Pierce 1.5, 1.7-1.11, 1.14, 1.16-1.21
Pierce worked problems 9.1-9.2, 9.1, 9.2, 9.8, 9.9, 9.11, 9.19-9.22
Pierce worked problem 10.1, 10.9, 10.11, 10.15
Pierce worked problem 17.1, 17.1, 17.2, 17.4, 17.9, 17.17-17.20
See practice Quizzes #1 on Blackboard
*5. List the three traditional subdivisions of genetics and summarize what each covers.
Transmission genetics: inheritance of genes from one generation to the next, gene mapping.
Molecular genetics: structure, organization, and function of genes at a molecular level.
Population genetics: genes and changes in genes in populations.
7. When and where did agriculture first arise? What role did genetics play in the development of the first domesticated plants and animals?
Agriculture first arose 10,000 to 12,000 years ago in the area now referred to as the Middle East (i.e., Turkey, Iran, Iraq, Syria, Jordan, and Israel). Early farmers selectively bred individual wild plants or animals that had useful characteristics with others that had similar useful traits. The farmers then selected for offspring that contained those useful features. Early farmers did not completely understand genetics, but they clearly understood that breeding individual plants or animals with desirable traits would lead to offspring that contained these same traits. This selective breeding led to the development of domesticated plants and animals.
*8. Outline the notion of pangenesis and explain how it differs from the germ-plasm theory.
Pangenesis theorizes that information originating from all parts of the body is carried through the reproductive organs to the embryo at conception. Pangenesis allows changes in parts of the body to then be conveyed to the reproductive organs and to the next generation. The germ-plasm theory, in contrast, states that the reproductive cells possess all of the information required to make the complete body; the rest of the body contributes no information to the next generation.
*9. What does the concept of the inheritance of acquired characteristics propose and how is it related to the theory of pangenesis?
The theory of inheritance of acquired characteristics postulates that traits acquired during one’s lifetime can be transmitted to offspring. It developed from pangenesis, which postulates that information from all parts of one’s body is transmitted to the next generation. Thus, for example, learning acquired in the brain or larger arm muscles developed through exercise could be transmitted to offspring.
*10. What is preformationism? What does it have to say about how traits are inherited?
Preformationism is the theory that the adult form is already preformed in the sperm or the egg. All traits thus would be inherited from only one parent, either the father or the mother, depending on whether the homunculus (the preformed miniature adult) resided in the sperm or the egg.
11. Define blending inheritance and contrast it with preformationism.
The theory of blending inheritance proposes that the egg and sperm from two parents contains material that blends upon conception, influencing the development of the offspring. This theory indicates that the offspring is an equal blend of the two parents. In preformationism, the offspring inherits all of its traits from one parent.
*14. Who first discovered the basic principles that laid the foundation for our modern understanding of heredity?
Gregor Mendel
*16. Briefly define the following terms: (a) gene; (b) allele; (c) chromosome; (d) DNA;
(e) RNA; (f) genetics; (g) genotype; (h) phenotype; (i) mutation; (j) evolution.
(a) Gene: the fundamental unit of heredity, a unit of information that determines an inherited characteristic
(b) Allele: a form of the gene
(c) Chromosome: a structure consisting of DNA and associated proteins that carries a linear array of genes
(d) DNA: deoxyribonucleic acid, the molecule that encodes genetic information through the sequence of bases A, C, G, and T
(e) RNA: ribonucleic acid; encodes genetic information through the sequence of bases A, C, G, and U
(f) Genetics: the science of heredity
(g) Genotype: the genetic information that an individual possesses that determines a trait
(h) Phenotype: a trait expressed by an individual
(i) Mutation: heritable alteration in the genotype of the individual, brought about by permanent alteration in the DNA
(j) Evolution: genetic change in a species or population
17. What are the two basic cell types (from a structural perspective) and how do they differ?
The two basic cell types are prokaryotic and eukaryotic. Prokaryotic cells do not have a nucleus, and their chromosomes are found within the cytoplasm. They do not possess membrane-bound cell organelles. Eukaryotic cells possess a nucleus and membrane-bound cell organelles.
18. Outline the relations between genes, DNA, and chromosomes.
Genes are composed of DNA nucleotide sequences and are located on the chromosomes.
APPLICATION QUESTIONS AND PROBLEMS
*19. Genetics is said to be both a very old science and a very young science. Explain what is meant by this statement.
Genetics is old in the sense that hereditary principles have been applied at least since the beginning of agriculture and the domestication of plants and animals. It is very young in the sense that the fundamental principles were not uncovered until Mendel’s time, and the advent of molecular biology and recombinant DNA has revolutionized genetics.
20. Match the theory or concept on the left with the correct description on the right.
Preformationism b a. each reproductive cell contains a complete set of genetic information
Pangenesis d b. all traits inherited from one parent
Germplasm theory a c. genetic information may be altered by use of a feature
Inheritance of acquired characteristics c d. different genetic information occurs in cells of different tissues
*21. For each of the following genetic topics, indicate whether it focuses on transmission genetics, molecular genetics, or population genetics.
a. Analysis of pedigrees to determine the probability of someone inheriting a trait
Transmission genetics
b. Study of the genetic history of people on a small island to determine why a genetic form of asthma is so prevalent on the island.
Population genetics
c. The influence of nonrandom mating on the distribution of genotypes among a group of animals.
Population genetics
d. Examination of the nucleotide sequences found at the ends of chromosomes.
Molecular genetics
e. Mechanisms that ensure a high degree of accuracy during DNA replication.
Molecular genetics
f. Study of how the inheritance of traits encoded by genes on sex chromosomes (sex-linked traits) differs from the inheritance of traits encoded by genes on nonsex chromosomes (autosomal traits).
Transmission genetics
COMPREHENSION QUESTIONS
*1. List the different types of chromosome mutations and define each.
Chromosome rearrangements:
Deletion: loss of a portion of a chromosome.
Duplication: addition of an extra copy of a portion of a chromosome.
Inversion: a portion of the chromosome is reversed in orientation.
Translocation: a portion of one chromosome becomes incorporated into a different (nonhomologous) chromosome.
Aneupoloidy: loss or gain of one or more chromosomes so that the chromosome number deviates from 2n or the normal euploid complement.
Polyploidy: Gain of entire sets of chromosomes so the chromosome number changes from 2n to 3n (triploid), 4n (tetraploid), and so on.
*2. Why do extra copies of genes sometimes cause drastic phenotypic effects?
The expression of some genes is balanced with the expression of other genes; the ratios of their gene products, usually proteins, must be maintained within a narrow range for proper cell function. Extra copies of one of these genes cause that gene to be expressed at proportionately higher levels, thereby upsetting the balance of gene products.
8. Explain why recombination is suppressed in individuals heterozygous for paracentric and pericentric inversions.
A crossover within a paracentric inversion produces a dicentric and an acentric recombinant chromatid. The acentric fragment is lost, and the dicentric fragment breaks, resulting in chromatids with large deletions that lead to nonviable gametes or embryonic lethality. A crossover within a pericentric inversion produces recombinant chromatids that have duplications or deletions. Again, gametes with these recombinant chromatids do not lead to viable progeny.
*9. How do translocations produce phenotypic effects?
Like inversions, translocations can produce phenotypic effects if the translocation breakpoint disrupts a gene or if a gene near the breakpoint is altered in its expression because of relocation to a different chromosomal environment (a position effect).
11. What is a Robertsonian translocation?
The long arms of two acrocentric chromosomes are joined to a common centromere through translocation, resulting in a large metacentric chromosome and a very small chromosome with two very short arms. The very small chromosome may be lost.
*19. Which types of chromosome mutations:
(a) increase the amount of genetic material on a particular chromosome?
Duplications
(b) increase the amount of genetic material for all chromosomes?
Polyploidy
(c) decrease the amount of genetic material on a particular chromosome?
Deletions
(d) change the position of DNA sequences on a single chromosome without changing the amount of genetic material?
Inversions
(e) move DNA from one chromosome to a nonhomologous chromosome?
Translocations
*20. A chromosome has the following segments, where · represents the centromere.
A B · C D E F G
What types of chromosome mutations are required to change this chromosome into each of the following chromosomes? (In some cases, more than one chromosome mutation may be required.)
(a) A B A B · C D E F G: Tandem duplication of AB.
(b) A B · C D E A B F G: Displaced duplication of AB.
(c) A B · C F E D G: Paracentric inversion of DEF.
(d) A · C D E F G: Deletion of B.
(e) A B · C D E: Deletion of FG.
(f) A B · E D C F G: Paracentric inversion of CDE.
(g) C · B A D E F G: Pericentric inversion of ABC.
(h) A B · C F E D F E D G: Duplication and inversion of DEF.
(i) A B · C D E F C D F E G: Duplication of CDEF, inversion of EF.
*21. A chromosome initially has the following segments:
A B · C D E F G
Draw and label the chromosome that would result from each of the following mutations:
(a) Tandem duplication of DEF: A B · C D E F D E F G.
(b) Displaced duplication of DEF: A B · C D E F G D E F.
(c) Deletion of FG: A B · C D E.
(d) Deletion of CD: A B · E F G.
(e) Paracentric inversion that includes DEFG: A B · C G F E D.
(f) Pericentric inversion of BCDE: A E D C · B F G.
22. The following diagrams represent two nonhomologous chromosomes:
A B · C D E F G
R S · T U V W X
What type of chromosome mutation would produce the following chromosomes?
(a) A B · C D
R S · T U V W X E F G
Nonreciprocal translocation of E F G
(b) A U V B · C D E F G
R S · T W X
Nonreciprocal translocation of U V
(c) A B · T U V F G
R S · C D E W X
Reciprocal translocation of C D E and T U V
(d) A B · C W G
R S · T U V D E F X
Reciprocal translocation of D E F and W
*1. What three general characteristics must the genetic material possess?
(1) The genetic material must contain complex information.
(2) The genetic material must replicate or be replicated faithfully.
(3) The genetic material must encode the phenotype.
9. How does a purine differ from a pyrimidine? What purines and pyrimidines are found in DNA and RNA?
A purine consists of a six-sided ring attached to a five-sided ring. A pyrimidine consists of only a six-sided ring. In both DNA and RNA, the purines found are adenine and guanine. DNA and RNA differ in their pyrimidine content. The pyrimidine cytosine is found in both RNA and DNA. However, DNA contains the pyrimidine thymine, whereas RNA contains the pyrimidine uracil but not thymine.
11. Which bases are capable of forming hydrogen bonds with each another?
Adenine is capable of forming two hydrogen bonds with thymine. Guanine is capable of forming three hydrogen bonds with cytosine.
*15. What are the major transfers of genetic information?
The major transfers of genetic information are replication, transcription, and translation. These are the components of the central dogma of molecular biology.
*1. What is the difference between somatic mutations and germ-line mutations?
Germ-line mutations are found in the DNA of germ (reproductive) cells and may be passed to offspring. Somatic mutations are found in the DNA of an organism’s somatic tissue cells and cannot be passed to offspring.
*2. What is the difference between a transition and a transversion? Which type of base substitution is usually more common?
Transition mutations are base substitutions in which one purine (A or G) is changed to the other purine, or a pyrimidine (T or C) is changed to the other pyrimidine. Transversions are base substitutions in which a purine is changed to a pyrimidine or vice versa. Although transversions would seem to be statistically favored because there are eight possible transversions and only four possible transitions, about twice as many transition mutations are actually observed in the human genome.
4. What is the difference between a missense mutation and a nonsense mutation? A silent mutation and a neutral mutation?
A base substitution that changes the sequence and the meaning of a mRNA codon, resulting in a different amino acid being inserted into a protein, is called a missense mutation. Nonsense mutations occur when a mutation replaces a sense codon with a stop (or nonsense) codon.
A nucleotide substitution that changes the sequence of a mRNA codon but not the meaning is called a silent mutation. In neutral mutations, the sequence and the meaning of a mRNA codon are changed. However, the amino acid substitution has little or no effect on protein function.
9. How do insertions and deletions arise?
Strand slippage that occurs during DNA replication and unequal crossover events due to misalignment at repetitive sequences have been shown to cause deletions and additions of nucleotides to DNA molecules. Strand slippage results from the formation of small loops on either the template or the newly synthesized strand. If the loop forms on the template strand, then a deletion occurs. Loops formed on the newly synthesized strand result in insertions. If, during crossing over, a misalignment of the two strands at repetitive sequence occurs, then the resolution of the cross over will result in one DNA molecule containing an insertion and the other molecule containing a deletion.
*17. A codon that specifies the amino acid Gly undergoes a single-base substitution to become a nonsense mutation. In accord with the genetic code given in Figure 15.12, is this mutation a transition or a transversion? At which position of the codon does the mutation occur?
By examining the four codons that encode for Gly, GGU, GGC, GGA, and GGG, and the three nonsense codons, UGA, UAA, and UAG, we can determine that only one of the Gly codons, GGA, could be mutated to a nonsense codon by the single substitution of a U for a G at the first position:
GGA à UGA
Because uracil is a pyrimidine and guanine is a purine, the mutation is a transversion.
*18. (a) If a single transition occurs in a codon that specifies Phe, what amino acids could be specified by the mutated sequence?
Two codons can encode for Phe, UUU, and UUC. A single transition could occur at each of the positions of the codon resulting in different meanings.
| Original codon |
Mutated codon (amino acid encoded) |
|
UUU |
CUU (Leu), UCU (Ser), UUC (Phe) |
|
UUC |
CUC (Ser), UCU (Ser), UUU (Ser) |
(b) If a single transversion occurs in a codon that specifies Phe, what amino acids could be specified by the mutated sequence?
| Original codon |
Mutated codon (amino acid encoded) |
|
UUU |
AUU (Ile), UAU (Tyr), UUA (Leu), GUU (Val), UGU (Cys), UUG (Leu) |
|
UUC |
AUC (Ile), UAC (Tyr), UUA (Leu), GUC (Val), UGC (Cys), UUG (Leu) |
(c) If a single transition occurs in a codon that specifies Leu, what amino acids could be specified by the mutated sequence?
| Original codon |
Mutated codon (amino acid encoded) |
|
CUU |
UUU (Phe), CCU (Pro), CUC (Leu) |
|
CUC |
UUC (Phe), CCC (Pro), CUG (Leu) |
|
CUA |
UUA (Leu), CCA (Pro), CUG (Leu) |
|
CUG |
UUG (Leu), CCG (Pro), CUA (Leu) |
|
UUG |
CUG (Leu), UCG (Ser), UUA (Ser) |
|
UUA |
CUA (Leu), UCG (Ser), UUG (Leu) |
(d) If a single transversion occurs in a codon that specifies Leu, what amino acids could be specified by the mutated sequence?
| Original codon |
Mutated codon (amino acid encoded) |
|
UUA |
AUA (Met), UAA (Stop), UUU (Phe), GUA (Val), UGA (Stop), UUC (Phe) |
|
UUG |
AUG (Met), UAG (Stop), UUU (Phe), GUG (Val), UGG (Trp), UUC (Phe) |
|
CUU |
GUU (Val), CGU (Arg), CUG (Leu), AUU (Ile), UAU (Tyr), UUA (Leu) |
|
CUC |
AUC (Ile), CAC (His), CUA (Leu), GUC (Val), CGC (Arg), CUG (Leu) |
|
CUA |
AUA (Ile), CAA (Gln), CUC (Leu), GUA (Val), CGA (Arg), CUG (Leu) |
|
CUG |
AUG (Met), CAG (Gln), CUC (Leu), GUG (Val), CGG (Arg), CUU (Leu) |
19. Hemoglobin is a complex protein that contains four polypeptide chains. The normal hemoglobin found in adults—called adult hemoglobin—consists of two α and two β polypeptide chains, which are encoded by different loci. Sickle cell hemoglobin, which causes sickle cell anemia, arises from a mutation in the β chain of adult hemoglobin. Adult hemoglobin and sickle cell hemoglobin differ in a single amino acid: the sixth amino acid from one end in adult hemoglobin is glutamic acid, whereas sickle cell hemoglobin has valine at this position. After consulting the genetic code provided in Figure 15.12, indicate the type and location of the mutation that gave rise to sickle-cell anemia.
There are two possible codons for glutamic acid, GAA and GAG. Single base substitutions at the second position in both codons can produce codons that encode valine:
GAA--------> GUA (Val)
GAG--------> GUG (Val).
Both substitutions are transversions. However, in the gene encoding the β chain of hemoglobin, the GAG codon is the wild-type codon and the mutated GUG codon results in the sickle-cell phenotype.
*20. The following nucleotide sequence is found on the template strand of DNA. First, determine the amino acids of the protein encoded by this sequence by using the genetic code provided in Figure 15.12. Then give the altered amino acid sequence of the protein that will be found in each of the following mutations.
Sequence of DNA template: 3'–TAC TGG CCG TTA GTT GAT ATA ACT–5'
Nucleotide number à 1 24
mRNA sequence: 5'–AUG ACC GGC AAU CAA CUA UAU UGA–3'
amino acid sequence: Amino–Met Thr Gly Asn Gln Leu Tyr Stop–Carboxyl
(a) Mutant 1: A transition at nucleotide 11.
The transition results in the substitution of Ser for Asn.
original sequence: 3'–TAC TGG CCG TTA GTT GAT ATA ACT–5'
mutated sequence: 3'–TAC TGG CCG TCA GTT GAT ATA ACT–5'
mRNA sequence: 5'–AUG ACC GGC AGU CAA CUA UAU UGA–3'
amino acids: Amino–Met Thr Gly Ser Gln Leu Tyr Stop–Carboxyl
(b) Mutant 2: A transition at nucleotide 13.
The transition results in the formation of a UAA nonsense codon.
original sequence: 3'–TAC TGG CCG TTA GTT GAT ATA ACT–5'
mutated sequence: 3'–TAC TGG CCG TTA ATT GAT ATA ACT–5'
mRNA sequence: 5'–AUG ACC GGC AAU UAA CUA UAU UGA–3'
amino acid sequence: Amino–Met Thr Gly Asn STOP–Carboxyl
(c) Mutant 3: A one-nucleotide deletion at nucleotide 7.
The one-nucleotide deletion results in a frameshift mutation.
original sequence: 3'–TAC TGG CCG TTA GTT GAT ATA ACT–5'
mutated sequence: 3'–TAC TGG CGT TAG TTG ATA TAA CT–5'
mRNA sequence: 5'–AUG ACC GCA GUC AAC UAU AUU GA–3'
amino acids: Amino–Met Thr Ala Ile Asn Tyr Ile –Carboxyl
(d) Mutant 4: A Tà A transversion at nucleotide 15.
The transversion results in the substitution of His for Gln in the protein.
original sequence: 3'–TAC TGG CCG TTA GTT GAT ATA ACT–5'
mutated sequence: 3'–TAC TGG CCG TTA GTA GAT ATA ACT–5'
mRNA sequence: 5'–AUG ACC GGC AAU CAU CUA UAU UGA–3'
amino acids: Amino–Met Thr Gly Asn His Leu Tyr Stop–Carboxyl
or
mutated sequence: 3'–TAC TGG CCG TTA GTG GAT ATA ACT–5'
mRNA sequence: 5'–AUG ACC GGC AAU CAC CUA UAU UGA–3'
amino acids: Amino–Met Thr Gly Asn His Leu Tyr Stop–Carboxyl
(e) Mutant 5: An addition of TGG after nucleotide 6.
The addition of the three nucleotides results in the addition of Thr to the amino acid sequence of the protein.
original sequence: 3'–TAC TGG CCG TTA GTT GAT ATA ACT–5'
mutated sequence:3'–TAC TGG TGG CCG TTA GTT GAT ATA ACT–5'
mRNA sequence: 5'–AUG ACC ACC GGC AAU CAA CUA UAU UGA–3'
amino acids: Amino–Met Thr Thr Gly Asn Gln Leu Tyr Stop–Carboxyl
(f) Mutant 6: A transition at nucleotide 9.
The protein retains the original amino acid sequence.
original sequence: 3'–TAC TGG CCG TTA GTT GAT ATA ACT–5'
mutated sequence: 3'–TAC TGG CCA TTA GTT GAT ATA ACT–5'
mRNA Sequence: 5'–AUG ACC GGU AAU CAA CUA UAU UGA –3'
amino acids: Amino–Met Thr Gly Asn Gln Leu Tyr Stop–Carboxyl