Mastering Biology Chapter 17 PDF

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Mastering Biology Chapter 17 ABIO121 Assignment. ...

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Mastering Biology Chapter 17 Assignment No. 7 Item 1 The central paradigm of biochemistry holds that information flows from DNA to RNA to protein. The process of making protein from the mRNA is called translation. Translation is carried out by the ribosome, which binds to the mRNA and binds tRNA, which recognizes the codons on the mRNA and brings the appropriate amino acid with it. The ribosome forms the peptide bond between the new amino acid and the growing peptide chain.

The process of translation, or protein synthesis, is a crucial part of the maintenance of living organisms. Proteins are constantly in use and will break down eventually, so new ones must always be available. If protein synthesis breaks down or stops, then the organism dies. Item 2 DNA polymerase is very accurate and rarely makes a mistake in DNA replication. Occasionally, however, an error in replication, known as a point mutation, is introduced. There are two general categories of point mutations—frameshift mutations (also called base-pair insertions or base-pair deletions) and base substitution mutations (shown in the diagram).

Base substitution mutations and Frameshift mutations If a segment of DNA were replicated without any errors, the replicated strand would have the following sequence of nucleotides: 5' - ACTACGTGA - 3'

Sort the following replicated DNA sequences by the type of point mutation each contains (frameshift, base substitution, or neither), as compared to the correct sequence shown above.

A base substitution mutation can occur if the DNA polymerase inserts the wrong nucleotide base as it synthesizes a new strand of DNA. A frameshift mutation can occur if the DNA polymerase leaves out a nucleotide or adds an extra nucleotide to the sequence. Certain forms of cancer occur because of mutations in DNA sequences that are located in so-called mutational hotspots. These hotspots are locations in the DNA sequence where mutations occur more often than in other places. __________________________________________________________________________________ Types of base substitution mutations When a base substitution mutation occurs, one nucleotide in a replicating DNA sequence is substituted for another, which results in the production of a mutant strand of DNA. The result of the mutation depends on how the substituted nucleotide base alters the string of amino acids coded by the mutant DNA. The three types of base substitution mutations are nonsense mutations, missense mutations, and silent mutations. Each type is defined by how it affects protein synthesis.

Point mutations in DNA sequences can profoundly affect protein synthesis, or they can have no effect at all. Point mutations can be beneficial to an organism but are more commonly neutral or harmful. Silent mutations are errors in DNA replication that do not cause a change in the amino acid sequence A nonsense mutation is a point mutation in a sequence of DNA that results in a premature stop codon (in this example leads to codon UGA and UAG) – shortens the proteome A missense mutation is a point mutation in which a single nucleotide change results in a codon that codes for a different amino acid A frameshift mutation is a type of mutation involving the insertion or deletion of a nucleotide in which the number of deleted base pairs is not divisible by three. "Divisible by three" is important because the cell reads a gene in groups of three bases. Each group of three bases corresponds to one of 20 different amino acids used to build a protein. If a mutation disrupts this reading frame, then the entire DNA sequence following the mutation will be read incorrectly. Severity of point mutations 1.

Generally speaking, which of the following mutations would most severely affect the protein coded for by a gene? - A frameshift deletion at the beginning of the gene - A frameshift mutation at the beginning of a gene would affect every codon after the point where the mutation occurred. During protein synthesis, incorrect amino acids would be inserted from the point where the frameshift mutation occurred on; the resulting protein would most probably be nonfunctional. For this reason, a frameshift mutation at the beginning of a gene is generally the most severe type of mutation. Item 3

Translation is the mRNA-directed synthesis of polypeptides. In translation, the information encoded in a sequence of RNA nucleotides is converted into a sequence of amino acids according to the genetic code. Translation also includes the first stage of targeting proteins to their eventual cellular location. tRNA interactions with mRNA and the ribosome

Ribosomes provide the scaffolding on which tRNAs interact with mRNA during translation of an mRNA sequence to a chain of amino acids. A ribosome has three binding sites, each of which has a distinct function in the tRNA-mRNA interactions

During translation, new amino acids are added one at a time to the growing polypeptide chain. The addition of each new amino acid involves three steps: Binding of the charged tRNA to the A site. This step requires correct base-pairing between the codon on the mRNA and the anticodon on the tRNA. Formation of the new peptide bond. In the process, the polypeptide chain is transferred from the tRNA in the P site to the amino acid on the tRNA in the A site. Movement of the mRNA through the ribosome. In this step, the discharged tRNA shifts to the E site (where it is released) and the tRNA carrying the growing polypeptide shifts to the P site.

Predicting the effect of a point mutation The diagram below shows an mRNA molecule that encodes a protein with 202 amino acids. The start and stop codons are highlighted, and a portion of the nucleotide sequence in the early part of the molecule is shown in

detail. At position 35, a single base-pair substitution in the DNA has changed what would have been a uracil (U) in the mRNA to an adenine (A).

1. Based on the genetic code chart above, which of the following would be the result of this single basepair substitution? - a nonsense mutation resulting in early termination of translation - The effect of a single base substitution depends on the new codon formed by the substitution. To identify the new codon, it is first necessary to determine the reading frame for the amino acid sequence. The first codon starts with base 1, the second codon with base 4, the third with base 7, and so on. - In this problem, the codon that contains the single base substitution begins with base 34. The original codon (UUA, which encodes the amino acid leucine) is converted by the single base substitution to UAA, which is a stop codon. This will cause premature termination of translation, also called a nonsense mutation. Protein targeting pathways The DNA in a cell’s nucleus encodes proteins that are eventually targeted to every membrane and compartment in the cell, as well as proteins that are targeted for secretion from the cell. For example, consider these two proteins:  

Phosphofructokinase (PFK) is an enzyme that functions in the cytoplasm during glycolysis. Insulin, a protein that regulates blood sugar levels, is secreted from specialized pancreatic cells.

Assume that you can track the cellular locations of these two proteins from the time that translation is complete until the proteins reach their final destinations. For each protein, identify its targeting pathway: the sequence of cellular locations in which the protein is found from when translation is complete until it reaches its final (functional) destination. (Note that if an organelle is listed in a pathway, not in the membrane that surrounds the organelle.)

There are two general targeting pathways for nuclear-encoded proteins in eukaryotic cells.  

Proteins that will ultimately function in the cytoplasm (PFK, for example) are translated on free cytoplasmic ribosomes and released directly into the cytoplasm. Proteins that are destined for the membranes or compartments of the endomembrane system, as well as proteins that will be secreted from the cell (insulin, for example), are translated on ribosomes that are bound to the rough ER.

For proteins translated on rough ER, the proteins are found in one of two places at the end of translation. If a protein is targeted to a membrane of the endomembrane system, it will be in the ER membrane. If a protein is targeted to the interior of an organelle in the endomembrane system or to the exterior of the cell, it will be in the lumen of the rough ER. From the rough ER (membrane or lumen), these non-cytoplasmic proteins move to the Golgi apparatus for processing and sorting before being sent to their final destinations. Item 4 In the process of transcription, the genetic information encoded in the sequence of bases that makes up a gene is “transcribed,” or copied in the same language, into a strand of RNA bases. The enzyme that catalyzes this reaction is called an RNA polymerase. In eukaryotes, before the resulting strand (called pre-mRNA) leaves the nucleus, it is processed in several ways. The product of this processing is the mRNA that functions as the template for protein synthesis outside the nucleus. Before beginning this tutorial, watch the Transcription and RNA Processing animations. Pay particular attention to the base pairing that occurs during transcription and the various steps involved in RNA processing There are three principles to keep in mind when predicting the sequence of the mRNA produced by transcription of a particular DNA sequence. 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 steps of transcription and RNA processing

Once RNA polymerase II is bound to the promoter region of a gene, transcription of the template strand begins.   

Soon after transcription begins, the 5' end of the RNA transcript is capped by addition of a modified guanine nucleotide. As transcription continues, introns are spliced out of the RNA transcript. After transcription ends, a poly-A tail (chain of adenine nucleotides) is added to the 3' end of the RNA transcript.

Only after all these steps have taken place is the mRNA complete and capable of exiting the nucleus. Once in the cytoplasm, the mRNA can participate in translation. You labeled 5 of 6 targets incorrectly. For target (a), consider that before any modification of the RNA transcript can occur, there must be at least a partially completed RNA transcript to modify. Item 5 1. Which statement is correct concerning the function(s) of the 5’ cap and the 3’ poly(A) tail of eukaryotic mRNAs? -

Both structures serve as recognition signals for the translational machinery and extend the life span of the mRNA. Experiments have shown that mRNAs with a cap and a tail last longer and produce more proteins when introduced into cells. Item 6



During elongation, the RNA strand is extended in a 5' to 3' direction. Item 7

1. Where does translation take place? -

Ribosome

2. Which nucleic acid is translated to make a protein? -

mRNA mRNA is the message that is translated to make a protein

3. Which of the following processes is an example of a post-translational modification? -

Phosphorylation

-

Enzymes can phosphorylate proteins to alter their activity.

4. Which of the following steps occurs last in the initiation phase of translation? -

The large ribosomal subunit joins the complex. This step occurs after the 5’ mRNA is bound by the ribosome and the start codon is bound by an aminoacyl tRNA.

5. At which site do new aminoacyl tRNAs enter the ribosome during elongation? -

A-Site This is the site at which new aminoacyl tRNAs that are complementary to the mRNA codon enter the ribosome.

6. What is meant by translocation? -

The ribosome slides one codon down the mRNA. Translocation is the process by which the ribosome slides down the mRNA so a new cycle of elongation can begin.

7. True or false. A tRNA with an anticodon complementary to the stop codon catalyzes the reaction by which translation is terminated. -

False There are no tRNAs complementary to the three stop codons; termination occurs when release factors recognize the stop codon in the A-site and catalyze the release of the polypeptide from the tRNA in the P-site Item 8

1. There should be a strong positive correlation between the rate of protein synthesis and _____. -

the number of ribosomes

Item 9

1. The direction of synthesis of an RNA transcript is _____. -

5’  3’ Nucleotides are added to the 3' end of RNA.

Item 10

1. The wobble hypothesis explains the _____. -

ability of some tRNAs to read more than one codon Item 11

Chromosomal mutations are changes in the normal structure or number of chromosomes.  

Changes in chromosome structure can result from errors in meiosis or from exposure to radiation or other damaging agents. Certain changes in chromosome number can result from nondisjunction during either meiosis or mitosis.

Both structural mutations and nondisjunction can play a role in trisomy 21, commonly known as Down syndrome.

A deletion is the loss of part of a chromosomal segment. A duplication is the repetition of a segment. The repeated segment may be located next to the original or at a different location, and its orientation may be the same as the original or the reverse. An inversion is the removal of a segment followed by its reinsertion into the same chromosome in the reverse orientation. A translocation is the transfer of a segment to a nonhomologous chromosome. Translocations may be reciprocal (two nonhomologous chromosomes exchange segments) or nonreciprocal (one chromosome transfers a segment without receiving one). Nondisjunction Suppose a diploid cell with three pairs of homologous chromosomes (2 n = 6) enters meiosis.

How many chromosomes will the resulting gametes have in each of the following cases?

If one chromosome pair undergoes nondisjunction in meiosis I, half the gametes will have an extra chromosome (n +1), and half will be missing a chromosome ( n – 1). If all chromosome pairs undergo nondisjunction in meiosis I, half the gametes will have twice the normal haploid number of chromosomes (2n), and half will have no chromosomes. If one chromosome undergoes nondisjunction in meiosis II, half the gametes will have the normal haploid number of chromosomes (n), one-quarter will have an extra chromosome (n +1), and one-quarter will be missing a chromosome (n – 1). If all chromosomes undergo nondisjunction in meiosis II, half the gametes will have the normal haploid number of chromosomes (n), one-quarter will have twice the haploid number (2 n), and one-quarter will have no chromosomes. Trisomy 21 Down syndrome is caused by trisomy 21, the presence of three copies of chromosome 21. The extra copy usually results from nondisjunction during meiosis. In some cases, however, the extra copy results from a translocation of most of chromosome 21 onto chromosome 14. A person who has had such a translocation in his or her gamete-producing cells is a carrier of familial Down syndrome. The carrier is normal because he or she still has two copies of all the essential genes on chromosome 21, despite the translocation. However, the same may not be true for the carrier’s offspring. The diagram shows the six possible gametes that a carrier of familial Down syndrome could produce.

A carrier of familial Down syndrome has two copies of chromosome 21 and a normal phenotype. However, one of those copies has been translocated to another chromosome, often chromosome 14. Some of the carrier’s gametes will contain both the normal and the translocated chromosome 21. If one of those gametes fuses with a gamete from a person with a normal karyotype, a zygote with trisomy 21 will result. Item 12

1. During RNA processing a(n) _____ is added to the 5' end of the RNA. -

modified guanine nucleotide

2. During RNA processing a(n) _____ is added to the 3' end of the RNA. -

a long string of adenine nucleotides

3. Spliceosomes are composed of _____. -

snRNPs and other proteins

4. The RNA segments joined to one another by spliceosomes are _____. -

Exons

5. Translation occurs in the _____. -

Cytoplasm Ribosomes, the sites of translation, are found in the cytoplasm. Item 13

1. The micrograph shows a DNA-mRNA hybrid. If the noncoding regions of the gene did not exist, what would the micrograph look like? -

All of the loops would be missing.

-

If there were no introns (noncoding regions of DNA not present in mRNA) then the DNA and mRNA would match up exactly when they base pair with each other. There would be no loops, which correspond to the noncoding DNA. Item 14

Put the following events of transcription in chronological order. 1. RNA polymerase holoenzyme binds to the promoter region. 2. The double helix of DNA is unwound, breaking hydrogen bonds between complementary strands. 3. Sigma binds to RNA polymerase. 4. Sigma is released. 5. Transcription begins. Answer: 3, 1, 2, 5, 4 Item 15

1. What is the process called that converts the genetic information stored in DNA to an RNA copy? -

Transcription DNA is transcribed to give an RNA copy.

2. Transcription begins at a promoter. What is a promoter? -

A site in DNA that recruits the RNA Polymerase in order to begin transcription. This is the site where the RNA polymerase must bind to initiate transcription.

3. What determines which base is to be added to an RNA strand during transcription? - Base pairing between the DNA template strand and the RNA nucleotides - Transcription involves the formation of an RNA strand that is complementary to the DNA template strand. 4. Which of the following terms best describes the relationship between the newly synthesized RNA molecule and the DNA template strand? -

Complementary Because the template strand determines the nucleotides to be added to the RNA strand, using the same complementarity rules of the DNA, they will be complementary to each other.

5. What happens to RNA polymerase II after it has completed transcription of a gene? -

It is free to bind to another promoter and begin transcription. The enzyme is free to transcribe other genes in the cell. Item 16

In translation, a cell reads an mRNA message and assembles a polypeptide accordingly.

Here's how translation occurs. As an mRNA strand slides though a ribosome, triplets of RNA bases spell out the amino acid sequence of a polypeptide. It is the job of transfer RNA molecules to match RNA bases with the correct amino acids. Click on each component shown here to find out more about them.

We can divide translation, the process of building a polypeptide, into three stages— initiation, elongation, and termination. Translation is initiated when the small ribosomal subunit binds to the leader at the 5' end of the mRNA molecule. The anticodon of the initiator tRNA binds to the start codon,

AUG. The initiator tRNA always bears the amino acid methionine. Proteins called initiation factors help bring the mRNA, the initiator tRNA, and the small ...