BIO 304 Test 3 HW Questions

Which of the following single base pair subsitutions is a transversion mutation?

A:T to C:G One way to keep them straight is to remember that the larger bases (A and G) have the smaller name (purines) whereas the smaller bases (C and T) have the longer name (pyrimidines). Transistions replace one purine with another or one pyrimidine with another whereas transversions replace a purine with one of the pyrimidines or a pyrimidine with one of the purines.

Which of the following events could result in a frameshift mutation?

Base deletion A base deletion would shorten the DNA sequence and change the reading frame of the mRNA.

Classify the nature of the mutation in colony 5.

adenine auxotrophic mutation

The inversion of a portion of a chromosome that does not contain a centromere

paracentric inversion

Which of the following is not a valid definition of a mutation rate?

# mutations/gene/generation # mutations/chromosome/year # mutations/gene/gamete *all of the above are valid ways to express a mutation rate* none of the above are used to calculate mutation rate Mutation rates are presented in lots of different ways, but you need three key components to show a rate. First, you need a way to detect the number of mutations that occurred (we'll count bacterial colonies in the Ames Test and yeast colonies in lab to do this). The second is to focus you measure of mutatoin on a unit of DNA (gene or chromosome - we'll foucs on genes using the Ames Test and in lab). Third, you must measure over a unit of time that can be defined in any way that is convenient. Time (minutes/hours/days/years) or reproductive units (generations/gametes) are common. How many heritable, dominant neurofibromatosis mutations are created each generation in humans? About 1/4000 NF genes/generation. You could also write that 0.00025/gene/generation. That's a lot!

Suppose a diploid cell with three pairs of homologous chromosomes (2n=6) enters meiosis. How many chromosomes will the resulting gametes have if there is nondisjunction of all three chromosomes in one daughter cell in meiosis II?

0, 3, or 6

If a plant is an autotetraploid with the genotype A1A1A2A2, what is the frequency of offspring with the genotype A1A1A1A1 following self-fertilization of the plant?

1/36

Suppose a diploid cell with three pairs of homologous chromosomes (2n=6) enters meiosis. How many chromosomes will the resulting gametes have if there is nondisjunction of one chromosome pair in meiosis I?

2 or 4

Because of lenient rules for third-base wobble in translation, the minimum number of different tRNAs required in mammalian mitochondria is __________.

22

Suppose that the transient rare guanine tautomer shifted back to the common guanine tautomer prior to a second round of replication. Which DNA sequence(s) would be present in the sister chromatids after this second round of replication? Select all that apply.

5' - C A A - 3'3' - G T T - 5 5' - T A A - 3'3' - A T T - 5' After the first round of DNA replication, the DNA sequence of the sister chromatids will be: (Note: G* is the rare guanine tautomer.)5' - CAA - 3'3' - GTT - 5'5' - TAA - 3'3' - G*TT - 5'After transient rare guanine tautomer shifted back to the common guanine tautomer prior to a second round of replication, the DNA sequence of the sister chromatids will be: (Note: Italics represent the template strand used during the second DNA replication.)5' - TAA - 3'3' - ATT - 5'5' - CAA - 3'3' - GTT - 5' (Note that this sister chromatid is present after the first replication and after the second replication.)Now a permanent mutation has been incorporated into one of the sister chromatids as a result of a tautomeric shift.

A fragment of a wild-type polypeptide is sequenced for seven amino acids. The same polypeptide region is sequenced in four mutants.Wild-type N . . . Thr-His-Ser-Gly-Leu-Lys-Ala . . . C polypeptideMutant 1 N . . . Thr-His-Ser-Val-Leu-Lys-Ala . . . CMutant 2 N. . . Thr-His-Ser-CMutant 3 N . . . Thr-Thr-Leu-Asp-CMutant 4 N . . . Thr-Gln-Leu-Trp-Ile-Glu-Gly . . . C Determine the wild-type mRNA sequence.

5'-A-C-N-C-A-C-U-C-U-G-G-A-U-U-G-A-A-G-G-C-N-3'

Two unrelated diploid plant species of grass are chosen to create new hybrid varieties. Species A has 18 chromosomes and species B has 28 chromosomes. One hybrid variety was created by fusion of a diploid gamete from species A and a haploid gamete from species B. Which of the following statements best describes what is expected to be seen in cells at metaphase of meiosis I in the hybrid?

9 bivalants and 14 univalents

A chromosome contains the following gene order:A B C D • E F G HWhich of the following rearrangements represents a paracentric inversion?

A C B D • E F G H Submit

Which of the following crosses will produce progeny whose germ cells will produce an unpaired loop at metaphase of meiosis I?

A cross between a strain homozygous for a chromosomal deletion and a wild type strain. A cross betwween a strain homozygous for a chromosomal duplication and a wild type strain. A cross between a strain homozygous for a reciprocal translocation and a wild type strain. A cross between a strain homozygous for an inversion and a wild type strain.all the answers are correct *Only answers (a) and (b) are correct.* Only answers (c) and (d) are correct. A strain heterozygous for either a large deletion or a large duplication will align all homologous sequnces on the two chromosomes, leaving the "extra" segment as an unpaired loop. You should be able to determine whether the unpaired loop is on the wild type chromosome or the mutant chromosome in the progeny of the crosses described in (a) and (b).

Which of the following is an accurate description what what would be seen at metaphase of mitosis in the F1 hybrid created by mating fly line 1 and 2.

A normal alignment of all chromosomes without synapsis of homologs.

Which of the following is an accurate description of what would be seen at metaphase I of meiosis in F1 hybrid progeny created by mating fly line 1 and 2?

A synapsed pair of homologs showing an inversion loop, with the centromere included in the loop. Inversion loops are just one of many types of structures that help illustrate that synapsis during meiosis I is an active process that will occur if it can, even when the homologous segments are inverted relative to one another as in this example, or if they are on different chromosomes, as is the case for cells heterozygous for a reciprocal translocation. What about the unpaired loop answer? That is a structure seen in meiosis I - what does that mean about the dosage of genes located in the unpaired region?

The amino acid changes noted in Parts A-C are the result of changes in DNA sequence. For each of these amino acid changes, evaluate whether a single base change in the given codons could account for the change in the amino acid. For which codon(s) could a single base change account for this amino acid change?Lysine to Asparagine

AAG AAA There are two codons for lysine (AAA, AAG) and two codons for asparagine (AAC, AAU). Since positions #1 and #2 are identical in the two sets of codons, then a single base change in the #3 position of the lysine codons can result in either asparagine codon (AAA/G -> AAC/U).

An F1 hybrid is created by mating fly line 1 with fly line 2. A single crossover occurs in the region between genes G and F in a meiotic cell in the hybrid. Which of the following is the best description of the genotypes of the four chromatids in this cell after the crossover is completed but before anaphase I begins? NOTE: the position of centromeres is indicated by an asterisk (*).

ABHGF*EDMN; abde*fghmn; ABHGf*edba; nmhgF*EDMN This question REALLY tests your understanding of meiosis, synapsis and recombination because the crossover occurs within the inverted region. If you are good a drawing the chromosomes aligned with an inversion loop and you can use that structure to detemine the genotypes of the recombinant chromatids, then you are awesome (see figure 13.18 for help with this). If not, then I'll show you a trick to help you work out the results of crossing over within the inverted segments in class.

An F1 hybrid is created by mating fly line 1 with fly line 2. A single crossover occurs in the region between genes A and B in a meiotic cell in the hybrid. Which of the following is the best description of the genotypes of the four chromatids in this cell after the crossover is completed but before anaphase I begins? NOTE: the position of centromeres is indicated by an asterisk (*).

ABHGF*EDMN; abde*fghmn; Abde*fghmn; aBHGF*EDMN In this case, the crossover is not within the region that is inverted in one homolog relative to the other, so there are no abnormal meiotic products. We'll work out problems like this in class and I can show you a trick that helps you leave out the inversion loop in the drawing.

Which of the following fly lines would be expected to be sterile?

All the progeny of the F1 hybrid. Fly line 2 Fly line 1 An F1 hybrid of fly lines 1 and 2 *None of the above are correct.* That's right, inversions do not make individuals sterile. They do affect the frequency that you will detect certain types of recombinant progeny (you'll have a chance to display your understanding of that concept below!).

__________________ can occur when there is hybridization between two closely related species, followed by chromosome doubling.

Allopolyploidy

Polyploidy that is produced by the duplication of chromosomes from a single genome.

Autopolyploidy

Which of the following statements regarding mutation rates is correct?

Average mutation rates vary among taxonomic groups.

Which structures would be seen in a triploid? Check all that apply.

B, H, F That's it! In triploids, there are three copies of each chromosome and some threesomes but not all synapse to form trivalents. When two of three are zipped up together to create a bivalent, then third stands alone as a univalent.

Which of the following statements regarding Nucleotide Excision Repair (NER) and Base Excision Repair (BER) is true?

Both NER and BER involve the removal of one or more damaged bases by a nuclease.

For which codon(s) could a single base change account for this amino acid change?Leucine to Glutamine

CUG CUA There are six codons for leucine (UUA, UUG, CUA, CUG, CUC, CUU) and two codons for glutamine (CAA, CAG). Only two leucine codons share two of the codon positions with glutamine. CUA shares two positions with CAA. CUG shares two positions with CAG. For both leucine codons, a base change from U to A will change them to glutamine codons.

In an allopolyploid organism, what is true regarding the fertility of interspecies hybrids?

Chromosome doubling and nondisjunction in gametocytes can lead to homologous chromosome pairing, disjunction, and fertile hybrids.

Which structure identifies the cell as being heterozygous for a reciprocal translocation?

E Good job! You identified the tetravalent structure that is seen in cells heterozgyous for a reciprocal translocation. Two of the chromosomes are normal in stucture and two have reciprocal translocations. All regions of all four chromosomes are paired with a homologous region.

All compounds that have been found to be mutagenic in the Ames test are also carcinogenic.

False The Ames test is used as a preliminary screening tool. Not all compounds that give a positive Ames test are carcinogenic.

Which type of mutation results in a string of altered amino acids at the end of the polypeptide product?

Frameshift

For which codon(s) could a single base change account for this amino acid change?Alanine to Phenylalanine

GCC GCA GCU GCG *none of these codons* There are four codons for alanine (GCA, GCU, GCG, GCC) and two codons for phenylalanine (UUC, UUU). Neither of the two sets have codons that share any two positions, therefore it is not possible for a single base change to result in a change from an alanine codon to a phenylalanine codon.

Which nucleotide will base‑pair with the enol form of 5‑bromouracil?

Guanine The enol form of 5‑bromouracil forms a base pair with guanine.

Consider the correct answers for questions Parts E, F, G and H. If you were asked to determine which of the chromatids produced by meiosis could be incorporated into gametes that would be viable, how would you make that determination?

I would check that one centromere and one copy of each gene was present on a chromatid to determine its viability. If there is only one centromere and there is one copy of each gene, then it is viable. For this and other questions on gamete viability, alwasys assume that chromatids that lack genes or have duplicated genes will lead to non-viable gametes because the "genes" in these drawings really represent large segments of the chromosome, and large duplications or deletions alter gene dosage to an extent that is almost always lethal. Note that, even though all the recombinant chromatids had one and only one centromere in this problem, counting the number of centromeres per chromatid could be relevant for other types of problems involving crossing over in inversion heterozygotes. Which type of F1 hybrid can produce chromatids that are dicentric or acentric at the end of metaphase I when there is a crossover in the inverted region? Compare figures 13.17 and 13.18 in your textbook.

Which of the following transposition events is most likely to result in a loss of function mutation?

Insertion of an IS element within the coding region of a gene Insertions of DNA elements within coding regions have the greatest potential for altering amino acid sequence and/or causing truncation of a polypeptide due to frameshift.

Chromosome 1 in humans is about 250 Mbs long. You discover an indvidual who has a normal version of chromosome 1 and a mutant version that is 260 Mbs long. Given that all you know is the size of chromosome 1, which of the following is NOT a possible cause of the mutation you discovered in chromosome 1?

It contains an inversion of a 10 Mb segment of chromosome 1. The only description that wouldn't work is the inverion. All the other mutations could result in an increase in the size of the chromosome. For example, there could have been a reciprocal translocation of chromosome 1 with chromosome 2 in which a 10 Mb piece of chromosome 1 was moved to chromosome 2 while a 20 Mb piece of chromosome 2 was moved to chromosome 1. This would give a net increase of 10 Mb in chromosome 1 length. This would also make a prediction concerning chromosome 2, which you could check.

Why are liver extracts used in the Ames test?

Liver enzymes may activate some innocuous compounds, making them mutagenic.

What type of mutation results in a single amino acid substitution?

Missense.

Which of the following are always encoded by the mitochondrial genome?

Mitochondrial rRNAs

The presence of a single chromosome instead of a homologous pair in a diploid individual.

Monosomy

Bypass (or translesion) DNA polymerases in E. coli are unique in their ability to do which of the following?

Replicate across DNA damage that stalls DNA pol III.

The fusion of two non-homologous chromosomes to form a single, larger chromosome

Robertsonian translocation

AluI is an example of which type of transposable element commonly found in the genome of humans?

SINE elements

The Ames test is used to determine the potential mutagenicity of test compounds by screening for new mutations in which of the following organisms?

Salmonella typhimurium

In eukaryotes, homologous recombination is initiated by ________.

Spo11 generating double-stranded DNA breaks

Spontaneous mutations can arise following DNA replication when DNA bases undergo a tautomeric shift. Tautomers have the same chemical composition, but the hydrogen atom is in a different position. This change can result in different hydrogen bonding configurations. Stable common tautomeric forms make up the standard base pairing rules (A-T and G-C). Rare unstable tautomers have altered base pairing rules, as shown in this figure. Use this partial DNA sequence (the original sequence with no mutations) to answer the following questions.5' - C A A - 3'3' - G T T - 5' Suppose that a transient tautomeric shift occurred in the guanine base to produce a rare tautomer in the partial DNA sequence just prior to a round of DNA replication. Which base would be added opposite this rare tautomer during DNA replication?

T When the replicative polymerase encounters this rare guanine tautomer, thymine is added to the newly synthesized strand instead of cytosine.

Which of the following statements concerning the relationship between the chromosomes in Fly l and Fly 2 is true?

The fly 1 chromosomes and the fly 2 chromosomes differ by a pericentric inversion.

The terms "univalent", "bivalent", "trivalent" and "tetravalent" refer to chromosome structures that can be seen at metaphase I of meiosis in certain types of cells. A single, unpaired chromosome is a univalent. Two synapsed (paired) chromosomes is a bivalent. Three synapsed chromosomes is a trivalent. Four synapsed chromosomes is a tetravent. Use this information to help you answer the questions below. Two unrelated diploid plant species of grass are chosen to create new hybrid varieties. Species A has 18 chromosomes and species B has 28 chromosomes. One hybrid variety has 23 chromosomes. Which of the following statements concerning this hybrid is false?

The hybrid could have been created by the fusion of one haploid gamete from each parent. Hybrid cells at metaphase of meiosis I are expected to have no bivalents. Hybrid cells at metaphase of meiosis I are expected to have 23 univalents. The hybrid is expected to be sterile. *None of the statements are false.*

The genetic distance beteen genes G and H has been studied extensively and shown to be 7 cM. A female F1 hybrid created by crossing Fly line 1 and Fly line 2 was crossed to a strain that is hg/hg and the frequency of progeny with recombinant genotype for G and H were identified. Which of the following is the most accurate statement of the expected result?

The recombination frequency will be signficiantly less than 7%. You remembered that inversions tend to suppress the production of recombinant chromosomes, so you would expect to see less than the frequency predicted by simply looking at the genetic distance.

What are the consequences of having pyrimidine dimers in DNA?

These dimers distort the DNA structure and result in errors during DNA replication.

Which type of DNA damage is repaired by the enzyme photolyase?

Thymine dimers.

Thymine dimers can be repaired by Photoreactivation Repair or Nucleotide Excision Repair.

True Both Photoreactivation Repair and Nucleotide Excision Repair will target UV-induced pyrimidine dimers in DNA.

Many chemicals are more mutagenic after being processed in the liver.

True Many potential mutagens are poorly mutagenic until passing through the liver

Tautomers of nucleotide bases are isomers that differ from each other in the location of one hydrogen atom in the molecule.

True Nucleotide tautomers differ only in the bonding location of one hydrogen atom.

Thymine dimers are most commonly caused by which of the following?

U.V. irradiation.

The diagram below shows two normal chromosomes in a cell. Letters represent major segments of the chromosomes. The following table illustrates some structural mutations that involve one or both of these chromosomes. Identify the type of mutation that has led to each result shown.

What type of chromosomal mutation occurred in line 2? duplication A deletion is the loss of part of a chromosomal segment. A duplication is the repetition of a segment. The repeated segment may be located next to the original or at a different location, and its orientation may be the same as the original or the reverse. An inversion is the removal of a segment followed by its reinsertion into the same chromosome in the reverse orientation. A translocation is the transfer of a segment to a nonhomologous chromosome. Translocations may be reciprocal (two nonhomologous chromosomes exchange segments) or nonreciprocal (one chromosome transfers a segment without receiving one).

You recently discovered four new chemical mutagens (#1 - #4), but you do not know what type(s) of mutations these chemicals induce.You exposed E. coli to each chemical individually and observed mutant phenotypes. You then exposed this bacteria to a second known mutagen to observe if the mutations from chemicals #1 - #4 could be reversed. Your data is presented below. Mutations Induced byMutations Reversed by ChemicalEMS Acridine Orange 5-Bromouracil #1NoYesNo #2YesNoYes #3YesYesYes #4NoNoNo

What type(s) of mutations are likely produced by chemical #1? frameshifts Mutations that are induced by chemical #1 could be reversed by mutagens that can create the same type of mutations. If chemical #1 induced frameshift mutations by making small deletions or insertions, then acridine orange (which also induces frameshift mutations) could restore the proper reading frame with a second mutation. Since acridine orange is the only mutagen that reversed the mutations, then it is likely that chemical #1 induces frameshift mutations.

Common baker's yeast (Saccharomyces cerevisiae) is normally grown at 37∘∘C, but it will grow actively at temperatures down to approximately 20∘∘C. A haploid culture of wild-type yeast is mutagenized with EMS. Cells from the mutagenized culture are spread on a complete-medium plate and grown at 25∘∘C. Six colonies (1 to 6) are selected from the original complete-medium plate and transferred to two fresh complete-medium plates. The new complete plates (shown below) are grown at 25∘∘C and 37∘∘C. Four replica plates are made onto minimal medium or minimal plus adenine from the 25∘∘C complete-medium plate. The new plates are grown at either 25∘∘C or 37∘∘C, as indicated in the diagram.

Which colonies are prototrophic and which are auxotrophic? prototrophic: 1, 2, 3, 6 auxotrophic: 4, 5

All of the following are characteristics of insertion sequences elements except _______.

a copy of the insertion sequence becomes integrated at a new location The insertion sequence itself, not a copy, is excised from its original location and transferred to a new one.

What results from unequal crossing over between homologous chromosomes?

a duplication on one chromosome a deletion on one chromosome a reciprocal translocation an inversion on both chromosomes *Both (a) and (b) occur but not (c) or (d).* answers (a), (b), (c) and (d) are all possible results.

Generally speaking, which of the following mutations would most severely affect the protein coded for by a gene?

a frameshift deletion at the beginning of the gene A frameshift mutation at the beginning of a gene would affect every codon after the point where the mutation occurred. During protein synthesis, incorrect amino acids would be inserted from the point where the frameshift mutation occurred on; the resulting protein would most probably be nonfunctional. For this reason, a frameshift mutation at the beginning of a gene is generally the most severe type of mutation.

A chromosome has the following segments, where * represents the centromere: AB*CDEFG. What type(s) of chromosome rearrangement(s) would produce the chromosome ADC*BEF?

a pericentric inversion and a deletion

A _______________ chromosome produced from a chromosome breakage will get lost from the cell during cell division.

acentric

How is homologous recombination initiated in eukaryotes?

by a double-strand break in one homolog

The role of transposase activity in IS element transposition includes _______.

cutting DNA at the target sequence Transposase makes staggered cuts at the site targeted for insertion.

The transmission of genes contained in mitochondrial and chloroplast genomes is referred to as

cytoplasmic inheritance

Which of the following spontaneous changes in DNA structure/sequence generally results from strand slippage?

expansion of trinucleotide repeat sequences

Which of the following refers to a mutation that causes a change from mutant phenotype to wild type phenotype.

forward mutation reversion (reverse) mutation second site reversion (suppressor) mutation nonsense mutation a, b and c are correct. b, c and d are correct. b and d are correct *b and c are correct* Reversion mutations are key to understanding the Ames test, which starts out with bacteria that cannot make histidine and therefore cannot grow on medium that lacks histidine (for example, minimal medium). It then selects for rare mutations that convert the his- ("histidine minus") mutants to his+ (histidine plus) by plating millions of cells on a plate that lacks histidine. Only the rare his+ reverse mutants ("revertants") can grow on medium lacking histidine. We'll do lots of work with the Ames test as a way of helping you understand mutagens and mutation frequency.

Identify the mutation that produces Mutant 3.

frameshift mutation

Identify the mutation that produces Mutant 4.

frameshift mutation

If a segment of DNA were replicated without any errors, the replicated strand would have the following sequence of nucleotides: 5' - ACTACGTGA - 3' Sort the following replicated DNA sequences by the type of point mutation each contains (frameshift, base substitution, or neither), as compared to the correct sequence shown above.

frameshift mutation: 5'-ACTTACGTGA-3' 5'-ACTCGTGA-3' base substitution mutation: 5'-ACTAAGTGA-3' 5'-ACTACGTGT-3' neither: none A base substitution mutation can occur if the DNA polymerase inserts the wrong nucleotide base as it synthesizes a new strand of DNA. A frameshift mutation can occur if the DNA polymerase leaves out a nucleotide or adds an extra nucleotide to the sequence. Certain forms of cancer occur because of mutations in DNA sequences that are located in so-called mutational hotspots. These hotspots are locations in the DNA sequence where mutations occur more often than in other places.

Which bacteria grow on the agar plate if the Ames test is positive?

his + prototrophs The bacteria used in the Ames test to evaluate mutagenicity are his− auxotrophs. If the Ames test is positive, these bacteria have reverted back to wild type and are his + prototrophs

When all of the alleles of a cytoplasmic organelle gene are identical within a cell this is referred to as

homoplasmy

What growth information is used to make these determinations?

information about medium content

A chromosome has the following segments, where * represents the centromere: AB*CDEFG. What type(s) of chromosome rearrangement(s) would produce the chromosome AB*CDFG?

interstitial deletion

Refer to the table in part A above. What kind of mutation occurred in the chromosome on line 4?

inversion

Which of the following statements is true of non-homologous end joining (NHEJ)?

it is a double-strand repair pathway it is error-prone it utilizes the sister chromatid as a template for repair it is error-free *both A and B*

Identify the mutation that produces Mutant 1.

missense mutation

What type of aneuploidy is responsible for Turner syndrome in humans?

monosomy XO

Which type of double-stranded break repair would you expect to be used by a cell in the G1 phase of the cell cycle?

non-homologous end joining

Identify the mutation that produces Mutant 2.

nonsense mutation

Suppose that the top strand is the coding (nontemplate) strand and the three bases shown represent a single in frame codon in a gene. What will be the effect of the tautomeric shift-induced mutation on the amino acid sequence?

nonsense mutation The codon changed from CAA to TAA as a result of a tautomeric shift-induced mutation. This point mutation results in a change from a glutamine codon to a stop codon, causing a nonsense mutation.

Part complete Suppose a diploid cell with three pairs of homologous chromosomes (2n=6) enters meiosis. How many chromosomes will the resulting gametes have if meiosis occurs normally?

only 3 Submit

Retrotransposons are unique among transposable elements in that they require the enzyme __________ for mobilization.

reverse transcriptase

Which of the following describes a mutation that changes mutant phenotype to wild type phenotype without changing the mutant genotype to a wild type genotype?

second site reversion mutation Suppressor mutation is a great name for these things because the second site reversion mutation "suppresses the expression" of the original mutation. These type of mutants are identified initially as reversion mutants (like the his+ revertants in the Ames test) but are later found to still contain the original mutation plus a second mutation (they have two mutations!). Sometimes the second mutation is in the same gene as the original mutation (intragenic reversion or intragenic suppressor) and sometimes it is found in a different gene (an extragenic supressor mutation). Figure 12.6c provides and example of a second site (extragenic) suppressor mutation. In that example, gene B is discovered as a suppressor of a gene A mutation, revealing that gene B has a function related to gene

You mate a male fly that is pure-breeding for a reciprocal tanslocation involving chromosomes #2 and #3 with a wild type female. Which of the following is an expected phenotype of the F1?

semi-sterile Semisterility is a term that refers to the reduced fertility seen in strains that are heterozygous for a reciprocal tanslocation. Only about half of the products of meiosis in these individuals are fertile. Check the terms adjacent I and alternate segregation and understand how they relate to this semisterile phenotype.

Which types of mutation are possible thanks to the redundant nature of the genetic code?

silent

When a base substitution mutation occurs, one nucleotide in a replicating DNA sequence is substituted for another, which results in the production of a mutant strand of DNA. The result of the mutation depends on how the substituted nucleotide base alters the string of amino acids coded by the mutant DNA. The three types of base substitution mutations are nonsense mutations, missense mutations, and silent mutations. Each type is defined by how it affects protein synthesis.

silent: C instead of A nonsense: T instead of G missense: A instead of G frameshift: Extra C Point mutations in DNA sequences can profoundly affect protein synthesis, or they can have no effect at all. Point mutations can be beneficial to an organism but are more commonly neutral or harmful.

Which of the following mutations can be spontaneous?

single base subsitution insertion mutation trinucleotide repeat expansion reverse mutations transition mutation all but one of the above *all of the above* Spontaneous mutation simply refers to a mutation that occurred in the absence of a mutagen. What's a mutagen? Something that increases the mutation rate (or frequency) above that of the spontaneous mutation rate (or frequency). Different genes (and different non-genic regions of chromosomes) have different mutation rates (hot-spots = region that undergoes a high rate of mutation), so the spontaneous mutation rate is specific to each gene (although most genes have similar mutation rates).

In the Ames Test, the appearance of his+ revertants in the presence of a non-mutagenic control compound indicates that _______.

some of the reversion mutations are not caused by the mutagen being tested His+ revertants on the control plate are the result of spontaneous mutation.

Which type of mutation converts a nucleotide to an alternative structure with the same composition but slightly different placement of hydrogen bonds with a rare, less stable form that causes base-pair mismatch?

tautomeric shift

Classify the nature of the mutation in colony 1.

temperature sensitive mutation

Classify the nature of the mutation in colony 2.

temperature sensitive mutation

What can you say about colony 4?

temperature sensitive mutation adenine auxotrophic mutation

The purpose of the Ames Test is to _______.

test the mutagenic effects of chemicals The Ames test detects whether a given chemical can cause a reversion mutation in his- bacteria.

Each of the following is a consequence of heterozygous reciprocal translocation except _______.

the segregation pattern of chromosomes at anaphase I Alternate segregation at anaphase I will result in the formation of genetically complete gametes.

Which of the following types of mutations could be due to a single base substitution?

transition transversion missense nonsense silent frameshift *all are correct but one* all are correct except for two answers

What type(s) of mutations are likely produced by chemical #2?

transitions Mutations that are induced by chemical #2 could be reversed by mutagens that can create the same type of mutations. Since both EMS and 5-Bromouracil reversed the mutations created by chemical #2, and these mutagens create transition mutations, it is likely that chemical #2 also induces transition mutations.For example if chemical #2 induced an adenine-to-guanine transition mutation that gave rise to a mutant phenotype, then EMS could induce a guanine-to-adenine transition mutation to reverse the mutant phenotype.

What type(s) of mutations are likely produced by chemical #3?

transitions and frameshifts Mutations that are induced by chemical #3 could be reversed by mutagens that can create the same type of mutations. Since all three mutagens reversed mutations created by chemical #3, it is likely that chemical #3 induces transition mutations as well as small deletions or insertions that result in frameshift mutations.

Refer to the table in part A above. What kind of mutation occurred in the chromosomes on line 6?

translocation

An individual with a chromosome pair that consists of one normal chromosome and one chromosome containing a segment of a non-homologous chromosome.

translocation heterozygote

What type(s) of mutations are likely produced by chemical #4?

transversions Mutations that are induced by chemical #4 could be reversed by mutagens that can create the same type of mutations. Since none of three mutagens reversed mutations created by chemical #4, it is likely that chemical #4 does not induce transition mutations or frameshift mutations.Chemical #4 does in fact induce mutations since exposing E. coli to chemical #4 resulted in observed mutant phenotypes. Because these mutant phenotypes were not reversed by any of the three mutagens, it is likely that chemical #4 induces a different type of mutation. The only remaining choice listed as a mutation type is transversions. In the future, a mutagen that induces transversion mutations could be used to determine if the chemical #4 mutations could be reversed.

A wild type protein has 100 amino acids, including 10 consecutive glutamines. A mutant version of this protein contains 160 amino acids and is like wild type except that it contains 60 consecutive glutamines. Which of the following terms is the best description of the mutation?

trinucleotide repeat expansion Trinucleotide repeat expansion can occur within the reading frame of a gene, leading to abnormal proteins that typically have dominant effects. Note that this type of mutation also can occur in non-protein coding sequences where it affects expression of the gene (an example of a regulatory mutation).

Mutations can also occur within the regulatory regions of genes. Fragile X is an example of a disorder caused by one such mutation by increasing the length of the long arm of the X chromosome and ultimately decreasing mRNA stability. What type of mutation causes Fragile X?

triplet-repeat expansion

Which of the following chromosomal abnormalities would most likely be caused by nondisjunction during meiosis?

trisomy

Heterozygous carriers of chromosome inversions or translocations

will be viable and completely fertile because no genes were deleted. may or may not exhibit phenotypic abnormalities. may be infertile due to complications during meiosis. Both A and B are correct. *Both B and C are correct.*