+= Constitutive because the repressor gene is mutant; therefore, no repressor protein is available
* I-OcZ+= Constitutive because the repressor protein is mutant, and the operator cannon bind the RNA polymerase
* I-OcZ+/F'O+= Constitutive because there is no functional repressor
* I+OcZ+/F'O+= Constitutive because there is a constitutive operator next to a normal Z gene. Remembering that this operator function in cis and is not influenced by the repressor protein, constitutive synthesis of β-galactosidase will occur
* IsO+Z+= Repressed because the product of the Is gene is insensitive to the inducer lactose and thus cannot be inactivated. The repressor will continually interact with the operator and shut off transcription regardless of the presence of absence of lactose
* IsO+Z+/F'O+= Repressed because the product of the Is gene is insensitive to the inducer lactose and thus cannot be inactivated. The repressor will continually interact with the operator and shit off transcription regardless of the presence of absence of lactose the fact that there is a normal I+ gene is of no consequence because once a repressor from Is binds to an operator, the presence of normal repressor molecules will make no difference

Predict the effect on the inducibility of the lac operon of a mutation that disrupts the function of (a) the crp gene, which encodes the CAP protein, and (b) the CAP-binding site within the promoter.

a.) absence of a functional crp would compromise the positive control exhibited by CAP

b.) without a CAP binding site there would be a reduction in the inducibility of the lac operon

Describe the role of attenuation in the regulation of tryptophan biosynthesis.

* attenuation functions to reduce the synthesis of tryptophan when it is in full supply. it does so by reducing transcription of the tryptophan operon
* the same phenomenon is observed when tryptophan activates the repressor to shut off transcription of the tryptophan operon

In this chapter, we focused on how eukaryotic genes are regulated at the transcriptional level. Along the way, we found many opportunities to consider the methods and reasoning by which much of this information was acquired. From the explanations given in the chapter,

(a) How do we know that promoter and enhancer sequences control the initiation of transcription in eukaryotes?

* mutations within promoters alter transcription efficiencies, whereas deletions alter the initiation point of transcription

In this chapter, we focused on how eukaryotic genes are regulated at the transcriptional level. Along the way, we found many opportunities to consider the methods and reasoning by which much of this information was acquired. From the explanations given in the chapter

(b) How do we know that the orientation of promoters relative to the transcription start site is important while enhancers are orientation independent?

enhancers are chromosomal elements that negatively influence transcription when deleted or altered by mutation. insertion of an enhancer by recombinant technology increases transcription. insertion of a silencer by recombinant technology decreased transcription

In this chapter, we focused on how eukaryotic genes are regulated at the transcriptional level. Along the way, we found many opportunities to consider the methods and reasoning by which much of this information was acquired. From the explanations given in the chapter,

(c) How do we know that eukaryotic transcription factors bind to DNA sequences at or near promoter regions?

transcription factors possess a DNA-binding domain that binds to DNA sequences and provides cis-regulation. deletions of promoters and surrounding regions suppress binding of transcription factors

In this chapter, we focused on how eukaryotic genes are regulated at the transcriptional level. Along the way, we found many opportunities to consider the methods and reasoning by which much of this information was acquired. From the explanations given in the chapter,

(d) How do we know that there is an association between disease susceptibility in humans and regulatory DNA sequences?

mutations allow room for increased susceptibility to diseases when it is not regulated

Review the Chapter Concepts list on page 393. All of these concepts relate to various ways in which transcription is regulated in eukaryotes. Write a short essay describing how cis-acting regulatory elements, activators, and chromatin modifiers are all coordinately involved in regulating transcription initiation.

* while transcription and translation are tightly coupled in bacteria, in eukaryotes these processes are spatially and temporarily separated and thus independently regulated
* gene expression is regulated at many steps in eukaryotes, including DNA packaging into chromatin, transcription, RNA processing, RNA export from the nucleus, RNA stability, RNA localization, translation, and posttranslational modifications
* chromatin remodeling, as well as modifications to DNA and histones, play important roles in regulating gene expression in eukaryotes
* eukaryotic transcription initiation requires the assembly of transcription regulatory proteins at cis-acting DNA sites known as promoters, enhancers, and silencers
* transcriptional activators and repressors influence the association of the general transcription factors into pre-initiation complexes at gene promoters, and they may influence chromatin remodeling or modifications
* the genome-wide identification of transcription factor binding sites and chromatin modifications in the human genome in a manner that is cell-type specific has proven to be a powerful tool for understanding transcriptional regulation and how it relates to disease

What features of eukaryotes provide additional opportunities for the regulation of gene expression compared to bacteria?

* separated processes of transcription and translation
* presence of chromatin
* opportunity for posttranscriptional and pre translational regulation

Provide a definition of chromatin remodeling, and give two examples of this phenomenon.

* chromatin remodeling: process of significant changes in chromatin organization, an example is SWI/SWF complex
* ex: nucleosome remodeling: alteration in methylation and interaction of DNA w/ histones, HATS & HDACS contribute to nucleosome remodeling

Explain how the addition of acetyl groups to histones leads to a weaker association of DNA in nucleosomes.

positively charged histones are able to interact with negatively charged phosphate groups of the DNA backbone. addition of acetyl groups to histone tails weakens this interaction by reducing the positive charge on histones

Compare the control of gene regulation in eukaryotes and bacteria at the level of initiation of transcription. How do the regulatory mechanisms work? What are the similarities and differences in these two types of organisms in terms of the specific components of the regulatory mechanisms?

* recognition of the promoter can be affected by activators and repressors. the component of a RNA polymerase holoenzyme recognizes the promoter in bacteria.
* similarities: cis-acting elements and trans-acting factors must interact for transcription to begin. Upstream of a transcribed gene are component called promoters that are necessary. recognition of the promoter can be affected by activators and repressors. although they have diverse structures, DNA loops have a role in the control of transcription initiation
* differences: the component of a RNA polymerase holoenzyme recognizes the promoter in bacteria. the specificity of transcription is controlled by the recognition of various promoter sequences by various subunits. Additionally, RNAs (attenuators, riboswitches, and sRNAs) can either allow as well as repress initiation, making transcription adapt to environmental or cellular conditions. Repressor proteins that cause DNA conformational adjustaments (repression loops) can prevent RNA polymerase from binding to the promoter. In eukaryotes, the chromatin structure may have to be altered in order to change how accessible the promoter is to transcriptional machinery. General transcription factors attract RNAP II to promoters. Additionally, enhancers and silencers are each bound by activators and repressors, respectively. Large DNA loops may be created, putting promoters and enhancers (or silencers) near one another, according to one theory.

Because the degree of DNA methylation appears to be a relatively reliable genetic marker for some forms of cancer, researchers have explored the possibility of altering DNA methylation as a form of cancer therapy. Initial studies indicate that while hypomethylation suppresses the formation of some tumors, other tumors thrive. Why would one expect different cancers to respond differently to either hypomethylation or hypermethylation therapies?

* hypomethylation of tumor suppressor genes or hypermethylation of proto-oncogenes leads to cancer. many cancerous cells are found to have both hypo and hyper-methylated
* therefore, depending on the type of gene, hyper or hypo mutations may either increase the expression of an oncogene or reduce the expression of a tumor-suppressor gene, and vice versa

Explain how a tissue-specific RNA-binding protein can lead to tissue-specific alternative splicing via splicing enhancers or splicing silencers.

cis-acting sequences that promote (splicing enhancers) or inhibit (splicing silencers) splicing are recognized by different classes of RNA-binding proteins (RBPs)

Nonsense-mediated decay is an mRNA surveillance pathway that eliminates mRNAs with premature stop codons. How does the cell distinguish between normal mRNAs and those with a premature stop?

normally, stop codons are located near the polymerase-A tail of the message or downstream of exon-exon junctions. Termination signals that do not meet the criteria are considered as premature.

In 1998, future Nobel laureates Andrew Fire and Craig Mello, and colleagues, published an article in Nature entitled, “Potent and Specific Genetic Interference by Double-Stranded RNA in Caenorhabditis elegans.” Explain how RNAi is both “potent and specific.”

* gene silencing occurs in a sequence-specific manner since RNAi depends on complementary base pairings
* RNAi is considered potent because only a few molecules are needed to degrade large amounts of mRNA

RNAi may be directed by small interfering RNAs (siRNAs) or microRNAs (miRNAs); how are these similar, and how are they different?

* siRNAs and miRNAs are the two main types of small RNA molecules, they are produced when Dicer cleaves double-stranded RNA
* siRNAs regulate the expression of genes, is double-stranded, non-coding RNA molecule, inhibits translation
* miRNAs are single-stranded, non-coding RNA molecules, involved in RNA silencing and post-transcriptional regulation of gene expression, and degrade mRNA

what percentage of total RNA in a cell is messenger RNA? Transfer RNA? Ribosomal RNA?

* the percent of messenger RNA (mRNA) in an RNA cell is 5% total
* total transfer RNA (tRNA) is 15% of RNA
* ribosomal RNA (rRNA) accounts for 80% of total RNA

Isoginkgetin is a substance isolated from the *Ginko biloba* tree that binds to and inhibits snRNPs.  If you treated a variety of cells with this compound, would this be more problematic for E. coli or human cells?  Explain.

If you treated a variety of cells with Isoginkgetin, this will be more problematic for human \n cells than E. coli cells. The reason for this is because E. coli cells don’t have the same amount \n of introns so intron splicing doesn’t happen nearly as often. Meanwhile, human cells are \n eukaryotic so a good portion of the genome is made of introns and splicing of these introns will \n occur via the spliceosome complex. But with an inhibitor that binds like Isoginkgetin, it will inhibit \n the spliceosome mechanisms which involve snRNPs so introns cannot be removed and exons \n won’t join.

given this DNA sequence, write down the product of transcription and translation.

5’ ACCTTGGCACATGAAACGTGGCTGGTAATGG 3’
3’ TGGAACCGTGTACTTTGCACCGACCATTACC 5’

* the product of transcription of RNA is ACCUUGGCACAUGAAACGUGGCUGGUAAUGG
* the codon groups from start codon to stop codon is AUGAAACGUGGCUGGUAA
* the product of translation is AUG = start codon/methionine, AAA = lysine, CGU = arginine, GGC = glycine, UGG = tryptophan, and UAA is the stop codon

How will transcription of a gene be affected if the promoter for that gene is deleted?  Answer either:  more transcription, less transcription, or the same amount of transcription.  Explain your answer.

If the promoter of a gene is deleted, there will be less transcription because there is no place for the RNA polymerase to bind to

Lac operon: If you delete the promoter of the lac repressor, will the lac operon be expressed when cells are grown in glucose? Explain.

If you delete the lacl promoter, than lacl is not transcribed. Thus no repressor is made and no repression occurs. Therefore it is not expressed.

Trp operon: If you delete the promoter of the Trp operon, will the Trp operon be expressed when there is very little tryptophan in the cell?  Explain.

No trp operon promoter, no transcription of the trp operon, regardless of the amount of tryptophan present. This means that trp won’t be expressed.