Miller and Levine Biology
Miller and Levine Biology
1st Edition
Joseph S. Levine, Kenneth R. Miller
ISBN: 9780328925124
Textbook solutions

All Solutions

Page 468: Assessment

Exercise 1
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DNA and RNA are both nucleic acids. Deoxyribonucleic acid or DNA is made of nucleotides which are combined with covalent bonds and form a double chain. Nucleotides consist of deoxyribose, the phosphate group, and a nitrogenous base (adenine, cytosine, thymine, and guanine). Ribonucleic acid (RNA) also consists of nucleotides that are bound with covalent bonds, but this type of nucleic acid has one strand. Its nucleotides are made of ribose, the phosphate group, and a nitrogenous base (adenine, cytosine, uracil, and guanine).
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d. a single strand and contains uracil
Exercise 2
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During transcription, an enzyme known as RNA polymerase binds to promoters of the DNA. Promoters are specific sequences of nucleotides that are present in many locations in the DNA molecule. They represent the starting point for transcription. RNA polymerase then breaks down hydrogen bonds between two strands of the DNA and attaches on one of them. A DNA strand is used as a template for new RNA molecule. In this way, pre-mRNA is made, which becomes mRNA after the introns are removed.
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c. DNA
Exercise 3
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DNA and RNA are both nucleic acids. Deoxyribonucleic acid or DNA is made of nucleotides which are combined with covalent bonds and form a double chain. Nucleotides consist of deoxyribose, the phosphate group, and a nitrogenous base (adenine, cytosine, thymine, and guanine). Ribonucleic acid (RNA) also consists of nucleotides that are bound with covalent bonds, but this type of nucleic acid has one strand. Its nucleotides are made of ribose, the phosphate group, and a nitrogenous base (adenine, cytosine, uracil, and guanine).
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c. phosphate groups, guanine, and cytosine
Exercise 4
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During transcription, an enzyme known as RNA polymerase binds to promoters of the DNA, which are specific sequences of nucleotides that are present in many locations in the DNA molecule. They represent the starting point for transcription. RNA polymerase then breaks down hydrogen bonds between two strands of the DNA and attaches on one of them. A DNA strand is used as a template for a new RNA molecule. In this way, pre-mRNA is made, which becomes mRNA after the introns are removed.
Exercise 5
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The function of RNA is in synthesizing proteins. We differ three types of this molecule:
1. Messenger RNA
2. Ribosomal RNA
3. Transfer RNA
During transcription, an enzyme known as RNA polymerase binds to promoters of the DNA. RNA polymerase then breaks down hydrogen bonds between two strands of the DNA and attaches on one of them. A DNA strand is used as a template for mRNA molecule. A mRNA molecule enters the ribosomes that consist of several types of rRNA and proteins. A tRNA attaches to mRNA molecule. The starting point in the translation process is AUG codon. Based on codons, tRNA with adequate anticodons adds amino acids in order to make polypeptide molecules. The translation process is completed when the stop codon is reached.
Exercise 6
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Introns and exons are parts of the DNA molecule in which exons represent coding, while introns are non-coding areas. During the transcription process, an RNA polymerase builds a pre-mRNA molecule that contains both of these regions. Since introns don’t contain information about protein synthesis, they are being removed, so a mRNA molecule consists only of exons.
Exercise 7
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The translation process begins when a mRNA molecule enters the ribosomes, where tRNA attaches to it. A mRNA molecule is built of codons, which are sequences of three nucleotides in a row, which determinate an amino acid. The starting point in the translation process is AUG codon. Based on these codons, tRNA with adequate anticodons adds amino acids in order to make polypeptide molecules. The translation process is completed when the stop codon is reached.
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b. the cell uses a messenger RNA code to make proteins
Exercise 8
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A mRNA molecule is built of codons, which are sequences of three nucleotides in a row, which determinate an amino acid. A mRNA molecule is always red codon by codon and in the same direction.
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a. three bases at a time in the same direction
Exercise 9
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An RNA polymerase binds to promoters and then breaks down hydrogen bonds between two strands of the DNA and attaches on one of them. A DNA strand is used as a template for a new RNA molecule. In this way, pre-mRNA is made, which becomes mRNA after the introns are removed.
A mRNA molecule enters the ribosomes, located in the cytoplasm. In these organelles, a transfer RNA attaches to a mRNA and the process of translation begins. The final product is a polypeptide chain.
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d. translate DNA into RNA
Exercise 10
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The genetic code is written in the molecules of DNA and RNA. DNA is made of four types of bases – adenine, cytosine, thymine, and guanine. However, RNA molecule consists of adenine, cytosine, uracil, and guanine. These four bases build codons in the mRNA based on which the polypeptides are built in the process of translation.
Exercise 11
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During transcription, a mRNA molecule is built based on a certain nucleotide sequence on the DNA molecule. A mRNA molecule now carries an information about protein synthesis. The next process is the translation. In this stage, codons of the mRNA determine the order of amino acids in a polypeptide chain.
Exercise 12
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Genetic is a science that explores genes and their inheritance, while molecular biology studies DNA and RNA at a molecular level, which includes processes of replication, transcription, translation. The connection between these sciences lies in the fact that genetic material is contained in DNA and RNA molecules.
Exercise 13
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Eucaryotic cells have three mechanisms that control transcription of genes – transcription factors, cell specialization, and RNA interference. Transcription factors are able to bind on the spot of DNA molecule right before gene starts and attract RNA polymerase. They can also help in unwrapping chromatin so that certain genes would be exposed. These two mechanisms promote the transcription process. However, transcription factors may also block particular genes.
Cell specialization means that certain genes are expressed in some types of cells and not the others.
RNA interference is a process in which a miRNA bind to its complementary region on the mRNA and damages it, which stops the transcription.
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c allows for cell specialization
Exercise 14
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Procaryotic cells have two mechanisms that control transcription of genes – promoter and operator. Promoter is a spot on the DNA where RNA polymerase attaches to begin transcription. An operator is located distally from promoter, but before a certain gene or a group of genes (operon).
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b a binding site for RNA polymerase
Exercise 15
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A lac operon is a group of three genes that are always transcribed together which code proteins needed for the metabolism of lactose. When lactose is absent from the outer environment, a repressor protein is bound to the operator and blocks the transcription. However, if lactose is present, it binds on a specific spot on the repressor protein, which changes its structure, so it falls off the operon site allowing RNA polymerase to transcribe lac operon.
Exercise 16
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Homeotic genes are included in the development and differentiation of cells and tissues. Because of extraordinary similarity among these genes in various organisms we can conclude about their shared past. A subdivision of homeotic genes is known as Hox genes. They define a body plan for a developing embryo.
Exercise 17
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Promoters are specific sequences of nucleotides that are present in many locations in the DNA molecule. They represent the starting point, from which a certain gene is being transcripted. Common promoter sequence in the DNA of eukaryotic cells is known as the TATA box because it consists of thymine and adenosine. The TATA-binding protein and transcription factors connect to the TATA box which facilitate attachment of RNA polymerase. Therefore, the transcription process can start.
Exercise 18
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Mutations in the level of genes are called point mutations because only one to several nucleotides are changed. These processes are known as deletion, insertion, and substitution. Deletion presents a loss of nucleotides which changes DNA sequence. When new nucleotides are embedded in the DNA chain, it is known as insertion. Substitution implies the process where new nucleotides are inserted while the ones that were present in that specific spot in a DNA molecule are deleted.
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c point mutation
Exercise 19
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Mutations can appear in chromosomes altering their number and structure. There are four types of mutations – deletion, duplication, inversion, and translocation. In deletion, a whole or one part of a chromosome is lost. Duplication presents an extra copy of a whole or one part of a chromosome. In an inversion, parts of a chromosome change order, while in translocation a part of one chromosome detaches and then connects to another.
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b lost
Exercise 20
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Mutagens are chemical or physical factors that are able to induce mutations in the genetic material. Some examples of chemical mutagens are tobaco smoke or pesticides, while ionizing and ultraviolet radiation are examples of physical mutagens.
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c mutagen
Exercise 21
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Mutations can appear in chromosomes altering their number and structure. There are four types of mutations – deletion, duplication, inversion, and translocation. In deletion, a whole or one part of a chromosome is lost. Duplication presents an extra copy of a whole or one part of a chromosome. In an inversion, parts of a chromosome change order, while in translocation a part of one chromosome detaches and then connects to another.
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Mutations can appear in chromosomes altering their number and structure. There are four types of mutations – deletion, duplication, inversion, and translocation. In an inversion, parts of a chromosome change order.
Exercise 22
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Mutations that occur in the genetic code can be neutral, harmful, or beneficial to an organism. A beneficial mutation improves the health of an organism. Some examples include mutations in bacterial genome that causes their resistance to antibiotics, or mutation that affects the CCR5 gene, that codes a protein found in the surface of leukocytes, which causes HIV resistance in human.
Exercise 23
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If a DNA sequence is ACCGTCACT, the complementary sequence of mRNA would be UGGCAGUGA. Based on the generic code diagram we can determine following amino acids – tryptophan (UGG), glutamine (CAG), and a stop codon (UGA).
Exercise 24
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One particular amino acid is usually determined by several codons. Tryptophan is defined by only one codon (UGG), as well as methionine (AUG). The codon for methionine is a start codon.

The table represents codon possibilities for alanine, valine, and leucine.

begin{center}
begin{tabular}{ c c }
AMINO ACID & CODON \
Alanine & GCU, GCC, GCA, GCG \
Valine & GUU, GUC, GUA, GUG \
Leucine & UUA, UUG, CUU, CUC, CUA, CUG \

end{tabular}
end{center}

Exercise 25
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The table represents codon possibilities for alanine and valine.

begin{center}
begin{tabular}{ c c }
AMINO ACID & CODON \
Alanine & GCU, GCC, GCA, GCG \
Valine & GUU, GUC, GUA, GUG \

end{tabular}
end{center}

As we can see from the table above, the first two bases of codons for both alanine and valine are constant, while any one of the four bases could be in the third position. Therefore, substitution mutation in the third nucleotide woudn’t alter the synthesis of alanine, nor valine.

Exercise 26
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Right before the start of the meiotic division, genetic material is being doubled. In prophase I, chromosomes are paired with the adequate homologous chromosome. In metaphase I, they are aligned in the center of the cell. During anaphase I, two sets of chromosomes move toward the opposite ends of the cell. In telophase I and cytokinesis, this cell divides into two cells. Nondisjunction of chromosomes occurs in anaphase I and the result of meiosis I will be two daughter cells with aneuploidy that will start the meiosis II.
Exercise 27
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Sodium nitrate is often added to the food. However, in our organism, sodium nitrate is converted to nitrous acid, which acts as a mutagen because of its ability to deaminate nitrogenous bases. If mutation act on the cytosine molecule of the DNA strand, it will be converted to uracil. During transcription, mRNA will contain adenine because of its compatibility with uracil, instead of guanine (cytosine-guanine). Therefore, this mutation will affect the chemical structure of the mRNA molecule, which could impact the process of translation.
Exercise 28
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The genome of eukaryotes is far more complex than in prokaryotes. For example, the genome of E. Coli contains only one chromosome and is about 4.6 Mb large, while the human genome contains 46 chromosomes, approximately 20 000 genes, and 3000 Mb of the DNA. Also, multicellular organisms have the need for cell specialization because it would be impossible that all the genes are expressed in all of the different cells. Therefore, eukaryotic cells have three mechanisms that control transcription of genes – transcription factors, cell specialization, and RNA interference.
Exercise 29
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Mutations can appear in chromosomes altering their number and structure. There are four types of mutations – deletion, duplication, inversion, and translocation.
If a mutation occurs during meiotic cell division, that may result in both, functional and defective gametes. If a defective gamete merges with a gamete from the other parent, the mutation is passed on the zygote. However, if a zygote is formed from normal gamete cells, the mutation will not be inherited.
The mutation that occurs during mitotic division of somatic cells can’t be passed on the offspring, because gamete cells aren’t affected by this deviation.
Therefore, we can conclude that both claims are correct.
Exercise 31
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A cell interprets the genetic code through codons in the mRNA, which are sequences of three nucleotides in a row. If we know that one part of the mRNA is AAC, that doesn’t necessarily have to be a codon because we don’t know the beginning of the nucleotide sequence and the end of the previous codon. There is a higher probability that these three nucleotides are parts of two codons. Therefore, asparagine that is coded by AAC doesn’t have to appear in the polypeptide chain.
Exercise 32
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If a DNA sequence is CAC, the complementary region of mRNA will be GUG which is a codon for valine. If the mutation occurs and changes the sequence to CAT, the mRNA will be GUA which is also a codon for valine. There will be no change in the amino acid despite this mutation.
Exercise 33
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Technical terms for transcription and translation can relate to their everyday meanings. In genetics, transcription is a process in which DNA strand is copied to the complementary sequence of nucleotides of mRNA molecule. Translation is a process of decoding mRNA three bases at a time and making a polypeptide chain, which is an expression of nucleotide sequences to another language – amino acids.
Exercise 34
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Mutations in the level of genes are called point mutations because only one to several nucleotides are changed. These processes are known as deletion, insertion, and substitution. Deletion presents a loss of nucleotides which changes DNA sequence. When new nucleotides are embedded in the DNA chain, it is known as insertion. In both of these cases, a whole sequence of nucleotides from the mutation point is displaced because amino acids that build a protein are determined by codons, which are made by three nucleotides in a row. For example, the nucleotide sequence is:

$text{textcolor{#19804f}{UUU}}$$text{textcolor{#4257b2}{AAC}}$$text{textcolor{#19804f}{UGU}}$$text{textcolor{#4257b2}{UCU}}$$text{textcolor{#19804f}{GUC}}$$text{textcolor{#c34632}{UAG}}$AG

$text{textcolor{#19804f}{Phenylalanine}}$-$text{textcolor{#4257b2}{Asparagine}}$-$text{textcolor{#19804f}{Cystein}}$-$text{textcolor{#4257b2}{Serine}}$-$text{textcolor{#19804f}{Valine}}$-$text{textcolor{#c34632}{Stop Codon}}$

If a nucleotide is added right before the begining of the sequence, the codons will be completely changed into:

$text{textcolor{#19804f}{AUU}}$$text{textcolor{#c34632}{UAA}}$CUGUUCUGUCUAGAG

$text{textcolor{#19804f}{Isoleucine}}$-$text{textcolor{#c34632}{Stop Codon}}$

If the first nucleotide is deleted, the sequence will be:

$text{textcolor{#19804f}{UUA}}$$text{textcolor{#4257b2}{ACU}}$$text{textcolor{#19804f}{GUU}}$$text{textcolor{#4257b2}{CUG}}$$text{textcolor{#19804f}{UCU}}$$text{textcolor{#4257b2}{AGA}}$G

$text{textcolor{#19804f}{Leucine}}$-$text{textcolor{#4257b2}{Threonine}}$-$text{textcolor{#19804f}{Valine}}$-$text{textcolor{#4257b2}{Leucine}}$-$text{textcolor{#19804f}{Serine}}$-$text{textcolor{#4257b2}{Arginine}}$

Exercise 35
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In a double-stranded RNA molecule the percentage of adenine and uracil, as well as guanine and cytosine, is almost the same because these bases are complementary. Therefore, virus A most probably has a double-stranded RNA molecule.
Exercise 36
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Based on the values of cytosine and guanine we can assume that the virus B probably has a double-stranded RNA molecule. In a double-stranded RNA molecule the percentage of adenine and uracil, as well as guanine and cytosine, is almost the same because these bases are complementary.
G = 17,5%
C = 17,6%
A + U + G + C = 100%
A + U = 100% – 17,5% – 17,6%
A + U = 64,9%
A = U = 32,45%
The percentage of adenine, as well as uracil, is approximately 32%.
Exercise 37
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Erwin Chargaff first noticed that the percentage of adenine and thymine, as well as guanine and cytosine, is almost the same in every DNA molecule. This is known as the Chargaff’s rule.
Based on the values of cytosine and guanine we can assume that the viruses A and B probably have a double-stranded RNA molecule. In a double-stranded RNA molecule the percentage of adenine and uracil, as well as guanine and cytosine, is almost the same because these bases are complementary. However, values of adenine and uracil differ in virus C and virus D has an RNA molecule in which there is a difference in the percentage of guanine and cytosine. We can conclude that these two viruses probably contain single-stranded RNA molecules.
Exercise 38
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Mutations that occur in the genetic code can be neutral, harmful, or beneficial to an organism. A beneficial mutation improves the health of an organism. Some examples include mutations in bacterial genome that causes their resistance to antibiotics, or mutation that affects the CCR5 gene, that codes a protein found in the surface of leukocytes, which causes HIV resistance in human. Polyploidy is another example of beneficial mutations, in which a plant’s genome contains more than two sets of chromosomes. That makes these plants bigger and stronger. However, there are far more examples of harmful mutations. Some examples of chemical mutagens are tobacco smoke or pesticides, while ionizing and ultraviolet radiation are examples of physical mutagens. Ultraviolet radiation may cause skin cancer, as well as tobacco smoke, can induce lung cancer.
Exercise 39
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Homeotic genes are included in the development and differentiation of cells and tissues. Because of extraordinary similarity among these genes in various organisms we can conclude about their shared past. A subdivision of homeotic genes is known as Hox genes. They define a body plan for a developing embryo.
Exercise 40
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Epigenetics studies changes in phenotype which are heritable but do not involve changes in the nucleotide sequence of the DNA molecule. These changes could be caused by methylation of nucleotide bases, chromatin remodeling, histone modification, as well as non-coding RNA mechanisms. These processes don’t alter the nucleotide sequence but they affect gene expression. Some environmental factors can induce heritable changes in phenotype, while genotype remains the same. Early prenatal exposure to certain substances may cause disease in adulthood, an example is exposure of pregnant women to pollution and higher incidence of asthma in their children. Pollution can affect methyl groups in the DNA of an adult, leading to the appearance of neurodegenerative diseases. However, vitamin B has a protective role in this specific case. Aging and certain diet can also affect gene expression.
Various cells that make the same tissue have similar gene expression patterns. However, if certain factors are able to induce disease and alter the expression of genes, the knowledge of epigenetics can serve us for diagnostic purposes. The expression patterns of cancer cells can tell us about their origin, which is crucial for establishing the stage of cancer and choosing the right therapy treatment.
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