BLY-302 Ch. 3 & Ch. 18

Assume that a mistake is made during DNA replication in a bacterium and a G is inserted into the newly synthesized DNA strand opposite a T in the template DNA strand. If this mistake is not repaired before the next round of DNA replication, what mutation will eventually result?

A-to-G base substitution

A couple has six daughters and is expecting a seventh child. What is the probability that this child will be male? A. 1/2 B. 1/4 C. 1/16 D. 1/64 E. 1/128

A. 1/2

How many different types of gametes could be produced by the second flower? A. 2 B. 5 C. 10 D. 4 E. 1

A. 2

Which of the following was NOT one of Mendel's conclusions based on his monohybrid crosses? A. Genes are carried on chromosomes. B. Alleles exist in pairs. C. Alleles segregate equally into gametes. D. One allele can mask the expression of the other allele.

A. Genes are carried on chromosomes.

A polypeptide has the following amino acid sequence: Met-Ala-Gln-Arg-Glu-Leu. This polypeptide was mutated to produce the following mutant sequence: Met-Ala-Gln-Gly-Glu-Leu. Which describes the MOST likely type of mutation that occurred?

B. missense mutation

In humans, oculocutaneous (OCA) albinism is a collection of autosomal recessive disorders characterized by an absence of the pigment melanin in skin, hair, and eyes. That is, normal pigmentation (𝐴) is dominant over albinism (𝑎) . For this question, assume the phenotype is determined by a single gene with two alleles. If two people have normal pigmentation, what possible phenotypes may be observed in their offspring? A. albinism only B. normal pigmentation only C. normal pigmentation or albinism D. marbling of albino and normal pigmentation

C. normal pigmentation or albinism

Which of these is a step in the process of assessing carcinogenesis in a chemical? A. The bacteria are plated on a medium that incorporates histidine. B. The bacteria are incubated, and the resulting colonies that appear have undergone his- to his+ mutation. C. None of the bacteria are mixed with the chemical to be tested for mutagenic activity. D. Bacterial strains are mixed with liver enzymes that have the ability to convert compounds into potential mutagens.

D. Bacterial strains are mixed with liver enzymes that have the ability to convert compounds into potential mutagens.

Transposable elements are found: A. mainly in higher plants B. mainly in animals, particularly in mammals C. mainly in eukaryotes D. mainly in prokaryotes E. in practically all organisms

in practically all organisms

A geneticist is studying a mutation in a population of turtles that causes their shells to become extremely brittle. She determines the mutation is caused by the loss of two nucleotides in the coding region of a gene. Upon studying the mutant protein that is produced, she observes that it is 312 amino acids in length, as compared to the normal protein that is 588 amino acids in length. This mutant protein can no longer carry out its normal function of assisting in the stiffening of a turtle's shell. Which of the following could NOT describe this mutation?

transversion

Which of the crosses paired homozygous dominant dragonflies? A. cross 1 B. cross 2 C. cross 3 D. crosses 1 & 2 E. crosses 2 & 3

cross 3

While doing fieldwork in Madagascar, you discover a new dragonfly species that has either red (R) or clear (r) wings. Initial crosses indicate that R is dominant to r. You perform three crosses using three different sets of parents with unknown genotype and observe the following data: Cross: (1) -> Phenotypes: (72 red-winged, 24 clear-winged) Cross: (2) -> Phenotypes: (12 red-winged, 12 clear-winged) Cross: (3) -> Phenotypes: (96 red-winged) A. cross 1 B. cross 2 C. cross 3 D. crosses 1 & 2 E. crosses 2 & 3

cross 3

Insertion or removal of one or more nucleotide base pairs in DNA within a gene often results in a ___________ mutation.

frameshift

Is it better to have a large sample size in genetic crosses? A. Yes, larger sample sizes can produce phenotypic ratios closer to expected ratios. B. No, larger samples sizes result in greater deviations from expected ratios. C. Yes, larger samples sizes create more degrees of freedom. D. No, larger sizes lead to more errors in counting.

A. Yes, larger sample sizes can produce phenotypic ratios closer to expected ratios.

A phenotypically normal man has phenotypically normal parents, but he has a sister who has cystic fibrosis caused by a recessive mutant allele. What is the genotype of each parent? Use C for normal and c for disease form.

Both parents are Cc

Two parents are both heterozygous for the autosomal recessive allele for albinism. What is the probability that their first child will be an albino, their second child will be an albino, and their third child will have normal pigmentation in that order? A. 61/64 B. 3/64 C. 1/64 D. 9/64

B. 3/64

In mice, an allele for apricot eyes (a) is recessive to an allele for brown eyes (a+). At an independently assorting locus, an allele for tan (t) coat color is recessive to an allele for black (t+) coat color. A mouse that is homozygous for brown eyes and black coat color is crossed with a mouse having apricot eyes and a tan coat. The resulting F1 are intercrossed to produce the F2. In a litter of eight F2 mice, what is the probability that exactly two will have apricot eyes and tan coats? [Use three decimal places for the answer.] - Probability?

- Probability: 0.074

Transposons are movable elements of DNA that insert into new positions within the genome. A transposon may disrupt normal gene function, confer an evolutionary advantage, or have no effect on the organism. Select all scenarios in which a transposon has positively contributed to the evolution of the host's genome. - insertion of a transposon into the gene encoding Factor VIII, an essential blood‑clotting protein - aquisition of a transposon carrying an adhesin gene into the Lactobacillus casei genome - acquisition of a transposon that disrupts the promoter element of a breast cancer tumor‑suppressor gene - acquisition of a transposon carrying a gene resistant to penicillin into a bacterium - insertion of a transposon encoding a heat shock protein into the genome of a plant

- aquisition of a transposon carrying an adhesin gene into the Lactobacillus casei genome - acquisition of a transposon carrying a gene resistant to penicillin into a bacterium - insertion of a transposon encoding a heat shock protein into the genome of a plant

The given DNA non‑template sequence (coding sequence) is transcribed from 5' to 3'. Use the sequence to determine the type of mutation and the type of base substitutions that apply to each scenario. Place only one statement for each scenario. 5' A T G A C C G A A C G C T T G 3' - A thymine substituted for nucleotide 6? - A cytosine substituted for nucleotide 9? - An adenine substituted for nucleotide 14? Answer bank: - silent and transition - nonsense and transition - missense and transversion - silent and transversion - missense and transition - nonsense and transversion

A thymine substituted for nucleotide 6: - silent and transition A cytosine substituted for nucleotide 9: - missense and transversion An adenine substituted for nucleotide 14: - nonsense and transversion

In a plant species, an Aa Bb Cc plant is crossed with an Aa Bb cc plant. Assuming that all three genes assort independently, what is the probability that an offspring will be phenotypically identical to either of the parents with respect to the three genes? A. 9/64 B. 9/16 C. 9/32 D. 27/64

B. 9/16

Expansion of triplets of nucleotides are "fixed" into the DNA strand by which of the processes listed? A. DNA repair synthesis B. DNA replication and DNA repair synthesis C. DNA recombination D. DNA replication and DNA recombination E. DNA replication

B. DNA replication and DNA repair synthesis

A mutation that alters the wild-type phenotype is called a _____ mutation. A. silent B. forward C. reverse D. nonsense

B. forward

In pea plants, plant height is controlled by a single autosomal dominant gene. Tall plants (H) are dominant to short plants (h). In a cross of two tall heterozygous plants, which phenotype ratio is expected from the resulting offspring? A. 1:2:1 B. 9:3:3:1 C. 3:1 D. 1:1

C. 3:1

You perform a testcross by crossing an Aa Bb Cc individual with an aa bb cc individual. How many phenotypic classes of offspring do you expect from this cross? A. 2 B. 4 C. 8 D. 16

C. 8

In a monohybrid cross, the offspring from the parents in the P generation are called which of these? A. F2 generation B. backcross progeny C. F1 generation D. testcross progeny

C. F1 generation

A DNA strand of E. coli is being used for a molecular study by a genetics assistant. The assistant understands that the rate of spontaneous depyrimidination of thymine is 50%. Given the sequence shown, what would NOT be the resulting DNA strand once replicated? Assume that cytosine is incorrectly incorporated into depyrimidinic sites. DNA: TACTGTATATTGTCATTCAT A. TACTGTATATGGGCAGGCAG B. GACGGGAGAGTGTCATTCAT C. GACGGGATAGTGTCATGCAG D. TACTGTAGAGGGTCAGTCAG

C. GACGGGATAGTGTCATGCAG

In base-excision repair, there are several steps that need to be taken. What is the correct order of those steps? I. The deoxyribose sugar is removed from the backbone. II. DNA polymerase adds nucleotides to the 3'-OH group. III. DNA glycosylase recognizes and removes a damaged base, producing an AP site. IV. The nick in the backbone is sealed by DNA ligase. V. AP endonuclease cleaves the phosphodiester bond on the 5' side of the AP site. A. V, III, I, II, IV B. III, V, II, I, IV C. III, V, I, II, IV D. II, III, V, I, IV

C. III, V, I, II, IV

A mutation that changes a mutant phenotype back into a wild type is called a _____ mutation. A. nonsense B. forward C. reverse D. silent

C. reverse

Which insertion or deletion mutation would likely NOT lead to a frameshift mutation? A. a single base pair B. two base pairs C. six base pairs D. seven base pairs

C. six base pairs

A mutation that can affect the individual in which it occurs but not that individual's offspring is a _____ mutation. A. missense B. silent C. somatic D. neutral

C. somatic

Mutations that occur under normal conditions are considered: A. incorporated errors. B. adaptive mutations. C. spontaneous mutations. D. induced mutations.

C. spontaneous mutations.

What is the probability of a given gamete produced by the fifth flower having three recessive alleles? A. 0.333 B. 1 C. 0.25 D. 0 E. 0.5

D. 0

What is the probability of a given gamete produced by the first flower having three recessive alleles? A. 0.5 B. 1 C. 0.333 D. 0.25 E. 0

D. 0.25

What is the probability of a given gamete produced by the second flower having three recessive alleles? A. 1 B. 0.25 C. 0 D. 0.5 E. 0.333

D. 0.5

Given the sequence shown, what would be the resulting mutant sequence if it undergoes alkylation of guanine? DNA: 5'-AGTCCGATTAGCCCGTAATT-3' A. 3'-TTAATGCCCGATTAGCCTGA-5' B. 3'- AATTTAATTAGTTTATAATT -5' C. 5'-AATTTAATTAGTTTATAATT-3' D. 3'-TTAGGTTAATCGGGTATTAA-5'

D. 3'-TTAGGTTAATCGGGTATTAA-5'

If the offspring resulting from a self‑cross of the second flower were permitted to randomly interbreed, what would the phenotypic ratio be in the offspring of this intercross? A. 1:3 ratio of red flowers with smooth petals to red flowers with rough petals. B. All offspring would have identical phenotypes. C. 1:1 ratio of red flowers with smooth petals to red flowers with rough petals. D. 3:1 ratio of red flowers with smooth petals to red flowers with rough petals. E. This question is impossible to answer with the information provided.

D. 3:1 ratio of red flowers with smooth petals to red flowers with rough petals.

Insertion of additional repeats in repetitive segments of DNA often is a result of what? A. transformation B. chromosome rearrangements C. translation errors D. transcription slippage E. replication slippage

E. replication slippage

In pea plants, the allele for red flower color, F, is completely dominant to the allele for white flower color, f. Complete the Punnett square showing the genotypes possible among the offspring when two heterozygous individuals are crossed. Use the information from the Punnett square to answer the second question. In this cross between two heterozygous pea plants, what are the chances that an offspring with red flowers will be produced? A. 75% B. 3% C. 25% D. 50%

FF / Ff --------- Ff / ff A. 75%

Huntington's disease is an inherited autosomal dominant disorder that can affect both men and women. Imagine a couple has had seven children, and later in life, the husband develops Huntington's disease. He is tested and it is discovered he is heterozygous for the disease allele, Hh. The wife is also genetically tested for the Huntington's disease allele, and her test results show she is unaffected, hh. PART 1: What is the percent probability that the first child of this couple will have Huntington's disease? - Probability? _______% PART 2: What is the probability that two of the seven children will have Huntington's disease? - Probability? _______%

PART 1: - Probability: 50% PART 2: - Probability: 16.41%

Classify each definition or example as a somatic mutation, gametic (germ line) mutation, or both. - Somatic mutation? - Gametic mutation? - Both? Answer bank: - A man receives a pelvic X-ray. Nine months later, his child is born with a chromosomal abnormality. - The mutation originates in the reproductive cells of an individual and affects their offspring. - A particular tobacco leaf becomes discolored due to a mutation halfway through the life of the plant. - Mutations can be caused by an alteration in the DNA sequence. - The mutation affects only the individual in which the mutation occurs and is not passed on to the progeny.

Somatic mutation: - The mutation affects only the individual in which the mutation occurs and is not passed on to the progeny. - A particular tobacco leaf becomes discolored due to a mutation halfway through the life of the plant. Gametic mutation: - The mutation originates in the reproductive cells of an individual and affects their offspring. - A man receives a pelvic X-ray. Nine months later, his child is born with a chromosomal abnormality. Both: - Mutations can be caused by an alteration in the DNA sequence.

A template strand of DNA contains the nucleotide sequence. Use the codon table to translate the amino acid sequence for each mutation. Match the translated amino acid sequence to the appropriate mutation. All amino acid sequences are written in the amino‑to‑carboxyl direction. - Transition at nucleotide 11? - Transition at nucleotide 13? - Deletion at nucleotide 7? - Transversion of T to A at nucleotide 15? - Addition of TGG after nucleotide 6? - Transition at nucleotide 9? Answer bank: - Met Thr Ala Ile Asn Tyr Ile - Met Thr Thr Gly Asn Gln Leu Tyr - Met Thr Gly Ser Gln Leu Tyr - Met Thr Gly Asn Gln Leu Tyr - Met Thr Gly Asn His Leu Tyr - Met Thr Gly Asn

Transition at nucleotide 11: - Met Thr Gly Ser Gln Leu Tyr Transition at nucleotide 13: - Met Thr Gly Asn Deletion at nucleotide 7: - Met Thr Ala Ile Asn Tyr Ile Transversion of T to A at nucleotide 15: - Met Thr Gly Asn His Leu Tyr Addition of TGG after nucleotide 6: - Met Thr Thr Gly Asn Gln Leu Tyr Transition at nucleotide 9: - Met Thr Gly Asn Gln Leu Tyr

In cats, curled ears result from an allele, Cu, that is dominant over an allele, cu, for normal ears. Black color results from an independently assorting allele, G, that is dominant over an allele for gray, g. A gray cat homozygous for curled ears is mated with a homozygous black cat with normal ears. All the F1 cats are black and have curled ears. What phenotypes and proportions are expected from the two crosses? - Two of the F1 cats mate? - An F1 cat mates with a stray cat that is gray and possesses normal ears? Answer bank: - All black cats with curled ears - 9/16 black cats, curled ears; 3/16 black cats, normal ears; 3/16 gray cats, curled ears; and 1/16 gray cats, normal ears. - 1/4 black cats, curled ears; 1/4 black cats, normal ears; 1/4 gray cats, curled ears; 1/4 gray cats, normal ears. - All gray cats; normal ears - 9/16 gray cats, normal ears; 3/16 gray cats, curled ears; 3/16 black cats, normal ears; and 1/16 black cats, curled ears.

Two of the F1 cats mate: - 9/16 black cats, curled ears; 3/16 black cats, normal ears; 3/16 gray cats, curled ears; and 1/16 gray cats, normal ears. An F1 cat mates with a stray cat that is gray and possesses normal ears: - 1/4 black cats, curled ears; 1/4 black cats, normal ears; 1/4 gray cats, curled ears; 1/4 gray cats, normal ears.

Gregor Mendel carried out a cross between two pea plants by taking pollen from a plant that was homozygous for round seeds and dusting the pollen onto the stigma of a plant homozygous for wrinkled seeds. Which of the following would be the reciprocal cross that Mendel had carried out for this experiment?

stigma of a plant homozygous for round seed pollinated with pollen from a plant homozygous for wrinkled-seed plant

A mutation that changes the amino acid sequence of a protein without altering its function is called a(n): A. neutral mutation. B. nonsense mutation. C. silent mutation. D. intragenic suppressor mutation.

A. neutral mutation.

The repair of a broken DNA molecule by using identical or nearly identical genetic information contained in a sister chromatid is called: A. translesion correction. B. nonhomologous end joining. C. chromatid fusion. D. homologous recombination.

D. homologous recombination.

Which of these DNA repair mechanisms is most likely to correct double-strand DNA breaks in eukaryotes? A. base-excision repair B. nucleotide-excision repair C. mismatch repair D. nonhomologous end-joining repair

D. nonhomologous end-joining repair

How many different types of gametes could be produced by the sixth flower? A. 10 B. 1 C. 5 D. 4 E. 2

D. 4

A cell with mutated DNA may not be able to produce proteins accurately. Which situations can lead to mutations in the DNA? [Select all that apply.] - uncorrected DNA replication errors - excess pyruvate from glycolysis - chemicals in some pesticides - mistakes in mRNA transcription - ultraviolet radiation exposure

- uncorrected DNA replication errors - chemicals in some pesticides - ultraviolet radiation exposure

Breaks in double-stranded DNA molecules are most likely to result from exposure to: A. UV light. B. ionizing radiation. C. visible light. D. alkylating agents.

B. ionizing radiation.

What is one reason Gregor Mendel and Thomas Hunt Morgan were so successful in their studies of genetics? A. They only studied mutations that were beneficial. B. They understood exactly what a gene was and how to look for its function. C. They used populations of organisms that had different alleles for particular traits. D. They used model organisms with identical genetic backgrounds.

C. They used populations of organisms that had different alleles for particular traits.

In an intercross of a flower with many rough petals with a second flower with a few smooth petals, what proportion of the offspring would you expect to have few rough petals? A. 1 / 16 B. 0 C. This is impossible to predict without more information. D. 1 / 4 E. 3 / 16

C. This is impossible to predict without more information.

How can mutations be beneficial to populations? A. Only organisms with mutations can reproduce successfully. B. Mutations will cause disease in less fit organisms, thus taking those organisms out of the genetic pool. C. Every mutation causes an organism to become more suited to their environment. D. Mutations are the source of genetic diversity.

D. Mutations are the source of genetic diversity.

Which of these is NOT true in the generation of flanking direct repeats? A. Short flanking direct repeats are present on both sides of transposable elements and are of constant length. B. Short flanking direct repeats are generated in the process of transposition as they are not part of the transposable element. C. The cuts that create flanking direct repeats are typically staggered. D. The cuts on either side leave long single-stranded pieces of DNA which, through transcription, create flanking direct repeats.

D. The cuts on either side leave long single-stranded pieces of DNA which, through transcription, create flanking direct repeats.

Use the image to observe the results of a cross between a tall pea plant and a short pea plant. What phenotypes and proportions will be produced for the two crosses? - Tall F1 progeny backcrossed to the short parent? - Tall F1 progeny backcrossed to the tall parent? Answer bank: - All short - All tall - one-half tall and one-half short - one-forth tall and three-fourths short - three-fourths tall and one-fourth short

Tall F1 progeny backcrossed to the short parent: - one-half tall and one-half short Tall F1 progeny backcrossed to the tall parent: - three-fourths tall and one-fourth short

After using a chemical mutagen to generate mutations in a DNA sequence, scientists noted a mutation from C to T at the 10th position within the coding region of a gene. This mutation led to a change of proline into serine at the fourth position in the resulting peptide. Using this information and the sequences listed, select all the types of mutations that occurred. - transversion - missense - transition - induced - point - nonsense - spontaneous

- missense - transition - induced - point

At the end of your biology class, your professor asks you to develop a project to determine the genotype of a plant with red flowers. Red petal color (R) is dominant to pink flower color (r). To accomplish this task, you cross the plant with the unknown genotype with heterozygous red‑flowered plants. A partially filled Punnett square is provided. Which ratios are valid predictions of flower colors in the offspring? A. 3 red : 1 pink flowers B. all pink flowers C. all red flowers D. 1 red : 3 pink flowers

A. 3 red : 1 pink flowers C. all red flowers

An interesting mutation is discovered in a species of wasp where the DNA sequence of the genome is unknown. In order to try to determine the molecular nature of this mutation, a geneticist treats mutant wasps with the chemical 5-bromouracil, a chemical mutagen that induces both types of transition mutations at high frequencies, and hydroxylamine, which induces only GC-to-AT transitions. Neither chemical mutagen caused the back mutation or reversion frequencies to increase beyond the low spontaneous values. Which of these types of mutations is the least likely to be the original wasp mutation? A. A-to-G B. A-to-T C. +1 frameshift D. -1 frameshift

A. A-to-G

The probability of rolling a single die and obtaining a two is one-sixth [1/6]. What is the probability of rolling two dice and obtaining a two on one and a five on the other? A. one thirty-sixth [1/36] B. one-third [1/3] C. one-sixth [1/6] D. one-ninth [1/9]

A. one thirty-sixth [1/36]

DNA damage can occur as a result of exposure to chemicals or ultraviolet radiation. What happens during nucleotide excision repair of damaged DNA? A. Enzymes delete an incorrect DNA sequence, pull the gap closed, and join the bases on either side of the gap together. B. Enzymes open the DNA strand, remove a segment of DNA from the strand that contains the damage, and resynthesize the correct DNA sequence. C. Enzymes detect an incorrect DNA base pairing in a DNA strand, remove the incorrect base, and insert the correct base. D. Enzymes cut out the damaged gene, copy the same gene from the homologous chromosome, and insert the copy.

B. Enzymes open the DNA strand, remove a segment of DNA from the strand that contains the damage, and resynthesize the correct DNA sequence.

Which of these is a disadvantage of mutations in populations? A. Mutations are the source of genetic diversity. B. Mutations can be the source of many diseases and disorders. C. Mutations allow for offspring to further the line of inheritance by the principles of evolution. D. Mutations allow organisms to adapt to environmental changes.

B. Mutations can be the source of many diseases and disorders.

What generally causes thymine dimers to form in a strand of DNA, and why are thymine dimers a problem? A. Smoking often makes thymine dimers form, increasing the chance for mutations that lead to lung cancer. B. Ultraviolet light can cause thymine dimers, potentially creating a mutation that could lead to cancer. C. Cancer often causes thymine dimers, which prevent nucleotide excision repair enzymes from working. D. Thymine dimers, which are often caused by exposure to sunlight, prevent cells from replicating DNA.

B. Ultraviolet light can cause thymine dimers, potentially creating a mutation that could lead to cancer.

In fruit flies, long wings (W) are dominant over short wings (w), and red eyes (R) are dominant over orange eyes (r). Each individual possesses two alleles for each trait. If a fly that is homozygous dominant for both traits is crossed with a fly that is homozygous recessive for both traits, what is the predicted genotype of the offspring? A. WwRr, wwRr, Wwrr, and wwrr B. WwRr C. long wings and red eyes D. short wings and orange eyes E. WWRR

B. WwRr

A mutagen that causes the loss of an amino group from a nucleotide is: A. ethylmethylsufonate (EMS). B. nitrous acid. C. 5-bromouracil (5BU). D. hydroxylamine.

B. nitrous acid.

Classify each mutation example as definitely positive, definitely negative, or most likely neutral. Each category has two examples. - Beneficial mutations? - Neutral mutations? - Harmful mutations? Answer bank: - A mutation produces bristled-antennae variants in moths. - A mutation confers antibiotic resistance in bacteria. - A mutation induces chlorophyll deficiency in plants. - A mutation causes Alzheimer's disease in humans. - A mutation leads to malarial resistance in humans. - A mutation results in red hair color in humans.

Beneficial mutations: - A mutation confers antibiotic resistance in bacteria. - A mutation leads to malarial resistance in humans. Neutral mutations: - A mutation produces bristled-antennae variants in moths. - A mutation results in red hair color in humans. Harmful mutations: - A mutation induces chlorophyll deficiency in plants. - A mutation causes Alzheimer's disease in humans.

A geneticist discovers an obese mouse in his laboratory colony. He breeds this obese mouse with a normal mouse. All the F1 mice from this cross are normal in size. When he interbreeds two F1 mice, eight of the F2 mice are normal in size and two are obese. The geneticist then intercrosses two of his obese mice, and he finds that all of the progeny from this cross are obese. These results lead the geneticist to conclude that obesity in mice results from a recessive allele. A second geneticist at a different university also discovers an obese mouse in her laboratory colony. She carries out the same crosses as the first geneticist and obtains the same results. She also concludes that obesity in mice results from a recessive allele. One day, the two geneticists meet at a genetics conference, learn of each others experiments, and decide to exchange mice. They both find that, when they cross two obese mice from the different laboratories, all the offspring are normal. However, when they cross two obese mice from the same laboratory, all the offspring are obese. Which option best explains their results? A. The obesity allele identified by the first geneticist is recessive to the obesity allele identified by the second geneticist. B. The two obesity alleles are codominant with the combination of the two alleles producing a normal phenotype. C. The two obesity alleles are recessive to the wild-type alleles but are located at different loci. D. The obesity allele identified by the first geneticist is dominant to the obesity allele identified by the second geneticist.

C. The two obesity alleles are recessive to the wild-type alleles but are located at different loci.

Germ-line mutations: A. only affect somatic tissue. B. cannot be passed down to offspring. C. are changes in the DNA of gametes. D. never occur in eukaryotes.

C. are changes in the DNA of gametes.

A nonsense mutation cannot be reversed by hydroxylamine exposure. This is because: A. hydroxylamine removes an amino group from the nucleotide. B. hydroxylamine will convert the codon to a different amino acid. C. hydroxylamine can only induce a CG-to-TA transition. D. hydroxylamine can only induce a TA-to-CG transition.

C. hydroxylamine can only induce a CG-to-TA transition.

What mechanism is likely responsible for most replication errors? A. strand slippage B. incorporated error C. mispairings through wobble D. tautomeric shifts

C. mispairings through wobble

Which of the flowers shown would be suitable as a tester strain (homozygous recessive) for all five loci? A. the second flower B. the third flower C. none of the flowers D. the fourth flower E. the first flower

C. none of the flowers

A transition is the substitution of a purine for a _____ or a pyrimidine for a _____. A transversion is a substitution of a pyrimidine for a _____ or a purine for a _____. A. pyrimidine; purine; purine; pyrimidine B. purine; pyrimidine; pyrimidine; purine C. purine; pyrimidine; purine; pyrimidine D. pyrimidine; purine; pyrimidine; purine

C. purine; pyrimidine; purine; pyrimidine

How many different types of gametes could be produced by the fifth flower? A. 10 B. 2 C. 1 D. 4 E. 5

D. 4

How many different types of gametes could be produced by the first flower? A. 10 B. 2 C. 1 D. 4 E. 5

D. 4

Assume that an Aa animal produces four sperm from the same meiosis. A geneticist examines each of the four products of this meiosis and finds that one of them carries the a allele, one carries the A allele, one carries both the A and a alleles, and one has neither allele. This result is not consistent with which genetic concept? A. principle of independent assortment B. conditional probability C. addition rule D. principle of segregation

D. principle of segregation

The table contains the DNA sequence for a segment of the human insulin gene and the same DNA sequence with a mutation. Original sequence: ATG GAA TAA AGC CCT TGA ACC AGC Mutated sequence: ATG GAA TAA AGG CCT TGA ACC AGC Which type of mutation occurred in the original sequence to generate the mutated sequence? A. deletion B. insertion C. inversion D. substitution

D. substitution

Match the statements with the correct types of DNA repair systems. Not all statements will be used. - Nucleotide excision? - Direct? - Mismatch? Answer bank: - The emergency repair system initiated when DNA damage is extreme. - Adenine binds to cytosine in a newly synthesized DNA strand. - Enzyme complexes remove bulky DNA lesions, creating a single-strand gap in DNA. - The methylation of guanine is corrected prior to DNA replication. - DNA is scanned for the removal and replacement of a few incorrect base pairs. - Alkyltransferase removes methylated guanine.

Nucleotide excision: - DNA is scanned for the removal and replacement of a few incorrect base pairs. - Enzyme complexes remove bulky DNA lesions, creating a single-strand gap in DNA. Direct: - Alkyltransferase removes methylated guanine. - The methylation of guanine is corrected prior to DNA replication. Mismatch: - Adenine binds to cytosine in a newly synthesized DNA strand.

Phenylketonuria (PKU) is a disease that results from a recessive gene. Two normal parents produce a child with PKU. PART 1: What is the probability that a sperm from the father will contain the PKU allele? A. 1/2 B. 2/3 C. 1/4 D. 3/4 E. 1/3 PART 2: What is the probability that an egg from the mother will contain the PKU allele? A. 1/2 B. 1/4 C. 2/3 D. 1/3 E. 3/4 PART 3: What is the probability that their next child will have PKU? A. 2/3 B. 3/4 C. 1/2 D. 1/4 E. 1/3 PART 4: What is the probability that their next child will be heterozygous for the PKU gene? A. 1/3 B. 1/4 C. 1/2 D. 2/3 E. 3/4

PART 1: A. 1/2 PART 2: A. 1/2 PART 3: D. 1/4 PART 4: C. 1/2

An alteration in the nucleotide sequence of a gene can alter the gene product. Each sentence is made up of three‑letter words, representing mRNA codons. THE FAT CAT SAT THE CAR WAS RED THE DOG RAN The same sentences are written with a substitution, deletion, or insertion, much as a gene may have a substitution, deletion, or insertion mutation. Identify which sentence represents a substitution, which represents a deletion, and which represents an insertion. - Substitution? - Deletion? - Insertion? Answer bank: - THE DGR AN - THE CAT WAS RED - THE FAT ACA TSA T

Substitution: - THE CAT WAS RED Deletion: - THE DGR AN Insertion: - THE FAT ACA TSA T

Why was the pea plant an ideal plant for Mendel to use?

simple traits that are easy to identify

Two gene loci, A and B, assort independently, and alleles A and B are dominant over alleles a and b. What is the probability of producing an AABB zygote from a cross of AaBb x AaBb?

1/16

How could two genetically identical oak trees have different heights when tree height is a heritable trait? A. The height of the trees may differ if one of them received more sunlight, water, and fertilizer than the other. B. The seeds that gave rise to these two trees were generated from two different sources of pollen. C. Genes do not contribute to the limits on the height of oak trees. D. Independent assortment of alleles for height during meiosis can produce offspring with varying degrees of height at maturity.

A. The height of the trees may differ if one of them received more sunlight, water, and fertilizer than the other.

What are transposable elements that transpose through an RNA intermediate called? A. retrotransposons B. composite transposons C. noncomposite transposons D. insertion sequences

A. retrotransposons

Which of the following base changes in DNA is an example of a transition? A. A to C B. G to C C. C to A D. A to G E. A to T

D. A to G

Which of the following crosses would produce a 1:1 ratio of phenotypes in the next generation? A. AA x AA B. AA x aa C. Aa x Aa D. Aa x aa E. aa x aa

D. Aa x aa

Red (A) is dominant to white (a). Two red individuals are crossed, and some white progeny are produced. What are the most likely genotypes of the two red parents of this cross? A. AA × Aa B. Aa × aa C. AA × aa D. Aa × Aa

D. Aa × Aa

Which of the following statements about silent mutations is FALSE? A. Silent mutations arise due to the redundancy of the genetic code. B. Silent mutations never have phenotypic effects. C. Silent mutations do not change the identity of an amino acid. D. Silent mutations are typically gain-of-unction mutations.

D. Silent mutations are typically gain-of-unction mutations.

Which of the following statements CORRECTLY describes nonsense mutations? A. They cause a nonfunctional amino acid to replace a functional amino acid. B. They change the nucleotide sequence of a gene but do not change the sequence of the resulting protein. C. They result in the insertion or deletion of a small number of nucleotides to the DNA. D. They convert a codon for a particular amino acid within a gene into a stop codon. E. They cannot revert to wild type.

D. They convert a codon for a particular amino acid within a gene into a stop codon.

Long stretches of which amino acid are highly toxic to cells? A. glycine B. guanine C. alanine D. glutamine E. tyrosine

D. glutamine

If two heterozygous Aa plants are crossed with each other, what will be the genotypic ratio found in the offspring? A. 1:1 B. 3:1 C. 1:1:1:1 D. 2:1 E. 1:2:1

E. 1:2:1

Of the following ideas postulated by Gregor Mendel, which one requires at least two genes to be demonstrated? A. One of two alleles from each parent is randomly transmitted to offspring. B. Some alleles are recessive and are masked by dominant alleles. C. Traits are not determined by blending. D. Traits are controlled by discrete units. E. Genes assort independently in diploids.

E. Genes assort independently in diploids.

How do germ-line mutations differ from somatic mutations?

Germ-line mutations can be passed on to offspring, while somatic mutations cannot.

Genes come in different versions called:

alleles

Ultraviolet light causes what type of DNA lesion?

pyrimidine dimers such as T = T

Trinucleotide repeat expansions involve all of the items except: A. All of the items are involved. B. replication. C. hairpins. D. triplets of base pairs. E. repetitive DNA segments.

A. All of the items are involved.

Which of the statements can be concluded from Gregor Mendel's experiments with pea plants? [Select all that apply.] A. The inheritance of alleles of one gene does not affect the inheritance of alleles of another gene. B. Parents with dominant phenotypes always have offspring with dominant phenotypes. C. Offspring inherit two alleles per gene from each parent. D. Two recessive alleles are necessary for a recessive phenotype.

A. The inheritance of alleles of one gene does not affect the inheritance of alleles of another gene. D. Two recessive alleles are necessary for a recessive phenotype.

Mary is evaluating DNA structure for a genetics assignment. From a strand of DNA, she was able to back track the original DNA from its mutated state. Which of these conclusions would be FALSE concerning the models? [The illustration on the left shows a DNA sugar phosphate backbone running from 5 prime at the top to 3 prime at the bottom. The pentose sugars have bases on the C1 and phosphate molecules on the C4. The first sugar molecule at the top has pyrimidine labeled 'X' attached to it. The sugar below it has purine labeled 'Y' attached. The bottom sugar molecule has a purine attached to it. There is a red arrow pointing to a second illustration. In the next illustration, the DNA sugar phosphate molecule has three sugar bases with the first sugar having nothing attached, the second has a pyrimidine labeled 'Y' attached, and the third with a pyrimidine base attached.] A. The mutation observed with the loss of object X can either be guanine or adenine. B. The mutation observed can be classified as depyrimidination. C. Depyrimidination changes occur at a lower rate than depurination. D. The mutation exhibited is one of the most common forms of a spontaneous chemical mutation.

A. The mutation observed with the loss of object X can either be guanine or adenine.

Which of these is NOT a possible function of silent mutations? A. production of gene product that stimulates growth factor B. altering rate of protein synthesis C. influencing binding of miRNA sequences to mRNA D. altering nucleotides that serve as binding sites for regulatory proteins

A. production of gene product that stimulates growth factor

During transposition, a retrotransposon may be transcribed into RNA to produce double-stranded DNA by which process? A. reverse transcription B. resolution C. forward methylation D. back mutation

A. reverse transcription

Use the Mutation interactive to answer the question. In the lower half of the interactive, change the template DNA strand such that the sequence: ...AACCTG... is changed to: ...AACTTG... Which of the terms correctly describes this type of mutation? A. substitution B. deletion C. duplication D. insertion If this mutation occurred in a protein‑coding gene, how would it be classified? A. missense mutation B. silent mutation C. nonsense mutation D. frameshift mutation

A. substitution C. nonsense mutation

A chi-square value is determined by calculating the _____ of all the squared differences between observed and expected and divide by the expected values. A. sum B. phenotypes C. multiples D. alleles

A. sum

Hydroxylamine converts cytosine (C) into hydroxylaminocytosine, which forms base pairs with adenine. Based on this information, all mutations induced by hydroxylamine should be _____ mutations. A. transition B. nonsense C. transversion D. frameshift

A. transition

How many different types of gametes could be produced by the fourth flower? A. 2 B. 4 C. 1 D. 10 E. 5

C. 1

How many different types of gametes could be produced by a flower that was heterozygous at all five loci? A. 1024 B. 1 C. 32 D. 4 E. 2

C. 32

How many different types of gametes could be produced by the third flower? A. 2 B. 1 C. 4 D. 10 E. 5

C. 4

In general, all of these statements are true concerning DNA repair mechanisms, EXCEPT: A. they always correctly repair mutations. B. they repair all mistakes in the cell. C. they require a template strand of DNA only. D. they are present in most organisms.

C. they require a template strand of DNA only.

Joel is assessing the presence of mutations in model DNA strands. He captures the DNA sequence shown. What would the resulting RNA complement be if undergoing the mutations shown? Note that the mutations occur in the order of position. DNA: 5'-TCAATAGATGTTTTACCGGGTTAC-3' Mutation: insertions at positions 4, 7, and 13; deletions at 2, 9, and 24 A. 5'-GAAGCCTCGTAACAACATCTATTG-3' B. 5'-GUAACCCGGUAAAACAUCUAUUGA-3' C. 5'-GAGACUCGGUGAAAACAUCUAUUG-3' D. 5'-GAAGCCUCGUAACAACAUCUAUUG-3'

D. 5'-GAAGCCUCGUAACAACAUCUAUUG-3'

How did Mendel use self‑pollination and cross‑pollination techniques in his experiments with flower color to observe the basic patterns of inheritance? A. By cross‑pollinating a parental generation of plants with same‑colored flowers and allowing the F1 generation to self‑pollinate, Mendel observed the basic patterns of inheritance in the F2 generation. B. By allowing a parental generation of plants to self‑pollinate and cross‑pollinating the F1 generation, Mendel observed the basic patterns of inheritance in the F2 generation. C. By cross‑pollinating a parental generation of plants with different colored flowers, Mendel was able to observe the basic patterns of inheritance in the F1 generation. D. By cross‑pollinating a parental generation of plants with different‑colored flowers and allowing the F1 generation to self‑pollinate, Mendel observed the basic patterns of inheritance in the F2 generation.

D. By cross‑pollinating a parental generation of plants with different‑colored flowers and allowing the F1 generation to self‑pollinate, Mendel observed the basic patterns of inheritance in the F2 generation.

At the molecular level, what is the difference between the R allele resulting in round seed and the r allele resulting in wrinkled seed in the garden pea? A. The two alleles differ by only one nucleotide in their DNA sequence, and this difference occurs near the end of the gene. B. The R allele produces a protein involved in removing water from the seed, and the r allele produces a protein involved in producing starch in the seed. C. The R allele is located on chromosome five, and the r allele is located on chromosome six. D. The r allele contains an extra 800 base pairs of DNA that disrupt the coding region of the gene.

D. The r allele contains an extra 800 base pairs of DNA that disrupt the coding region of the gene.

How is a true breeding round‑seeded pea plant different from a hybrid round‑seeded pea plant? A. They have a different genotype and phenotype. B. They have the same genotype and phenotype. C. They have the same genotype but different phenotypes. D. They have the same phenotype but different genotypes.

D. They have the same phenotype but different genotypes.

What is a transposable element? A. an enzyme that removes chunks of DNA from the genome B. an RNA molecule that can insert into the genome C. a protein that can transform other proteins D. a segment of DNA that can move around in the genome

D. a segment of DNA that can move around in the genome

Which of the examples is not one of the basic types of gene mutations? A. all are basic types of mutations B. insertions of base pairs C. deletions of base pairs D. base‑pair rearrangements E. base‑pair substitutions

D. base‑pair rearrangements

Which type of mutation usually has the smallest effect on the function of the protein encoded by the mutated gene? A. insertion of base pairs B. they all have no effect C. they all have the same effect D. base‑pair substitutions E. deletion of base pairs

D. base‑pair substitutions

Which of these is NOT a factor affecting mutation rates? A. probability of repair when DNA alterations take place B. probability that a mutation is detected C. frequency at which DNA changes take place D. changes in the environment

D. changes in the environment

The formation of chemical bonds between nucleotides on the same strand of DNA resulting from UV light exposure may cause all of these except: A. thymine-cytosine dimers. B. pyrimidine dimers. C. cytosine dimers. D. guanine dimers.

D. guanine dimers.

Which type of DNA mutation results in a change in the reading frame of an mRNA? A. substitution of one nucleotide with another B. duplication of a single codon C. substitution of one codon for another D. insertion of a single nucleotide

D. insertion of a single nucleotide

The genetic principle that states that alleles at one locus do not affect the segregation of alleles at another locus is called the A. principle of segregation. B. concept of dominance. C. chromosome theory of heredity. D. principle of independent assortment.

D. principle of independent assortment.

White (w) coat color in guinea pigs is recessive to black (W). In 1909, W. E. Castle and J. C. Phillips transplanted an ovary from a black guinea pig into a white female whose ovaries had been removed. They then mated this white female with a white male. All the offspring from the mating were black in color (W. E. Castle and J. C. Phillips, 1909, Science 30:312-313). PART 1: Which statement explains the results of this cross? A. The eggs from the ovary of the black coat guinea pig were likely fertilized by a black‑coated male guinea pig before the transfer. B. The black coats likely arose from mutations that occurred from the transplanting of the ovaries. C. The white female was heterozygous for the black trait, which explains how the offspring inherited the trait for black coats. D. The researchers failed to remove the original ovaries of the white female before adding the ovaries from the black guinea pig. E. The color of the offspring was determined by genes in the transplanted ovary, not the genes of the female who gave birth. PART 2: Select the genotype of the offspring from this cross. A. ww B. WW C. Ww D. Both WW and Ww are likely genotypes. E. Both WW and ww are likely genotypes. PART 3: Select the conclusions that are indicated by this experiment regarding the validity of the pangenesis and the germ‑plasm theories discussed in Chapter 1. [Select all that apply.] - The experiment indicates nothing regarding the validity of either the pangenesis or germ‑plasm theory. - Because no white guinea pigs were produced, no white coat alleles traveled to the ovary and into the gametes of the white female, thus indicating pangenesis did not occur. - The male white guinea pig likely had a greater contribution to the phenotypes of the offspring, thus invalidating the pangenesis theory. - The production of black guinea pig offspring suggests that the allele for black coat color was passed along to the offspring from the transplanted ovary, thus supporting the germ‑plasm theory.

PART 1: E. The color of the offspring was determined by genes in the transplanted ovary, not the genes of the female who gave birth. PART 2: C. Ww PART 3: - Because no white guinea pigs were produced, no white coat alleles traveled to the ovary and into the gametes of the white female, thus indicating pangenesis did not occur. - The production of black guinea pig offspring suggests that the allele for black coat color was passed along to the offspring from the transplanted ovary, thus supporting the germ‑plasm theory.

Use the interactive Punnett Squares, Level 2 to answer the questions. Suppose the parents indicated in the interactive produced 192 peas . Determine the expected number of yellow and wrinkled offspring. PART 1: - expected number of yellow and wrinkled offspring? _______ Now, suppose one of the pea plants is homozygous recessive for both genes, resulting in the cross Yy Rr × yy rr. Calculate the percentage of yellow and round offspring. PART 2: - yellow and round offspring? _______% PART 3: Suppose one of the pea plants is homozygous recessive for seed color, resulting in the cross Yy Rr × yy Rr. Select the correct phenotypic ratio for the offspring. A. 6 yellow, round : 6 yellow, wrinkled : 2 green, round : 2 green, wrinkled B. 9 yellow, round : 3 yellow, wrinkled : 3 green, round : 1 green, wrinkled C. 6 yellow, round : 2 yellow, wrinkled : 6 green, round : 2 green, wrinkled D. 4 yellow, round : 4 yellow, wrinkled : 4 green, round : 4 green, wrinkled

PART 1: - 36 PART 2: - 25% PART 3: C. 6 yellow, round : 2 yellow, wrinkled : 6 green, round : 2 green, wrinkled

Alleles A and a are located on a pair of metacentric chromosomes. Alleles B and b are located on a pair of acrocentric chromosomes. A cross is made between individuals having the genotypes Aa Bb and aa bb. PART 1: Label the gametes of the aa bb parent with the chromosomes carrying the correct alleles. PART 2: Which gametes can the Aa Bb parent generate? - a b - A B - A b - a B PART 3: For example progeny resulting from this cross, match the chromosomes as they would appear in a cell at G1, G2, and metaphase of mitosis.

PART 1: 1. ab 2. ab 3. ab 4. ab PART 2: - a b - A B - A b - a B PART 3: G1: *image* G2: *image* Metaphase of Mitosis: *image*

In watermelons, bitter fruit (B) is dominant over sweet fruit (b), and yellow spots (S) are dominant over no spots (s). The genes for these two characteristics assort independently. A homozygous plant that has bitter fruit and yellow spots is crossed with a homozygous plant that has sweet fruit and no spots. The F1 are intercrossed to produce the F2. PART 1: What will be the phenotypic ratio in the F2? A. All sweet fruit with no spots B. All bitter fruit with yellow spots C. 1/2 bitter fruit, yellow spots and 1/2 sweet fruit, no spots D. 1/4 bitter fruit, yellow spots; 1/4 bitter fruit, no spots; 1/4 sweet fruit, yellow spots; and 1/4 sweet fruit, no spots E. 9/16 bitter fruit, yellow spots; 3/16 bitter fruit, no spots; 3/16 sweet fruit, yellow spots; and 1/16 sweet fruit, no spots PART 2: If an F1 plant is backcrossed with the bitter, yellow‑spotted parent, what phenotypes and proportions are expected in the offspring? A. 1/4 bitter fruit, yellow spots; 1/4 bitter fruit, no spots; 1/4 sweet fruit, yellow spots; and 1/4 sweet fruit, no spots B. 9/16 bitter fruit, yellow spots; 3/16 bitter fruit, no spots; 3/16 sweet fruit, yellow spots; and 1/16 sweet fruit, no spots C. All sweet fruit with no spots D. All bitter fruit with yellow spots E. 1/2 bitter fruit, yellow spots and 1/2 sweet fruit, no spots PART 3: If an F1 plant is backcrossed with the sweet, nonspotted parent, what phenotypes and proportions are expected in the offspring? A. All sweet fruit with no spots B. 1/4 bitter fruit, yellow spots; 1/4 bitter fruit, no spots; 1/4 sweet fruit, yellow spots; and 1/4 sweet fruit, no spots C. All bitter fruit with yellow spots D. 9/16 bitter fruit, yellow spots; 3/16 bitter fruit, no spots; 3/16 sweet fruit, yellow spots; and 1/16 sweet fruit, no spots E. 1/2 bitter fruit, yellow spots and 1/2 sweet fruit, no spots

PART 1: E. 9/16 bitter fruit, yellow spots; 3/16 bitter fruit, no spots; 3/16 sweet fruit, yellow spots; and 1/16 sweet fruit, no spots PART 2: D. All bitter fruit with yellow spots PART 3: B. 1/4 bitter fruit, yellow spots; 1/4 bitter fruit, no spots; 1/4 sweet fruit, yellow spots; and 1/4 sweet fruit, no spots