WEEK- 13-HW- Chapter 17 PDF

Title WEEK- 13-HW- Chapter 17
Course General Biology I
Institution Grand Canyon University
Pages 9
File Size 454.4 KB
File Type PDF
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Summary

HW Week 13...


Description

CHAPTER 17: MATCH THE FOLLOWING

Answe r p

f l

k i j b o q h

Term

Definition

1. Ribozyme 2. Aminoacyl-tRNA Synthetase 3. Signal Recognition Particle 4. RNA Polymerase 5. 6. 7. 8. 9. 10.

mRNA rRNA tRNA Introns Exons Post-Transcriptional Modification 11. Transcription Factors

c 12. GTP d g r a n e m s

13. Frameshift Mutation 14. Amino Acid 15. Silent Mutation 16. Nonsense Mutation 17. Missense Mutation 18. Heterochromatin 19. Euchromatin

a. Has no effect on the encoded protein because of the redundancy of the genetic code b. Carries amino acid to a strand of mRNA c. Transcription in eukaryotes requires which of the following in addition to RNA polymerase d. Energizes the formation of the initiation complex, using initiation factors e. Changes the codon so a wrong amino acid is put in f. Enzyme that joins an amino acid to the correct tRNA g. Could result from either an insertion or a deletion of a base h. Occurs in eukaryotes, but not prokaryotes i. Carries genetic instructions from the nucleus to the cytosol j. Serves as a workbench for translation of mRNA and has catalytic roles k. Enzyme that pries the two strands of DNA apart and catalyzes the addition of RNA nucleotides to a strand l. Adapter that brings the ribosome and growing polypeptide to a receptor protein built into the ER membrane m. Condensed DNA that is not transcribed n. Changes an amino acid codon into codon, thus terminating translation o. Segments of eukaryotic genes that code for functional products p. RNA molecule that functions as an enzyme q. Noncoding segments of mRNA r. Binds concalently to 3’ end of tRNA s. DNA that is transcribed

Central Dogma…ctd Use the below information to complete the following questions. Make sure to indicate the 5’ and 3’ ends whenever possible. The DNA sequence below contains one gene from a Eukaryotic cell. DNA Template: 3’-TACTGGTCCCCCTAAGCCTCAACAATT-5’ 5’ ATG 5’Cap: AUUATG Intron: AUUCGG

1. Use the above DNA sequence to complete gene expression.

A. Replicate by giving the complimentary DNA sequence. 5’ATGACCAGGGGGATTCGGAGTTGTTAA3’ B. Transcribe the DNA template into pre-mRNA. 5’AUGACCAGGGGGAUUCGGAGUUGUUAA3’ C. Complete RNA Processing by adding a 5’ cap, removing introns, and adding a poly-A tail. 5’AUUATGAUGACCAGGGGGAGUUGUUAA3’ D. Translate the mRNA into a polypeptide sequence. Met

Thr Arg

Gly

Ser Cys Stop

2. If this were a prokaryotic cell how would gene expression differ? (i.e What steps would we add or eliminate for prokaryotic cells) In prokaryotes, proteins are assembled directly from DNA because it does not contain ribosomes. RNA polymerases are involved only in initiation in eukaryotes. A. Where in a Eukaryotic cell does Replication occur? In a eukaryotic cell, replication occurs in nucleus during the S phase. B. Where in a Eukaryotic cell does Transcription occur? In a eukaryotic cell, transcription occurs in nucleus. C. Where in a Eukaryotic cell does Translation occur? In a eukaryotic cell, transition occurs in the cytoplasm. D. Where in a Prokaryotic cell does Replication occur? In a prokaryotic cell, replication occurs in the cytoplasm. E. Where in a Prokaryotic cell does Transcription occur? In a prokaryotic cell, transcription occurs in the cytoplasm simultaneously. F. Where in a Prokaryotic cell does Translation occur? In a prokaryotic cell, translation occurs in the cytoplasm

simultaneously.

3. Suppose a point mutation (underlined) shown below occurs. Wildtype DNA Template:

3’-TACTGGTCCCCCTAAGCCTCAACAATT-5’

Mutant DNA Template:

3’-TACTGGTCGCCCTAAGCCTCAACAATT-5’

A. What is the new pre-mRNA sequence? 5’AUGACCAGCGGGAUUCGGAGUUGUUGUUAA3’ B. What is the new polypeptide sequence? (Don’t forget to go through RNA processing) Met Thr Thr Gly Ser Stop C. What type of mutation is this? This is a nucleotide pair substitution mutation, causing a missense.

4. Suppose a point mutation (underlined) shown below occurs. Wildtype DNA Template:

3’-TACTGGTCCCCCTAAGCCTCAACAATT-5’

Mutant DNA Template:

3’-TACTGTTCCCCCTAAGCCTCAACAATT-5’

A. What is the new pre-mRNA sequence? 5’AUGACAAGGGGGAUUCGGAGUUGUUGUUAA3’ B. What is the new polypeptide sequence? (Don’t forget to go through RNA processing) Met Thr Arg Gly Ser Stop C. What type of mutation is this? This is a nucleotide pair substitution mutation, causing it to be silent.

1. What is the codon sequences for the anti-codons shown above? Which amino acids will bind to each tRNA that complements the codon? The codon sequences for the anti-codons shown is AAG. The amino acids that complements the codons is Lys. 2. The template strand of a gene includes this sequence: 3’TACTTGTCCGATATC-5’. It is mutated to 3’TACTTGTCCAATATC-5’. (A) For both wild-type and mutant sequences, determine the DNA compliment, the resulting mRNA, and the amino acid sequence each encodes.

Wild Type DNA : DNA Complement:

3’-TACTTGTCCGATATC-5’ 5’-ATGAACAGGCTATAG-3’

RNA Complement:

5’-AUGAACAGGCUAUAG-3’

Amino Acid Sequence:

Met Asn Arg Leu Stop

Mutant DNA:

3’-TACTTGTCCAATATC-5’.

DNA Complement:

5’ATGAACAGGTTATAG-3’

RNA Complement:

5’-AUGAACAGGUUAUAG-3’

Amino Acid Sequence:

Met Asn Arg Leu Stop

(B) What is the effect of the mutation on the amino acid sequence? This mutation is a nucleotide mutation which have a silent effect on the amino acid sequences. This means that there is a chance in the base pairing, but not the amino acid sequence did not change.

Listed are the key terms to understand in order to be able to answer questions over material covered in this chapter. Be sure to give definitions for each term as you study and use these terms in responses to your worksheets. -

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Transcription: DNA strand is copied into a new molecule of RNA Translation: mRNA to polypeptides during protein synthesis RNA polymerase: enzymes that responsible for coping a DNA sequence into an RNA sequence during transcription. Ribosome (A,P, E site): o A site: Holds the tRNA that carries the next amino acid to be added to the chain. E site: It is the exit site, where discharges tRNAs leave the ribosome. o P site: Holds the tRNA that carries the growing polypeptides.

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mRNA: messenger RNA, carries instructions from DNA in the nucleus to the ribosomes. tRNA: transfer RNA, transfers amino acid to the growing polypeptide into a ribosome. rRNA: ribosomal RNA, type of RNA that makes up part of the ribosome. Codon: in RNA, a three-base “word” that codes for one amino acid. Anti-codon: base-pairs with a complementary codon on mRNA. RNA processing: a sequence of events through which the primary transcript from a gene acquires its mature form.

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5’Cap: the 5’ ends received a modified nucleotide. 3’ Poly-A-tail: attached to the 3’ends. Splicing: removes introns and joins exons creating an mRNA molecule with a continuous coding sequence. Exon: expressed sequences Intron: intervening (cut out) sequences Spliceosome: consists of a variety of proteins and several small RNAs that recognize the splice sites. Promoter: the DNA sequence where RNA polymerase attaches to.

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TATA box: crucial in forming the initiation complex in eukaryotes. Initiation: the small ribosomal subunit binds to the start of the mRNA sequence. Then a tRNA carrying the amino acid methionine binds to a start codon of the mRNA sequence. Then, the large ribosomal subunit binds to form the complete initiation complex. Elongation: the ribosome continues to translate each codon in turn. Each corresponding amino acid is added until all of the codons are read. Termination: This occurs when the ribosome reaches a stop codon. AS the ribosomes recognized that translation is complete. The new protein is then released, and the translation complex comes apart.

1.

2.

3.

4.

5.

In eukaryotic cells, transcription cannot begin until A.

the two DNA strands have completely sepa-rated and exposed the promoter.

B.

several transcription factors have bound to the promoter.

C.

the 5’ caps are removed from the mRNA.

D.

the DNA introns are removed from the template.

Which of the following is not true of a codon? A.

It may code for the same amino acid as another codon.

B.

It never codes for more than one amino acid.

C.

It extends from one end of a tRNA molecule.

D.

It is the basic unit of the genetic code.

The anticodon of a particular tRNA molecule is A.

complementary to the corresponding mRNA codon.

B.

complementary to the corresponding triplet in rRNA.

C.

the part of tRNA that bonds to a specific amino acid.

D.

catalytic, making the tRNA a ribozyme.

Which of the following is not true of RNA processing? A.

Exons are cut out before mRNA leaves the nucleus.

B.

Nucleotides may be added at both ends of the RNA.

C.

Ribozymes may function in RNA splicing.

D.

RNA splicing can be catalyzed by spliceosomes.

Which component is not directly involved in translation? A.

GTP

B.

DNA

C.

tRNA

D.

ribosomes

1. Which of the following best describes euchromatin and gene expression? A. The DNA is tightly compact and transcription occurs easily. B. The DNA is loosely compact and transcription occurs easily. C. The DNA is loosely compact and transcription does not occur easily. D. The DNA is tightly compact and transcription does not occur easily. 2. Histone Acetylation results in which of the following? A. The DNA becomes more tightly compact and transcription will not occur. B. The DNA becomes loose and transcription will not occur. C. The DNA becomes more tightly compact and transcription will occur. D. The DNA becomes loose and transcription will occur. 3. Which statement best describes transcription factors? A. Transcription factors encourage transcription. B. Transcription factors inhibit transcription. C. Transcription factors encourage translation. D. Transcription factors inhibit translation. 4. All of the following are ways in which to control gene expression at the RNA level except. A. Splicing B. Alternative splicing C. siRNAs D. DNA methylation 5. The proteasome controls gene expression at which level. A. DNA B. RNA C. mRNA D. Protein 6. Muscle cells differ from nerve cells mainly because they A. express different genes. B. contain different genes. C. use different genetic codes. D. have unique ribosomes.

7. Which of the following is an example of post-transcriptional control of gene expression? A. the addition of methyl groups to cytosine bases of DNA B. the binding of transcription factors to a promoter C. the removal of introns and alternative splicing of exons D. gene amplification contributing to cancer 6. What would occur if the repressor of an inducible operon were mutated so it could not bind the operator? A.

irreversible binding of the repressor to the promoter

B.

reduced transcription of the operon’s genes

C.

buildup of a substrate for the pathway controlled by the operon

D.

continuous transcription of the operon’s genes...


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