3333 BIOL- genetics exam 2 hmwrk q's

In the exclusion zone at chernobyl, the wild life populations have rebounded in response to the absence of human activity following the nuclear accident in 1986. However, scientists have found that the rates of albinism (that could be caused by mutations in pigmentation pathways) are higher in the exclusion zone than in other parts of Ukraine. Explain why rates of albinism are higher in the exclusion zone and the mechanism that probably leads to higher rates of that phenotype.

-Albinism is caused by the loss of function of genes in the pathway that creates pigment melanin (the pigment that determines skin, fur, and eye color.) -Nuclear radiation is a form of ionizing radiation that causes induced mutations. Ionization radiation causes free radicals that can cause DNA damage. It can alter the bases (guanine oxidation), break phosphodiester bonds (single-strand and double strand breaks), disrupt the chromosome (deletions, translocations). -Sources of DNA damage are not just due to human influence. There are many mutagens in the natural environment.

Assuming independent assortment, what proportion of the offspring of the cross AaBbCcDd × AabbCCdd will have the aabbccdd genotype? 1/64 1/256 1/512 1/1024 0%

0% Rationale: The cc genotype is not possible, therefore there is a 0% chance of getting aabbccdd.

The autosomal (not X-linked) gene for brachydactyly, short fingers, is dominant to normal finger length. Assume that a female with brachydactyly in the heterozygous condition is married to a man with normal fingers. What is the probability that their first child will have brachydactyly? 3/4 1/2 1/8 1/4 2/3

1/2

The house fly, Musca domestica, has a haploid chromosome number of 6. How many chromatids should be present in a diploid, somatic, metaphase cell? 6 24 3 12 18

24

The ant, Myrmecia pilosula, is found in Australia and is named bulldog because of its aggressive behavior. It is particularly interesting because it carries all its genetic information in a single pair of chromosomes. In other words, 2n = 2. (Males are haploid and have just one chromosome.) A G1 somatic cell nucleus in a female diploid M. pilosula contains 2 picograms of DNA. How much DNA would be expected in a metaphase I cell of a female? 32 picograms 8 picograms 16 picograms 4 picograms Not enough information to tell

4 picograms

An organism with a diploid chromosome number of 46 (2n=46) will produce ________ combinations of chromosomes at the end of meiosis. 46 23 529 8388608 7.4 x 10^13

8388608 Rationale: Two raised to the 23rd power is 8388608.

Imprinting

A form of monoallelic expression. One allele is expressed while the other is not.

What is chromatin immunoprecipitation (ChIP)?

A technique used to identify DNA sequences bound to a protein of interest. This is done by cells being lysed and DNA being fragmented into smaller pieces. The protein of interest is purified using an antibody that recognized the protein. The DNA is then isolated and identified. DNA will copurify because chemically crosslinked to DNA.

The codon 5'-AAA-3' codes for the amino acid lysine. Which of the following mutations in this codon is a silent mutation? A. 5'-AAG-3' codes for lysine B. 5'-ATA-3' codes for isoleucine C. 5'-AAC-3' codes for asparagine D. 5'-CAA-3' codes for glutamine

A. 5'-AAG-3' codes for lysine

Which of the following statements about transcriptional regulation is false? A. A response element binds to an enhancer B. Mediator does not bind directly to DNA C. The function of an activator might be to recruit chromatin remodelers to make the DNA more accessible. D. The general transcription factors recruit RNA polymerase to the transcriptional start site.

A. A response element binds to an enhancer

Transcriptionally _______ genes can show lower levels of DNA methylation compared with transcriptionally _______ genes. A. active, inactive B, inactive, active C. suppressed, inactive D. constitutive, active E. Enhanced, active

A. Active, inactive Methylation block transcription factors from binding, thereby decreasing transcription. If a gene's promoter id hypomethylated, gene expression will likely increase. DNA methylases will respond to cell signals and add methyl groups to CpG residues. Removing the methyl groups involves a DNA repair mechanism called base excision repair (BER)

Which of the following statements is the most accurate in regards to mice pups who receive less maternal nurturing? A. As adults they have increased promoter methylation B. As adults they have increased ability to adapt to stress. C. As adults they have high levels of glucocorticoid receptor expression. D. As adults they are more likely to nurture their offspring. E. As adults their brains are heavily mutated.

A. As adults they have increased promoter methylation

Figure 2 demonstrates there is more acetylated H3K9 in HN rats. Why is this significant? A. HN rats have greater acetylation at the GR promoter, corresponding to the patterns of DNA methylation in figure 1 B. It supports the idea that HN rats have more DNA methylation. C. LN rats also have acetylated H3K9 at the GR promoter, causing the decreased GR expression in LN rats.. D. It supports the idea that HN rats have silenced GR expression.

A. HN rats have greater acetylation at the GR promoter, corresponding to the patterns of DNA methylation in figure 1

When would mutations arise in yeast cells to confer survival in the presence of toxic levels of hydrogen peroxide? A. Mutations are random and would have appeared prior to being exposed to hydrogen peroxide B. Mutations are adaptive and would have formed prior to being exposed to hydrogen peroxide. C. Mutations are adaptive and would have appeared after most of the cells died in the presence of hydrogen peroxide. D. Mutations are adaptive and would have appeared immediately in response to the hydrogen peroxide. E. Mutations are random but would not have formed until after the hydrogen peroxide was introduced.

A. Mutations are random and would have appeared prior to being exposed to hydrogen peroxide

If the protein Argonaut in RISC is not functioning, the which kind of posttranscriptional gene regulation is most likely affected? A. RNA interference B. ubiquitination C. shortening of the microRNAs polyA tail D. protein phosphorylation E. inhibition of ribosome assembly F. alternative splicing

A. RNA interference

If a promoter region is mutated in such a way that it can no longer be methylated, what would the most likely effect be? A. The gene linked to that promoter would be over-expressed B. The gene linked to that promoter would not undergo replication C. The gene linked to the promoter would be expressed at regular levels D. The gene linked to the promoter would still be expressed, but the protein should contain different amino acids E. The gene linked to the promoter would be under expressed

A. The gene linked to that promoter would be over-expressed

Which of the following statements about ChIP is false? A. The protein of interest is purified using an antibody that binds to the DNA sequence B. ChIP stands for chromatin immunoprecipitation C. The precipitated DNA could be sequenced or analyzed using some other method D. It is a lab technique used to identify DNA that is bound to a protein of interest. E. All of these statements are true.

A. The protein of interest is purified using an antibody that binds to the DNA sequence

How is methylation of DNA connected to the acetylation of histones and gene expression? Choose all that are correct. a. when DNA is not methylated but histones are acetylated, it promotes expression of the DNA. b. When DNA is not methylated and histones are deacetylated, it promotes expression of the DNA. c. When DNA is methylated and histones are deacetylated, it represses expression of the DNA. d. When DNA is methylated and histones are acetylated, it represses expression of the DNA. e. There is no relationship between DNA methylation, histone acetylation, and gene expression

A. When DNA is not methylated but histones are acetylated, it promotes expression of the DNA. C. When DNA is methylated and histones are deacetylated, it represses expression of the DNA. These combinations of changes result in the formation of the chromatin structure in the open conformation, which promotes gene expression, or in the closed conformation, which silences gene expression.

Which protein modification is most closely linked to protein degradation? A. ubiquitination B. acetylation C. phosphorylation D. glycosylation E. methylation

A. ubiquitination

During meiosis, chromosome number reduction takes place in ________. (Choose the BEST answer. Hint: which is reductional division?) Anaphase I Prophase I telophase I Anaphase II Metaphase II

Anaphase I Rationale: Reductionist division occurs during meiosis I, specifically anaphase I, because the homologous chromosomes are being pulled apart.

Cross fostering experiments were an integral part of the authors' experimental design. Which of the following statements is FALSE? A. The pups from LN moms were placed with HN moms and vice-versa. B. Cytosine methylation depends on who was the biological mother of the pup, not who reared the pup. C. GR methylation was likely removed during gamete formation. D. It supported the behavior was not solely genetic. E. All of these statements are true.

B. Cytosine methylation depends on who was the biological mother of the pup, not who reared the pup.

Which of the following would be an example of a cis acting eukaryotic gene regulatory element? A. TATA binding protein B. Enhancer C. RNA polymerase II D. General transcription factors

B. Enhancer

The recessive disease Friedreich's ataxia is caused by a trinucleotide repeat in intron 1 of the FXN gene. Studies have shown the expansion causes decreased FXN expression, leading to the disease phenotype. Scientists have proposed several models of how the mutant locus might silenced. Which of the following is the least likely to cause silencing of the mutant FXN locus? A. CpG methylation B. Histone acetyltransferase acetylating histones near the locus C. Mis-splicing of the FXN mRNA

B. Histone acetyltransferase acetylating histones near the locus

Transcriptionally __________________ genes can show lower levels of adjacent histone acetylation compared with transcriptionally __________ genes. A. Active, inactive B. Inactive, active C. Suppressed, inactive D, Constitutive, active E. Enhanced, active.

B. Inactive, active

In figure 1a, the shaded box around a portion of the sequence represents the NGF1-A binding site. What is the relationship of NGF1-A and GR? A. NGF1-A is a general transcription factor for GR B. NGF1-A is an activator of GR C. GR is an activator of NGF1-A D. NGF1-A methylates the GR promoter

B. NGF1-A is an activator of GR

Which of the following is the most common place to find DNA methylation? A. On guanines adjacent to other guanines B. ON cytosines when adjacent to guanines C. On cytosines when they are base paired to guanines D. on guanines when adjacent to cytosines E. on cytosines when adjacent to other cytosines

B. ON cytosines when adjacent to guanines

Which is the most common place to find DNA methylation? A. on guanines when adjacent to cytosines B. On cytosines adjacent to guanines C. on cytosines when adjacent to other cytosines D. on guanines when adjacent to other guanines E. on cytosines when they are base paired to guanines

B. On cytosines adjacent to guanines

The phenotype of vestigial (short) wings in Drosophila melanogaster is caused by an autosomal recessive mutant gene (vg) that assorts independently from an autosomal recessive gene for hairy (h) body. Assume that a cross is made between a fly with wild-type wings and a hairy body and a fly with vestigial wings and wild-type body hair. The wild-type F1 flies were crossed among each other to produce 1024 offspring. Which phenotypes would you expect among the 1024 offspring, and how many of each phenotype would you expect? (Hint: The parents must be truebreeding since the F1s are all phenotypically wild-type.) A. All wild type. B. Phenotypes: wild, vestigial, hairy, and vestigial hairy; Numbers expected: wild (576), vestigial (192), hairy (192), and vestigial hairy (64). C. Phenotypes: wild, vestigial, hairy, and vestigial hairy; Numbers expected: wild (192), vestigial (256), hairy (64), and vestigial hairy (192). D. All vestigial hairy. E. Phenotypes: wild, vestigial, hairy, and vestigial hairy; Numbers expected: wild (256), vestigial (256), hairy (256), and vestigial hairy (256).

B. Phenotypes: wild, vestigial, hairy, and vestigial hairy; Numbers expected: wild (576), vestigial (192), hairy (192), and vestigial hairy (64). Rationale: If the parents are true-breeding, then all the F1 progeny would be heterozygous at the two loci. Therefore, the F2s would display a 9:3:3:1 ratio. If we apply the ratio to 1024 F2 progeny, then we'd get the numbers stated in the correct answer.

The authors use trichostatin A (TSA) in some of their experiments. What does it do? A. TSA strips all methyl groups from the genome. B. TSA inhibits HDAC C. TSA deactivates DNA methyltransferases D. TSA inhibits NGFI-A.

B. TSA inhibits HDAC

Some animals like C. elegans do not have 5-methylcytosine, but do have methylated adenine. It is not known whether the methylated adenine plays the same role as methylated cytosine. If it does, which of the following would you predict? a. That Angelman syndrome patients would have methylated adenines in the promoter of their UBE3A. b. That removing the methyl groups from adenines in C. elegans would increase gene expression c. That removing the methyl groups from adenines in C. elegans would decrease gene expression d. That removing the methyl groups from adenines C. elegans would have no effect on gene expression e. I would predict both choice a and choice c to be correct.

B. That removing thee methyl groups from adenines in C. elegans would increase gene expression If methylated adenines serve the same function as 5-methylcytosine, then we expect that removal of the methyl groups to have the same effect on gene expression. Since methylation silences genes, removal of methyl groups would activate gene expression.

In regards to mRNA stability, what is the effect of decapping enzymes? A. The m7G cap of mRNAs is stabilized allowing mRNAs to be protected at the 5' end B. The m7G cap of mRNAs is removed causing mRNAs to be degraded from the 5' end. C. The polyadenylation of mRNAs is stabilized allowing mRNAs to be protected at the 3' end D. The splicing of mRNAs is blocked, which destabilizes mRNAs

B. The m7G cap of mRNAs is removed causing mRNAs to be degraded from the 5' end.

Tautomeric shifts ________. A. allow purines to bind to purines B. can cause missense mutations due to mispairings C. are likely to cause frameshifts D. are known to cause insertion or deletion mutations E. allow pyrimidines to bind to pyrimidines

B. can cause missense mutations due to mispairings

A class of mutations that results in multiple contiguous amino acid changes in proteins is likely to be which of the following? (Hint: this question from the textbook is asking what type of mutation is the most likely to cause many amino acids in a row to be changed in a protein.) A. transition B. frameshift C. transversion D. alkylating agent E. recombinant

B. frameshift

The sum of histone modifications and their interactions is called the A. protein code B. histone code C. expression code D. genetic code E. transcription code

B. histone code

A regulatory sequence of DNA that is 10,000 base pairs away from the gene it regulates is mutated. The result is that the gene being regulated is now expressed at a higher rate compared to when this regulatory sequence was not mutated. What would the wild-type sequence be called (before it was mutated)? A. Insulator B. silencer C. Zinc finger motic D. activator protein E. Enhancer

B. silencer If you mutate a silencer, gene expression will increase because wild-type function of a silencer is to repress gene expression.

Which of the following categories of mutations is not possible to pass to offspring? A. Silent B. Somatic C. Frameshift D. Induced E. X-linked

B. somatic Somatic mutations occur in somatic cells and are not passed to the enxt generation. Germ-line mutations occur in germ cells. Since these cells are used to produce offspring, mutations can be passed to offspring.

What is bisulfite sequencing? What does it do?

Bisulfite sequencing is a technique to identify methylated Cytosines by converting unmethylated C to U. Bisulfite does not convert methylated C.

How many different types of gametes can be formed by the genotype AaBbCc? A. 3 b. 4 c. 8 d. 16 e. 32

C. 8 - Determine the number of different heterozygous gene pairs (n) involved. 2^n is the number of different gametes that can be formed. So, 2^3 = 8 in this case.

Oxidants are molecules that can damage other molecules within the cell, including DNA, causing cell death or cancer. Oxidants can be toxins, such as venom or plant poisons, or toxicants, such as heavy metals or the chemicals in cigarette smoke. Antioxidants are molecules within the cell that deactivate oxidants. Antioxidants can come from a person's diet, such as some vitamins, or can be synthesized within the cell and can be small molecules or proteins. The human genome contains many genes that encode antioxidant proteins. The transcription of these genes is upregulated using antioxidant response elements (AREs) when damaging oxidants flood the cell. Isothiocyanates (ITCs) are found in broccoli and other leafy green vegetables. It is known that when ITCs are added to cells grown in culture, the amount of antioxidant proteins increases in the cells. As a nutrition scientist, you are interested in learning the mechanism behind ITCs. You hypothesize ITCs activate a protein which then binds to the ARE. You engineer (clone) a reporter construct using an ARE and luciferase and transfect the construct into HEK293 cells. What is the best prediction? A. Once activated, the protein will act as a suppressor B. If you mutate the ARE, relative fluorescence units will increase. C. Adding ITC to the system will increase relative fluorescence units (RFU) D. Adding estrogen to the system will cause relative fluorescence units (RFU) to increase E. All of these are sound predictions

C. Adding ITC to the system will increase relative fluorescence units (RFU)

Germ cell mutations ________. A. are of no consequence to the next generation B. affect somatic cell function in the current generation C. Are usually found in every cell of the next generation D. are the cause of testicular cancer E. are only seen in females

C. Are usually found in every cell of the next generation

If a mouse has a dominant phenotype (P-), how would you determine if it is homozygous (PP) or heterozygous (Pp)? A. Cross it to a homozygous dominant mouse. B. Cross it to a mouse with the dominant trait but a similarly unknown genotype C. Cross it to a mouse with the recessive trait D. Cross it to a heterozygous dominant mouse E. It cannot be determined.

C. Cross it to a mouse with the recessive trait. In order to identify an unknown genotype, one performs a testcross, which is a cross to a homozygous recessive individual. If there are many offspring, all with the dominant phenotype, the tested mouse is homozygous dominant. If some offspring have the recessive trait (half would be expected), the tested mouse is heterozygous.

Which of the following statements does not accurately complete this statement? An epigenetic trait A. Can be passed from mother cell to daughter cell by mitosis B. Can be passed to the next generation by meiosis C. Is a change in the DNA sequence D. Causes changes in gene expression E. All are accurate

C. Is a change in the DNA sequence. Epigenetic changes are heritable modifications of the DNA bases themselves or of the chromatin structure, which can affect gene expression.

The authors knew of two populations of rat mothers, each with different behaviors. Which of the following best describes high-nurturing rat mothers? A. Cannibalism B. Defensive responses to predators C. Licking and arched-back nursing D. Increased anxious behavior

C. Licking and arched-back nursing

While there are several mutations that give rise to cystic fibrosis, a common one results in a missing phenylalanine amino acid at position 508 of the CFTR protein. This is the result of _______. a. a frameshift b. a transition c. the deletion of a single codon in the gene d. Premature termination of translation e. no effect on protein structure, but on transcription

C. The deletion of a single codon in the gene Removal of three bases (a codon) results in a single amino acid being absent in the protein. Nothing before or after is altered. This mutation is commonly denoted deltaF508 (F stands for phenylalanine)

Figure 2 shows results of ChIP experiments. B-actin is a housekeeping gene and ER-a exon 1b promoter controls expression of the estrogen receptor. Why would the authors include these panels in their ChIP experiments? A. B-actin has increased expression in HN rats. B. ER-a exon 1b promoter was previously shown to be methylated in LN rats. C. They are controls because B-actin or the ER-a exon 1b promoter do not differ between the HN and LN rats D. The authors suspected B-actin and ER-a exon 1b to be changed in the LN and HN rats, too.

C. They are controls because B-actin or the ER-a exon 1b promoter do not differ between the HN and LN rats

What is one cause for the 21,000 protein encoding sequences in the human genome producing between 200,000 and 1,000,000 different proteins? A. Restriction enzymes B. SNP's C. alternative splicing D. CNV's E. CRISPR-Cas systems

C. alternative splicing

Sister chromatids separate in a. mitosis only b. mitosis and meiosis I c. Meiosis I only d. mitosis and meiosis II e. meiosis II only f. meiosis I and II

C. mitosis and meiosis II The sister chromatids are pulled apart and are then define as sperate chromosomes. This happens during anaphase II of meiosis II and during mitosis.

The discovery of RNA interference (RNAi) led to its use in biotechnology and medicine. Which of the following could not be done with RNA? A. to treat a disease characterized by the overexpression of a wild-type gene B. to treat a disease characterized by the buildup of a specific protein C. to treat a disease characterized by a null allele D. to study loss-of-function phenotypes of specific genes in a model research organism in a laboratory

C. to treat a disease characterized by a null allele

We describe chromosomes in an interphase nucleus as "organized spaghetti". A eukaryotic chromosome exists in its "___________", while the space in between chromosomes is the "______" A. enhancer, silencer B. euchromatin, heterochromatin C. Chromosome territory, interchromosomal domain D. Nucleolus, nuclear periphery

Chromosome territory, interchromosomal domain. C.

What is the probability that two parents who are heterozygous for the recessive trait of albinism will have two albino offspring? a. 1/2 b. 1/4 c. 1/8 d. 1/16 e. 1/32

D. 1/16. There is a 1/4 chance in each child, so 1/4 x 1/4 = 1/16.

Mrs. Phillips knows she is a carrier for the autosomal recessive disease hemochromatosis. Mr. Philips is affected with the disorder. The couple discovers that they are both carriers for cystic fibrosis, also an autosomal recessive disorder. What are the chances of them having a child with both disorders, assuming independent assortment of those genes? a. 0 b. 1/2 c. 3/4 d. 1/8

D. 1/8. Use the product rule. 1/2 chance of hemochromatosis and 1/4 chance of CF = 1/8 chance for both.

In a cross between two individuals BbGG x Bbgg, what ratio of phenotypes would be expected in the offspring if the two genes show independent assortment? a. 1:2:1 b. 9:3:3:1 c. 1:1 d. 3:1 e. 2:1

D. 3:1 You only need to consider the B gene since G isn't adding to variation in cross. 3/4 B- and 1/4 bb = 3:1

In order for the nonautonomous dissociation element (Ds) mutations in corn to remain in the same position A. the DNA must not be replicated. B. the Ds must contain the gene for transposition. C. an autonomous Ac element must be present. D. An autonomous Ac element must not be present E. None of these

D. An autonomous Ac element must not be present

Which of the following statements about bisulfite sequencing is false? A. Bisulfite cannot convert methylated cytosines to uracil. B. Sequencing is performed before and after bisulfite treatment. C. It is a method used to identify methylated cytosines. D. Following bisulfite treatment and sequencing, the sequence CGATG would be TGATG if the cytosine was methylated E. All of these statements are true.

D. Following bisulfite treatment and sequencing, the sequence CGATG would be TGATG if the cytosine was methylated

A cancer biologist is studying the epigenome of certain types of cancers. As a general rule, cancer results from cells that have lost the ability to control (stop) cell division due to mutations and epimutations. The scientists notice histone acetyltransferases (HATs) are upregulated causing genome-wide effects in many different types of cancers. Which of the following is the most likely to be observed as a consequence of HAT upregulation in cancer? A. Genes that inhibit cell division have increased expression. B. Genes that initiate cell division have decreased expression. C. The chromosome locus containing growth factor response genes is condensed. D. Genes that initiate cell division have increased expression. E. The chromosome locus containing the telomerase gene is condensed

D. Genes that initiate cell division have increased expression

Which gene did the authors predict was methylated in a nurture-dependent manner? A. NGFI-A B. H3K9 C. Estrogen receptor (ER) D. Glucocorticoid Receptor (GR) E. CpG residues

D. Glucocorticoid Receptor (GR)

A geneticist is studying a particular Drosophila mutant that produces a phenotype of "no wing" due to a mutation in a gene called wingless. She takes the mutant wingless and mutagenizes it. She hypothesizes that any fly that emerges with wings will be due to the original mutated gene reverting back to the wild-type allele. From this study, she identifies a fly that now has wings. When she sequences the wingless gene, she discovers that the original mutation is still there, and there are no other mutations in the wingless gene. However, a second gene at a separate locus is mutated. The second mutation is best classified as a(n) ________. A. conditional mutation B. haploinsufficiency mutation C. intragenic suppressor mutation D. intergenic suppressor mutation E. null mutation

D. Intergenic suppressor mutation

Crossing over occurs in a. mitosis only b. mitosis and meiosis I c. mitosis and meiosis II d. meiosis I only e. meiosis II only f. meiosis I and II

D. Meiosis 1 only Crossing over only happens during prophase I during meiosis I when nonsister chromatids of homologous chromosomes pair up. Crossing over is not part of mitosis.

Homologous chromosomes synapse during a. mitosis only b. mitosis and meiosis I c. mitosis and meiosis II d. meiosis I only e. meiosis II only f. meiosis I and II

D. Meiosis I only Homologous chromosomes pair and recombine during meiosis I. Sister chromatids sperate during meiosis II and mitosis.

Which of the following is explained by Mendel's law of independent assortment? a. The probability of getting allele wE or w+ in a gamete is just as likely if the parent is w+/wE b. A locus can have two variants in a diploid c. One allele can be dominant over another d. The traits of eye color (w+/wE) and body color (e+/eB) will be found in equal proportions in the offspring.

D. The traits of eye color (w+/wE) and body color (e+/eB) will be found in equal proportions in the offspring.

Which of the follwing about RNAi is true? a. RISC guides itself to mRNA target b. RISC uses gRNA to base pair with the shRNA c. RISC is not involved in the process d. gRNA guides RISC to the mRNA through complementary base pairing e. Exonucleases cause the miRNA to be expressed

D. gRNA guides RISC to the mRNA through complementary base pairing There are several pathways that utilize different gRNAs. In RNA interference, gRNA is processed miRNA or siRNA that are taken up by RISC. They complementary base pair with the target mRNA. If they have 100% complementarity, Argonaut will cut the mRNA. If they have some mismatched base pairs, Argonaut will not cut, but the mRNA will be sequestered in processing bodies (P-bodies) within the cytosol. Either way, the mRNA will not be used in translation.

Discuss which of the following is most likely to result in frameshift mutations? A. Alkylating agent, which changes the H bonding potentials within the base b. Tautomeric shift, which changes the base pairing capabilities of the base. c. Base analog, which resembles another base. d. Intercalating agent, which inserts in between the base pairs.

D. intercalating agent, which inserts itself between the base pairs. They wedge between base pairs in the double helix, causing extra space between the base pairs. This extra space can be filed in by random dNTPs by DNA polymerase during DNA replication or during DNA repair, resulting in insertion. The other choices generally resulting in mispairing, which can lead to base pair substitution mutations.

Which of the following can trigger the RNAi pathway? Choose ALL correct. a. tRNA b. rRNA c. snRNA d. siRNA e. mRNA F. miRNA

D. siRNA F. miRNA Although from different sources, siRNA and miRNA take on a specific hairpin length, which triggers the RNAi pathway. These RNAs are processed by Dicer and recognized by the RNA-inducing silencing complex (RISC). The processed RNAs (guide RNAs; gRNA_ then complementary base pair with the target mRNA. Argonaut, a subunit of RISC, is an endonuclease that cuts the target mRNA, causing it to be degraded. If there isn't a 100% complementarity between the gRNA and mRNA target, translation will be inhibited. Either way. gene expression is repressed.

The major mechanisms of epigenetics

DNA methylation, Histone acetylation, and RNAi.

Which of the following nucleotide changes is a transition mutation? A. Guanine to thymine B. Adenine to cytosine C. Guanine to cytosine D. Thymine to guanine E. Adenine to guanine

E. Adenine to guanine

Of the following, which is an example of epigenetic changes? A. methylation of DNA bases B. addition of acetyl groups to histones C. methylation of CpG islands D. imprinting of genes E. All are examples of epigenetic changes

E. All are examples of epigenetic changes. The major mechanisms of epigenetics include DNA methylation, histone acetylation, and RNAi. Imprinting involves DNA methylation and histone acetylation, so it is epigenetic mechanism.

Spontaneous mutation rates are greatly reduced by A. exposure to ionizing radiation. B. induced mutations C. base-modifying agents. D. the cell no longer needing a mutation E. DNA polymerase's 3'-5' exonuclease activity

E. DNA polymerase's 3'-5; exonuclease activity

Which of the following is an example of genomic imprinting in humans? A. One of the two X-chromosomes in females is randomly expressed and the other is repressed. B. A random pattern of autosomal allele inactivation is observed. C. The maternal and paternal alleles of a gene pair are both expressed. D. Human males have only one Y-chromosome and one X-chromosome. E. For some genes, only the paternal source allele is expressed, and for other genes, only the maternal sourced allele is expressed.

E. For some genes, only the paternal sourced allele is expressed, and for other genes, only the maternal sourced allele is expressed.

Which of the following is a potential effect of transposons in the genome? A. Gene repression by insertion into the promoter region B. Altering splice sites of the genes C. Deletion of genetic material due to sloppy transposition D. Creation of a null allele E. All of the above

E. all of the above

Which of the following is the target of the RNAi pathway? a. tRNA b. rRNA c. snRNA d. shRNA e. mRNA

E. mRNA mRNAs are targeted by miRNAs. When gRNAs are taken up by RISC, they base pair with the target mRNA, allowing Argonaut to cut the mRNA. The mRNA is then degraded by devoted exonucleases.

T/F. Mutations occur more when cell needs them

FALSE In nature, mutations occur regardless of selective pressure. They are random, not adaptive. Most are neutral. However, they can be selected for or against depending upon the advantage or disadvantage they confer in that environment. Mutations results when damaged DNA is replicated. The damage could be spontaneous or induced.

After S-phase, how many copies of each gene would you have in a diploid organism?

Four copies of each gene - there are four chromatids total for each chromosome

In the absence of mutation, are homologous chromosomes identical?

No, homologous chromosomes are passed from either parent, one from the paternal source and one from the maternal source.

Three traits in pea plants assort independently. Smooth (S) is dominant to wrinkled (s), yellow (Y) is dominant to green peas (y), and purple flowers (C) is dominant to white flowers (c). What are the expected PHENOTYPIC ratios of the offspring from the following cross? SsYyCc x ssYycc

SYC = 3/16 SYc = 3/16 SyC= 1/16 Syc = 1/16 sYC = 3/16 sYc = 3/16 syC = 1/16 syc = 1/16

What type of DNA sequence would a repressor bind to? A. TATA box B. silencer C. Insulator D. Enhancer E. GC box

Silencer, E.

Which of the following is true about the second meiotic division? Nondisjunction would lead to extra bivalents forming Synapsis occurring in the second meiotic division Tetrads pulling apart The products are 4 identical gametes. Sister chromatids are pulling apart

Sister chromatids are pulling apart

Mutation hot spots are locations in the genome that are more susceptible to mutation than other regions of the genome. For example, these can be areas of the genome where replication slippage is more likely to occur, such as in tandem repeats, or in euchromatin since it is more exposed to mutagens due to decondensation. In addition, guanine bases are prone to oxidation (loss of electrons). Discuss the validity of this statement: Mutation hot spots are consistent with the idea that mutations are random.

Statement is valid. Although location may not be random, mutations still not caused by "need" of a cell, meaning they are random in that they are not adaptive.

Why are you sometimes asked to wear a lead apron across your abdomen when the dentist takes x-rays of your mouth?

The apron blocks X-rays, which can induce DNA damage in the cells or your internals organs and/or germ line cells. If the damage is too much for the DNA repair pathways to repair, programmed cell death (apoptosis) will be triggered. If the damage is converted to mutation, cancer could result. If the damage occurs in your germ cells, you can pass the mutation to the next generation.

In a mutant strain of yeast, a tRNA that recognizes the codon 5'-CUG-3' has been altered so that it now recognizes the codon 5'-GUG-3'. Name a protein that when mutated could cause suppression of the tRNA mutant. Is the suppression in above q intragenic or intergenic?

The tRNA synthetase could also be mutated since the synthetase charges the tRNA. This is an intergeneic suppressor because a second (different) gene is mutated.

The pre-initiation complex (PIC) contains several proteins. What would be the direct consequence if the PIC failed to form? A. Replication would not be initiated B. Txn would not be initiated C. mRNA splicing would not be initiated. D. protein would not fold properly E. translation would not be initiated.

Transcription would not be initiated, B.

Consider two theoretical yeast transposable elements, A and B. Each contains an intron, and each transposes to a new location in the yeast genome. Suppose you then examine the transposable elements for the presence of the intron. In the new locations, you find that A has no introns, while B does. From these observation, what can you conclude about the mechanisms of transposition for A and B?

Transposable element A must have an RNA intermediate. Most likely, it is transcribed into RNA, the intron is spliced out, the RNA is reverse transcribed onto cDNA before inserting into the new site. B does not follow this mechanism. It does not have an RNA intermediate.

In a diploid individual who is heterozygous at a locus, how many copies of each allele would you fins following S phase?

Two copies of each allele - each homologous chromosome consists of two sister chromatids. Since the individual is heterozygous, the have two different alleles at that locus.

Mendel crossed two pea plants with round seeds. All seeds of the offspring were round. He then crossed a plant with round seeds to a plant with wrinkled seeds and all offspring had wrinkled seeds. Which of the following is true? wrinkled is dominant a mutation occurred the trait does not breed true round is dominant the plants he used were not true breeding

Wrinkled is dominant

A certain type of congenital deafness in humans is caused by a rare autosomal dominant gene. In a mating involving a deaf man and a deaf woman, could all the children have wild-type hearing? Yes, if it is assumed the parents are heterozygotes, it is possible that all of the children could have wild-type hearing. Yes, because traits assort independently. No, because it is dominant. Children always get the dominant alleles. Yes, because it must be recessive if it's rare. No, because children favor their parents.

Yes, if it is assumed the parents are heterozygotes, it is possible that all of the children could have wild-type hearing.

In the absence of mutation, are sister chromatids identical?

Yes, sister chromatids of a homologous chromosome result from DNA replication. As long as there are no mistakes during replication, they should have the same sequence.