Prentice Hall Biology (California)
Prentice Hall Biology (California)
1st Edition
Kenneth R. Miller, Levine
ISBN: 9780132013529
Textbook solutions

All Solutions

Page 315: Chapter 12 Assessment

Exercise 1
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A bacterial strain is capable of changing into another strain through the process of **transformation**. This is due to the acquisition of factors such as plasmid genes by bacteria.
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The process of transformation was demonstrated in one of Griffith’s experiments on bacterial cells. A harmless strain of bacteria was able to change into a harmful strain by acquiring the necessary plasmid genes from harmful bacterial cells that were heat-killed.
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C
Exercise 2
Solution 1
Solution 2
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The answer is d, or viruses.
D is correct because bacteriophages are particles of nucleic acids and proteins that infect other things. That is what bacteria do. Bacteriophages infect bacteria.
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d. viruses
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A bacteriophage is a type of **virus** that specifically targets bacterial cells. They are generally harmless towards eukaryotic cells as they only infect and transfer their viral genes into bacteria.
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When a bacteriophage infects a bacterial cell, the virus injects its genetic material into the host cell. The viral genes would then be replicated and/or expressed by the host’s machinery.
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D
Exercise 3
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The structure of nucleotides typically consist of three things. These include a 5-carbon sugar (ribose or deoxyribose), a nitrogen base (purine or pyrimidine), and a phosphate group.
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**Polymerase** is *not* a part of the structure of nucleotides. It is an enzyme that is used to incorporate nucleotides into a nucleic acid strand such as DNA or RNA during its synthesis.
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B
Exercise 4
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Prokaryotes lack nucleus. Since they do not possess membrane-bound organelles, they have their genetic material in the cytoplasm.
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C. cytoplasm
Exercise 5
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The diagram shows DNA replication, wherein the DNA molecule has produced two new strands. One strand serves as a template for the new strand.
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A. replication
Exercise 6
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DNA polymerase is an enzyme involved in DNA replication because it joins nucleotides to form a DNA molecule. It ensures that the new DNA strand is the exact copy of the original DNA.
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D. DNA polymerase
Exercise 7
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In the process of transcription, an information or a particular segment in the DNA is used as a template and copied into the RNA by an enzyme called RNA polymerase.
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B. transcription
Exercise 8
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Codons, the three-letter word in mRNA, specifies a particular amino acid that is added to the polypeptide chain.
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C. amino acid
Exercise 9
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Mutation occurs when there are changes in the structure of the genetic material of an organism. It is usually caused by the changes in the arrangement of base units that are found in the DNA.
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B. mutations
Exercise 10
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Exon is the term used to describe the part of a gene that is expressed during protein synthesis. They are spliced back together to produce the final mRNA.
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d. is made of mRNA
Exercise 11
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Here are the three critical functions of a gene:

1. Genetic information can be transferred from one generation to the next one.

2. Genes determine the heritable characteristics or traits of an organism.

3. Genes can be copied or replicated during cell division.

Exercise 12
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The structure of a nucleotide in DNA typically consists of three components. These include a 5-carbon sugar, a nitrogenous base, and a phosphate group.
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In DNA, the 5-carbon sugar is known as *deoxyribose*. On the other hand, the sugar in RNA is known as *ribose*. The difference between the two is the presence of a hydroxyl group.
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In DNA, the nitrogenous bases would include adenine, guanine, cytosine, and thymine. These bases are also found in RNA except that thymine is replaced by another base, uracil.
Exercise 13
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Chargaff’s rules explain that there is always an equal quantity of adenine and thymine bases, whereas there is also an equal quantity of guanine and cytosine bases in a DNA molecule.
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The premise presented by Chargaff’s rules implies that there is base pairing that is specific between A and T and between G and C. This conclusion contributed in the modeling of DNA.
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In fact, the DNA model presented by Watson and Crick had shown that the adenine and thymine bases, as well as the guanine and cytosine bases, are bound by hydrogen bonds.
Exercise 14
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Base pairing is a principle which explains how hydrogen bonds are formed in the nitrogenous bases. A hydrogen bond forms only between specific base pairs – cytosine and guanine, and adenine and thymine. Base pairing becomes involved in the DNA replication when old strands are replicated to form a new strand. Base pairing is important in ensuring that the new strand complements the other template in order to conserve the base sequence.
Exercise 15
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Prokaryotes lack nucleus. Since they do not possess membrane-bound organelles, they have their genetic material in the cytoplasm. Usually, prokaryotic cells have a single chromosome that holds the DNA. Their chromosomes are circular in shape and they form a loop since their DNA molecules are long.
Exercise 16
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In DNA replication, the DNA molecule produces two new strands, wherein each original strand serves as a template for the new strands. The molecules found in the new strand must be complementary to the original strand in order to follow the principle of base pairing and to ensure that the base sequence is maintained.
Exercise 17
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Eukaryotic chromosomes have a tightly coiled packaging that is enclosed in the nuclear membrane. In the chromosome, the DNA is coiled around proteins which are called histones, and together, they form a beadlike structure called nucleosome. In turn, the nucleosomes are tightly coiled to produce chromatin fibers, then these fibers are supercoiled to form chromosomes.
Exercise 18
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Exon is the term used to describe the part of a gene that is expressed during protein synthesis. They are spliced back together to produce the final mRNA. On the contrary, introns are the sequences of nucleotides that are not involved during the coding for proteins. They are usually cut out of RNA molecules before they become functional.
Exercise 19
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A sequence of DNA typically consist of several nucleotides. The cell “reads” the genetic information encoded by the nucleotides in terms of three-letter sequences called **codons**.
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In the DNA sequence, each codon consists of three letters (which correspond to each of the nitrogenous bases) that match a specific amino acid according to the universal genetic code.
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As the cell transcribes DNA into RNA, the RNA transcript is then translated in terms of codons that would encode a polypeptide. This is how cells express their DNA genes into a protein.
Exercise 20
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Similar to a codon, an **anticodon** is a three-letter sequence, but it is found in transfer RNA. An anticodon is complementary to a codon in an RNA sequence that matches with an amino acid.
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Each anticodon is associated with the amino acid that matches with the codon in RNA according to the genetic code. Therefore, it is also important that the anticodon sequence is correct.
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The anticodon allows the tRNA to match with the correct codon in the RNA transcript in order to incorporate the correct amino acids in the growing polypeptide chain during translation.
Exercise 21
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The messenger RNA codon should be complementary to the codon in the DNA molecule, except that thymine is replaced with uracil. If the DNA codon is CTA, then the mRNA codon is **GAU**.
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Next, the transfer RNA anticodon should be complementary to the mRNA codon that was transcribed. If the mRNA codon is GAU, then the transfer RNA anticodon should be **CUA**.
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As we can see, the transfer RNA anticodon is identical in sequence with the DNA codon. However, the difference is that the thymine is replaced by uracil in the transfer RNA anticodon.
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mRNA codon = GAU; tRNA anticodon = CUA
Exercise 22
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A gene encodes a certain protein that controls the characteristics of an organism. Proteins, which are enzymes, are responsible for catalyzing chemical reactions such as the production of hormones and regulating the pattern and rate of growth. The changes in the genetic material of a person alter the nature of the protein which may lead to changes in the phenotype of an organism.
Exercise 23
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. There are two basic categories of mutations: $text{textcolor{#c34632}{gene mutations}}$ and $text{textcolor{#19804f}{chromosomal mutations}}$.

Gene mutations cause $text{textcolor{#c34632}{changes in one gene}}$, while chromosomal mutations cause $text{textcolor{#19804f}{changes in whole chromosomes}}$. An example of a gene mutation is a $text{textcolor{#c34632}{substitution}}$ mutation, in which one base into a codon is changed to a different base. An example of a chromosomal mutation is a $text{textcolor{#19804f}{deletion}}$ mutation, in which all or part of a chromosome is lost.

Result
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See explanation.
Exercise 24
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A TATA box, which is a sequence found in the DNA, helps the RNA polymerase by marking where a genetic sequence can be read and decoded. It is positioned next to a promoter and indicates where the transcription begins.
Exercise 25
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An **operon** is known as the functional unit of genetic information in prokaryotes. It consists of a cluster of genes that are all related to a particular cellular process or pathway in the cell.
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The presence of operons in prokaryotes allows them to coordinate the expression of multiple related genes in response to the needs of the cell. This gives an advantage in cell survival.
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Furthermore, the expression of an operon is controlled by a single promoter sequence in DNA. Additionally, the expression of operons is also regulated by proteins such as repressors.
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One example of an operon is the *lac* operon, which functions in the production of enzymes that digest lactose. The expression of the *lac* operon is dependent on the presence of lactose.
Exercise 26
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In Griffith’s experiment, the rat survived when it was injected with harmless bacteria (rough colonies). In a separate experiment, the rat also survived even when it was injected with the heat-killed disease-causing bacteria (smooth colonies). Because of these findings, Griffith realized that neither of the two samples has caused the death of the rat so he decided to mix the heat-killed disease-causing bacteria and the harmless bacteria and injected it into the rat. When the rat died of pneumonia after receiving the mixture, he was surprised the there are no harmless bacteria in its lungs and instead, the heat-killed bacteria were inside it. It was an indication that the heat-killed bacteria transferred its ability to the harmless strain. Griffith used the term transformation to describe this process.
Exercise 27
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Watson and Crick’s DNA model represented the DNA structure by showing its double helix structure and its components. They shed light on the principle of base pairing, particularly on how hydrogen bonds hold the nitrogenous base-pairs together.
Exercise 28
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DNA replication can be similar when a person photocopies a document. In DNA replication, the original DNA serves as a template to produce new strands. Just like when a document is being photocopied, an original document is used to produce new copies. However, the major difference between DNA replication and photocopying a document is the process of obtaining the final product. The former gives us two new complementary strands. It doesn’t produce an exact copy of the original strand, but it gives a base sequence that complements the original strand, which in this case is the one that follows the principle of base pairing. As a result, it gives us a helix containing the old and the new strand. On the contrary, the latter produces an exact copy of the original document in which there are no changes made in terms of the content of both documents – the original and the photocopied.
Exercise 29
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Here are the rules of base pairing between DNA strands and mRNA:

1. cytosine (C) pairs with guanine (G)
2. since thymine is replaced by uracil in RNA, the base pairing for DNA and RNA becomes: adenine (A) pairs with uracil (U)

Considering the rules of base pairing shown above, the base pair for ACCGTCAC and TCGCACGT is UGGCAGUG and AGCGUGCA.

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The base pair for ACCGTCAC is UGGCAGUG, whereas TCGCACGT will have AGCGUGCA.
Exercise 30
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In eukaryotic cells, introns are usually removed from a primary RNA transcript in order to produce the mature RNA. Introns are noncoding sequences that do not code for protein.
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If an intron is not removed during the splicing of RNA, then the final mRNA transcript would not be translated properly. This would result in a dysfunctional or nonfunctional polypeptide.
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However, if the intron is able to function as an exon (which means that it can encode for a functional protein), then the mRNA transcript would be translated into a different type of protein.
Exercise 31
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Transcription: The word transcribe refers to the process of copying or writing down what was said.

How the technical term relates to the transcription in genetics: In biology, the process of transcription occurs when RNA is produced by copying the nucleotide bases of DNA. RNA polymerase serves as the initiator of this process by binding to a promoter. It is the one responsible for telling that the DNA molecule must use the original DNA strand to produce another copy of the nucleotide base into a strand of RNA.

Step 2
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Translation: The word translate means to express an information from an original language to another language.

How the technical term relates to the transcription in genetics: In biology, the term translation refers to the process in which a genetic code from the mRNA is translated by the ribosome in order to produce proteins. The language used by the mRNA is the genetic code called codon which is used in order to carry instructions for different amino acids. Since codon uses three letter words to represent an amino acid, translation is needed in order for the transfer RNA to receive the instructions and deliver the right amino acids and the anticodon.

Exercise 32
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Watson and Crick found the remaining pieces of evidence to solve the mystery behind the DNA structure by using Franklin’s X-ray pattern as their basis.

1. The X-shaped pattern suggests that the DNA is a helix since there are two strands that are twisted around each other which is like the coils in a spring.
2. The pattern also indicates that there is nitrogenous base found at the center of the molecule.

Because of these discoveries, Watson and Crick were able to construct the structure of the DNA model and they were able to explain how DNA carries information and how it can replicate itself.

However, Franklin’s X-ray pattern did not fully explain how the bonds between the two strands are formed so Watson and Crick decided to find the forces that hold the strands together. Eventually, they discovered that hydrogen bonds can only be formed when certain base pairs are joined together.

Result
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By using Franklin’s X-ray pattern as a basis, Watson and Crick were able to construct the structure of the DNA model and they were able to explain how DNA carries information and how it can replicate itself. However, Franklin’s X-ray pattern did not fully explain how the bonds between the two strands are formed so Watson and Crick decided to find the forces that hold the strands together. Eventually, they discovered that hydrogen bonds can only be formed when certain base pairs are joined together.
Exercise 33
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If a chromosomal mutation happens during meiosis, the mutation will be inherited by the organism’s offspring. This event may lead to genetic variation and even evolution. On the contrary, if a chromosomal mutation occurs during mitosis, it won’t be inherited by the offspring. As a result, the harmful effects are just limited to the organism.
Exercise 34
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Since the given RNA strand is still within the nucleus, it is likely that it has not undergone RNA splicing. This means that the RNA strand may still have noncoding introns that are present.
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Even though the sequence AAC is identified in the RNA strand, this does not mean that it would be encoded into an amino acid. This is because the sequence might be a part of an intron.
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In RNA splicing, introns are removed from the mRNA transcript to yield the mature mRNA. It is possible that the sequence AAC is removed due to it being part of the intron sequence.
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Hence, the amino acid asparagine (encoded by AAC) would not necessarily appear in the translated protein. There is still a possibility that this sequence is spliced out of the mRNA.
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