Genetics Quizes

Color blindness is caused by an X-linked recessive allele that is rare in humans. Suppose a man and a woman, both of whom have normal color vision and color-blind fathers, have a child. What is the chance that their first son will be color blind? A) 1/2 B) 2/3 C) 1/4 D) 0

A) 1/2

If you roll two dice, what is the probability that you will roll an even number on one and an odd number on the other? A) 1/2 B) 1/4 C) 1/3 D) 1/6

A) 1/2

Your sister is pregnant with triplets, each of which was produced by a different fertilization event. What is the chance that two of her babies are male and one is female? A) 1/4 B) 1/6 C) 5/8 D) 2/3 E) 3/4 F) 3/8

F) 3/8

This question is about a possible biochemical explanation for a variation of Mendel's 9:3:3:1 ratio. In the diagram that follows, compounds 1, 2, 3, and 4 have different colors, as do mixtures of these compounds. A and B are enzymes that catalyze the indicated steps of the pathway. Alleles A and B specify functional enzymes A and B, respectively; these are completely dominant to alleles a and b, which do not specify any of the corresponding enzyme. If functional enzyme is present, assume that the compound to the left of the arrow is converted completely to the compound to the right of the arrow. What phenotypic ratio would you expect among the progeny of a dihybrid cross of the form Aa Bb x Aa Bb? Enzymes A & B both needed to catalyze reaction indicated: Compound 1 --(Enz A + Enz B)--> Compound 2 A) 15:1 B) 9:3:4 C) 12:4 D) 12:3:1 E) 9:3:3:1 F) 9:7

F) 9:7

Three short sequence reads from a genomic library made by partial digestion with Sau 3AI and ligation into the Bam HI site of plasmid vectors are shown below. Do all three of the following 20 nt sequence reads generate a single contig? If so, how long is that contig? (For the purpose of this question, 8 nt of sequence overlap is sufficient for any two reads to form a contig.) read 1: 5' AAAGTCAACCTGTATTGGCC 3' read 2: 5' GCGCGCTATAGGCCAATACA 3' read 3: 5' TATAGCGCGCAATTTTTGGG 3' A) Yes; the contig is 35 bp long. B) No C) Yes; the contig is 20 bp long. D) Yes; the contig is 60 bp long. E) Yes; the contig is 46 bp long. F) Yes; the contig is 38 bp long. G) Yes; the contig is 40 bp long.

G) Yes; the contig is 40 bp long.

The G and H genes are linked and 10 map units apart. In the cross G h / g H × g h / g h what fraction of the progeny will be G H / g h ? A) 45% B) 5% C) 10% D) 40%

B) 5%

Following conjugation of an Hfr cell with an F− cell, A) the recipient becomes an Hfr. B) the recipient becomes F+. C) the donor becomes F−. D) both the donor and the recipient become F−. E) the recipient remains F−.

E) the recipient remains F−.

A male fruit fly has three pairs of autosomes as well as an X and a Y chromosome. None of the chromosomes undergoes crossing-over. What fraction of an individual male's gametes will contain only the chromosomes that it inherited from its mother? A) 1/2 B) 1/16 C) 1/4 D) 1/6

B) 1/16

In Drosophila, the recessive allele for forked bristles is sex-linked, and the recessive allele for purple eyes is autosomal. The wild-type alleles are for straight bristles and red eyes, respectively. A female, heterozygous for both pairs of alleles, is mated with a male with forked bristles and purple eyes. What percentage of the daughters will have forked bristles and purple eyes? A) 50% B) 25% C) 75% D) 100%

B) 25%

Several genes in humans in addition to the ABO gene (I ) give rise to recognizable antigens on the surface of red blood cells. The MNand Rh genes are two examples. The Rh locus can contain either a positive or a negative allele, with positive being dominant to negative. M and N are codominant alleles of the MNgene. Suppose a child's blood type is B MN Rh(neg). The mother's blood type is O N Rh(neg). Check all of the blood types below that the father could have. (Assume no involvement of the Bombay phenotype; everyone is HH.) - O M Rh(neg) - B MN Rh(pos) - A M Rh(pos) - O MN Rh(pos) - B MM Rh(pos) - B NN Rh(neg) - B MN Rh(neg)

- O M Rh(neg) - A M Rh(pos) - O MN Rh(pos) - B NN Rh(neg) - B MN Rh(neg)

Which of the following are steps in the generation of knockout mice using ES cells? - Introduce chimeric host blastocyst into a surrogate mother mouse. -Mix modified ES cells with ES cells containing a Cre gene. Mix modified ES cells with ES cells containing a Cre gene. -Cross chimeric mouse to a wild-type mouse to obtain a heterozygous knockout mouse. -Make a construct with genomic DNA corresponding to the ends of the gene to be knocked out on either end of a drug resistance marker gene. -Introduce heterozygous knockout ES cells to a host blastocyst. -Replace the viral genes with a therapeutic gene. -Introduce a gene-targeting construct into host blastocysts. -Introduce a gene-targeting construct to ES cells growing in a petri dish. -Add LTRs to the ends of a cDNA for the gene to be knocked out. -Apply selection for drug resistance marker. -Cross two heterozygous knockout mice to each other to generate a homozygous knockout. -Introduce a gene-targeting construct into packaging cells.

-Mix modified ES cells with ES cells containing a Cre gene. Mix modified ES cells with ES cells containing a Cre gene. -Cross chimeric mouse to a wild-type mouse to obtain a heterozygous knockout mouse. -Make a construct with genomic DNA corresponding to the ends of the gene to be knocked out on either end of a drug resistance marker gene. -Introduce heterozygous knockout ES cells to a host blastocyst. -Replace the viral genes with a therapeutic gene. -Introduce a gene-targeting construct into host blastocysts. -Introduce a gene-targeting construct to ES cells growing in a petri dish. -Add LTRs to the ends of a cDNA for the gene to be knocked out. -Apply selection for drug resistance marker. -Cross two heterozygous knockout mice to each other to generate a homozygous knockout. -Introduce a gene-targeting construct into packaging cells.

Put the following events in the order in which you do them when performing P element transformation. - Cross injected flies individually with w/wflies of the opposite sex. - Clone a transgene into a recombinant Pelement vector containing a white⁺ marker gene. - Inject w/w embryos with a mixture of two plasmids: a recombinant P element plasmid containing both your transgene and a white⁺ marker gene, and a helper plasmid containing the P element transposase gene. - Screen progeny of injected flies for red eyes. - Cross red-eyed siblings to each other to generate transgenic lines, each with the recombinant P element in a different genomic location.

1) Clone a transgene into a recombinant Pelement vector containing a white⁺ marker gene. 2) Inject w/w embryos with a mixture of two plasmids: a recombinant P element plasmid containing both your transgene and a white⁺ marker gene, and a helper plasmid containing the P element transposase gene. 3) Cross injected flies individually with w/wflies of the opposite sex. 4) Screen progeny of injected flies for red eyes. 5) Cross red-eyed siblings to each other to generate transgenic lines, each with the recombinant P element in a different genomic location.

Order the steps in the production of a transgenic mouse from first (1) to last (5). -Zygotes are harvested from a recently mated female mouse. -Linear copies of the transgene DNA are injected into a pronucleus of a fertilized egg. - Cells from mouse tails are tested for transgenic DNA. -Tandem copies of DNA integrate into a random location on a chromosome. - Mice that were injected as embryos are born from a host female.

1) Zygotes are harvested from a recently mated female mouse. 2) Linear copies of the transgene DNA are injected into a pronucleus of a fertilized egg. 3) Tandem copies of DNA integrate into a random location on a chromosome. 4) Mice that were injected as embryos are born from a host female. 5) Cells from mouse tails are tested for transgenic DNA.

Snails are either hermaphrodites (they make both eggs and sperm) or males. Hermaphrodites can self-fertilize, or they can mate with males. Dextrality (rightward shell coiling) and sinistrality (leftward shell coiling) are maternal effect phenotypes in snails. An autosomal gene controls these phenotypes: s+ (dextral) is dominant to s- (sinistral). A hermaphrodite with a sinistral phenotype self-fertilizes. What are the possible phenotypes of its progeny? (Check all that apply.) 1) All dextral. 2) All sinistral. 3) Some sinistral and some dextral. 4) She is sterile; no progeny survive.

2) All sinistral. 3) Some sinistral and some dextral.

The CRISPR/Cas9 system can cleave genomic DNA at sequences other than the desired target, a phenomenon referred to as off-target cleavage. To avoid off-target cleavage, sites targeted for mutation should appear only once in the genome. What is the minimum length of the part of an sgRNA molecule that binds DNA whose base pair sequence would be expected to be unique within the fruit fly genome? (Assume that the size of the fruit fly genome is 1.4 x 108 bp, that the base pair sequence is random, and that AT = GC.) A) 14 B) 16 C) 109 D) 15 E) 234

A) 14

Drosophila melanogaster has four pairs of chromosomes (including a pair of sex chromosomes). Sperm from this species are formed by a meiotic process in which homologous chromosomes pair and segregate but do not undergo crossing-over. How many genetically different kinds of sperm could be produced by a Drosophila melanogaster male (XY)? A) 16 B) 256 C) 64 D) 8 E) 4

A) 16

Three genes in fruit flies affect a particular trait, and one dominant allele of each gene is necessary to get a wild-type phenotype. What phenotypic ratios would you predict among the progeny if you crossed triply heterozygous flies? A) 27:37 B) 3:5 C) 27:64 D) 1:3 E) 1:7

A) 27:37

Wild-type grapes are green. Two pure-breeding parental strains of grape plants are crossed: green and yellow. The F1 have all green grapes, and the F2 plants are present in a 9 green : 7 yellow ratio. What are the genotypes of the yellow F2? A) A- bb, aa B-, and aa bb B) AA bb, aa BB, and aa bb only C) all Aa Bb D) A- bb, and aa B- only E) A- bb and aa bb only F) all aa bb

A) A- bb, aa B-, and aa bb

Which of the following is a step in transcription that requires complementary base pairing? A) Adding the correct ribonucleotide to the growing RNA chain. B) Adding the correct deoxyribonucleotide to the growing DNA chain. C) Adding the correct amino acid to the growing polypeptide chain. D) Binding of RNA polymerase to the promoter. E) Binding of the ribosome to the mRNA.

A) Adding the correct ribonucleotide to the growing RNA chain.

What is the difference between the function of DNA and the function of proteins? A) DNA stores genetic information and proteins perform most cellular functions. B) DNA stores genetic information in the order of nucleotides and proteins store genetic information in the order of amino acids. C) DNA performs most cellular functions and proteins store information. D) DNA provides structure to the cell and proteins act as enzymes.

A) DNA stores genetic information and proteins perform most cellular functions.

Thirteen different SSR loci are used by the FBI to perform DNA fingerprinting. If SNP loci were assayed instead to identify individuals, which of the following would be true? Check all that apply. A) Fewer than 13 SNP loci would need to be assayed. B) Many more than 13 SNP loci would need to be assayed. C) All SNP loci could be genotyped, just as all SSRs are, by electrophoresis of a PCR product. D) SNP loci, because they are usually biallelic, cannot be used to identify individuals. E) Using SNPs instead of SSRs, only carriers of disease alleles could be identified. F) Using SNPs instead of SSRs, only non-anonymous SNP loci could be assayed.

A) Fewer than 13 SNP loci would need to be assayed. B) Many more than 13 SNP loci would need to be assayed. E) Using SNPs instead of SSRs, only carriers of disease alleles could be identified. F) Using SNPs instead of SSRs, only non-anonymous SNP loci could be assayed.

The tryptophan biosynthesis pathway is shown below. Suppose you want to identify bacterial colonies that have trpA- mutations. Which of the following procedures will enable you to do so? A) Grow prototrophs in liquid minimal medium with a tryptophan supplement and a mutagen. Spread some bacteria on plates containing minimal medium + tryptophan and then screen individual colonies for their ability to grow on minimal medium; those colonies that cannot are Trp- auxotrophs. The trpA- mutants among the colonies can be identified in supplementation experiments as colonies that can grow only with a tryptophan supplement in the minimal medium - no intermediate in the pathway will support growth. All of the other Trp- auxotrophs will be able to grow on minimal medium supplemented with at least one of the pathway intermediates. B) Grow prototrophs in liquid minimal medium with a tryptophan supplement and a mutagen. Spread some bacteria on plates containing minimal medium + tryptophan and then screen individual colonies for their ability to grow on minimal medium; those that cannot are Trp- auxotrophs. The trpA- mutants among the colonies can be identified in supplementation experiments as colonies that can grow in minimal medium supplemented with any one of the intermediates in the tryptophan biosynthesis pathway. All of the other Trp- auxotrophs will be able to grow on minimal medium supplemented with only one of the intermediate compounds. C) Grow auxotrophs in liquid minimal medium with a tryptophan supplement and a mutagen. Spread some bacteria on plates containing minimal medium to select for individual colonies that are Trp+ revertants. All of the colonies that did not revert are trpA- mutants.

A) Grow prototrophs in liquid minimal medium with a tryptophan supplement and a mutagen. Spread some bacteria on plates containing minimal medium + tryptophan and then screen individual colonies for their ability to grow on minimal medium; those colonies that cannot are Trp- auxotrophs. The trpA- mutants among the colonies can be identified in supplementation experiments as colonies that can grow only with a tryptophan supplement in the minimal medium - no intermediate in the pathway will support growth. All of the other Trp- auxotrophs will be able to grow on minimal medium supplemented with at least one of the pathway intermediates

Human mitochondrial diseases usually display which of the following? (Check all that apply.) A) Maternal Inheritance B) Paternal inheritance C) Biparental Inheritance D) Variable expressivity E) Autosomal dominant inheritance F) Autosomal recessive inheritance G) X-linked inheritance

A) Maternal Inheritance G) X-linked inheritance

Which statement about the structure of proteins is false? A) Most proteins consist of two amino acid polymers attached by hydrogen bonding. B) The chemical properties of each amino acid are determined by the amino acid's side chain. C) Proteins are made of long strings 20 different common amino acids in different orders. D) The order of amino acids in a polypeptide determines the three-dimensional shape of the folded molecule.

A) Most proteins consist of two amino acid polymers attached by hydrogen bonding.

A couple have a child with Turner syndrome (XO). How could the aneuploid gamete that caused this condition have been generated? (Check all that apply.) A) Nondisjunction in male meiosis I B) Nondisjunction in male meiosis II C) Nondisjunction in female meiosis I D) Nondisjunction in female meiosis II

A) Nondisjunction in male meiosis I C) Nondisjunction in female meiosis I

Polyploid plants with an odd number of chromosome sets are usually sterile, while polyploid plants with an even number of chromosome sets are fertile. Why? A) Odd numbers of homologous chromosomes usually segregate in a manner that produces unbalanced gametes, while even numbers of homologous chromosomes segregate to produce balanced gametes. B) Polyploid cells with odd ploidies are aneuploid (unbalanced); polyploid cells with even ploidies are euploid (balanced). C) Crossovers between odd numbers of homologs result in chromosomal rearrangements and aneuploid gametes; crossovers involving even numbers of homologs result in normal chromosomes and euploid gametes. D) Odd polyploidies cannot be tolerated by germ-line cells and they die; even polyploidies make germ-line cells larger and more viable.

A) Odd numbers of homologous chromosomes usually segregate in a manner that produces unbalanced gametes, while even numbers of homologous chromosomes segregate to produce balanced gametes.

Which of the following statements is true of trpA− auxotrophs? A) Of all Trp− auxotrophs, only trpA− mutants cannot grow on minimal medium supplemented with any intermediate in the tryptophan biosynthesis pathway; trpA− auxotrophs require a tryptophan supplement. B) Of all Trp− auxotrophs, only trpA− mutants can grow on minimal medium supplemented with any intermediate in the tryptophan biosynthesis pathway. C) Trp− auxotrophs that are trpA− mutants can grow on minimal medium supplemented only with tryptophan or indole. D)Trp− auxotrophs that are trpA− mutants can grow on minimal medium supplemented only with tryptophan, indole, or indole-3-glycerinphosphate.

A) Of all Trp− auxotrophs, only trpA− mutants cannot grow on minimal medium supplemented with any intermediate in the tryptophan biosynthesis pathway; trpA− auxotrophs require a tryptophan supplement.

Genes A and B are so close together on the same chromosome that crossing-over occurs between Aand B in only 1 in a million meioses. An AA BB individual mates with an aa bb individual to produce an F1 that is A B / a b. If 100 progeny are observed, what genotypic ratio is expected in the gametes produced by the F1? A) Only parental 1 A B : 1 a b B) More parental A B and a b, and fewer recombinant A b and a B C) 1 A B : 1 A b : 1 a B : 1 a b D) More parental A b and a B, and fewer recombinant A B and a b

A) Only parental 1 A B : 1 a b

Which of the following statements regarding CpG islands are true? A) The C bases in CpG islands are always methylated. B) Methylation of the C bases in a CpG island silences the promoter of the gene. C) The C bases in CpG islands are never methylated. D) Genes with CpG islands do not have enhancers or promoters. E) Only imprinted genes have CpG islands.

A) The C bases in CpG islands are always methylated. B) Methylation of the C bases in a CpG island silences the promoter of the gene. D) Genes with CpG islands do not have enhancers or promoters. E) Only imprinted genes have CpG islands.

Which of the following is a structural feature shared by DNA, RNA, and proteins? A) The order of monomers is important for function. B) Two strands of monomers are connected by hydrogen bonding. C) Each sequence of monomers folds into a different three-dimnesional shape. D) Each molecule is a polymer of four different nucleotides.

A) The order of monomers is important for function.

Check all of the statements that describe accurately the result of having a splice acceptor or donor site mutation in a gene. A) The primary transcript will be longer than normal. B) The primary transcript will be shorter than normal. C) The mRNA may be longer than normal. D) The protein encoded by the mutant gene could have some incorrect amino acids. E) The protein encoded by the mutant gene could have more than the normal number of amino acids. F) The protein encoded by the mutant gene could have fewer than the normal number of amino acids. G) The mutant allele could be amorphic. H) The mutant allele could be neomorphic.

A) The primary transcript will be longer than normal. G) The mutant allele could be amorphic. H) The mutant allele could be neomorphic.

Which is true of enhancer DNA sequences? A) They contain TATA boxes. B) They can retain function if their nucleotide sequence is moved or inverted. C) They can increase gene transcription levels from the basal level. D) They can be more than 10 kilobases from the genes they regulate. E) They may bind to more than one transcription factor at the same time.

A) They contain TATA boxes. C) They can increase gene transcription levels from the basal level. D) They can be more than 10 kilobases from the genes they regulate. E) They may bind to more than one transcription factor at the same time.

Which are mechanisms by which bacterial species become drug resistant? A) Transformation by naturally-occurring plasmids that contain drug-resistance genes. B) Loss-of-function mutations in genes that encode efflux pump proteins. C) Hypermorphic mutations in genes that encode efflux pump proteins. D) Loss-of-function mutations in genes that encode pore proteins; pores allow fluid and small molecules to enter the cell. E) Hypermorphic mutations in genes that encode pore proteins resulting; pores allow fluid and small molecules to enter the cell. F) Mutations in genes that encode the targets of drug action that result in the target being more sensitive to the drug. G) Mutations in genes that encode the targets of drug action that result in the target being less sensitive to the drug.

A) Transformation by naturally-occurring plasmids that contain drug-resistance genes. B) Loss-of-function mutations in genes that encode efflux pump proteins. C) Hypermorphic mutations in genes that encode efflux pump proteins. E) Hypermorphic mutations in genes that encode pore proteins resulting; pores allow fluid and small molecules to enter the cell. F) Mutations in genes that encode the targets of drug action that result in the target being more sensitive to the drug. G) Mutations in genes that encode the targets of drug action that result in the target being less sensitive to the drug.

To create transgenic mice DNA is injected into... A) a pronucleus of a zygote. B) a pronucleus of one cell from an embryo. C) the cytoplasm of a zygote. D) the nucleus after the pronuclei have fused. E) a host blastocyst.

A) a pronucleus of a zygote.

If integration of DNA provided by pronuclear injection occurs after three cell divisions, the resulting mouse will... A) be a mosaic of cells with some cells containing the transgene and others not. B) contain the transgene in only somatic cells. C) will die D) not contain the transgene in any cells. E) contain the transgene in only germ cells.

A) be a mosaic of cells with some cells containing the transgene and others not.

An Hfr strain (Hfr a+ b+ c+ d+ e+ drugS ) was crossed with an F− strain (F− a− b− c− d− e− drugR ). After an hour, the closest gene to the F factor, a+, was selected for. Of the a+ cells, 30% are b+, 90% are c+, 60% are d+, 0% are e+, and 70% are f+. What can you conclude about the gene order? A) dbeacf B) dbeafc C) aebcfd D) acbfde E) acfbed

A) dbeacf

What types of gene mutations can result in proteins that are larger (longer polypeptide chains) than the protein produced by the normal allele? (check all that apply.) A) frameshift mutation B) silent mutation C) missense mutation D) substitution mutation E) deletion of 2 nt F) addition of 2 nt G) addition of 1 nt H) deletion of 1 nt I) splice site mutations

A) frameshift mutation B) silent mutation C) missense mutation F) addition of 2 nt G) addition of 1 nt

Two DNA fragments are separated by gel electrophoresis. Check all of the statements that are true of the DNA fragment that traveled less far in the gel. A) it's shorter than the other one B) it's longer than the other one C) its more densely negatively charged than the other one D) It is less densely negatively charged than the other one. E) It must have blunt ends. F) It must have sticky ends. G) It must have more GC base pairs than the other one. H) It must have more AT base pairs than the other one.

A) it's shorter than the other one B) it's longer than the other one C) its more densely negatively charged than the other one D) It is less densely negatively charged than the other one. E) It must have blunt ends. G) It must have more GC base pairs than the other one. H) It must have more AT base pairs than the other one.

The disease Leber congenital amaurosis (LCA), caused by loss-of-function mutation of a gene called RPE65, has been treated with gene therapy by injecting recombinant AAV vectors containing a normal copy of the RPE65 gene into their retinal epithelial cells. Patients receiving this therapy... A) may need to repeat the treatment as the viral DNA is degraded. B) will pass the mutation to their offspring. C) may acquire detrimental mutations due to integration of the virus. D) now have the normal RPE65 gene in all cells.

A) may need to repeat the treatment as the viral DNA is degraded.

The different tRNAs are produced by ... A) transcription of different tRNA genes in the genome B) different aminoacyl-tRNA synthetases that produce tRNAs from ribonucleotides in the cytoplasm C) alternatively splicing the primary transcript of a single tRNA gene D) translation of different tRNA genes in the genome E) RNA editing of transcripts of a single tRNA gene

A) transcription of different tRNA genes in the genome

Check all of the statements that are true about microRNAs (miRNAs). A) The human genome contains genes that produce miRNAs. B) miRNAs are only synthetic; they are produced when scientists introduce transgenes into organisms. These so-called RNA-interference transgenes have promoters at each end of a cDNA, and thus produce double-stranded RNA, which is then processed into miRNAs. C) The targets of miRNAs are mRNAs of a specific genes. D) A single miRNA targets all mRNAs. E) Ribosomes are the targets of miRNAs; miRNAs bind to ribosomes to prevent translation. F) miRNAs prevent translation of their target genes by binding to their mRNAs, which can result in mRNA degradation by the RISC complex. G) miRNAs bind to their target RNAs by complementary base pairing and hide the ribosome binding site, thus preventing translation. H) Some mRNAs produce both proteins and miRNAs; other mRNAs produce miRNAs only.

All of them

FISH analysis is likely to detect which type of change in DNA? A) Translocations B) Duplications C) Deletions D) Silent point mutations E) Frameshift point mutations

All of them

In the diagram of the human Igfr2 gene shown below, how might the ncRNA called Air control expression of the Igfr2 gene? (Choose all statements that are possible.) A) Air transcription prevents Igfr2 transcription. B) Air translation prevents Igfr2 translation. C) Methylation of the CpG island represses transcription of both Igfr2 and Air. D) The Air transcript binds to the Igfr2transcript through complementary base pairing. E) Transcription of Air encourages transcription of Igfr2 by attracting transcription factors and RNA polymerase to the region of both genes.

All of them

Interactions between an enhancer and a repressor initiate a series of events that result in a change in gene transcription. In what order can the events occur? (Some steps may not be used. Put the remaining steps in chronological order from first to last.) 1. Transcription of the gene cannot occur.2. A corepressor is recruited.3. The contacts between histones and DNA loosen and the histones eventually leave the promoter DNA.4. Histone tails may be methylated or deacetylated.5. The contacts between histones and DNA are stabilized, and the chromatin around the promoter becomes more compact.6. The repressor protein contacts Mediator protein or TBP. A) 2,6,4,5,1 B) 2,4,5,1 C) 6,4,3,1 D) 4,5,1,2

B) 2,4,5,1

A couple went to see a genetic counselor because each has a sibling with cystic fibrosis. Cystic fibrosis is caused by homozygosity for a recessive disease allele, and neither member of the couple nor any of their four parents is affected. What is the probability that their child will be a carrier of the disease allele? A) 1/4 B) 4/9 C) 1/18 D) 1/6 E) 1/8 F) 1/9

B) 4/9

A mutation creates a dominant negative allele of a particular gene. The normal allele of the gene encodes a protein that forms a trimer within the cell. If one or more of the subunits has the mutant structure, the entire trimeric protein is inactive. In a heterozygous cell, if the proteins of both alleles are present at the same levels, what percent of the trimers present in the cell will be inactive? A) 50% B) 87.5% C) 100% D) 12.5% E) 33%

B) 87.5%

Snails are either hermaphrodites (they make both eggs and sperm) or males. Hermaphrodites can self-fertilize, or they can mate with males. Dextrality (rightward shell coiling) and sinistrality (leftward shell coiling) are maternal effect phenotypes in snails. An autosomal gene controls these phenotypes: s+ (dextral) is dominant to s- (sinistral). A hermaphrodite with a sinistral phenotype self-fertilizes. What are the possible phenotypes of its progeny? (Check all that apply.) A) All dextral B) All sinistral C) Some sinistral and some dextral D) She is sterile; no progeny survive

B) All sinistral D) She is sterile; no progeny survive

In which of the following scenarios will Allele 1, a normally functional allele of a gene, be recessive to Allele 2 in a heterozygote? A) Allele 2 is non-functional and for this particular gene, two normally functional alleles are required in order to avoid having a mutant phenotype. B) Allele 2 is non-functional and for this particular gene, as for most genes in diploid organisms, only one normally functional allele is required in order to avoid having a mutant phenotype. C) Allele 2 is abnormal; it makes an abnormal gene product that causes a mutant phenotype even though one dose of the normal gene product made by Allele 1 is present.

B) Allele 2 is non-functional and for this particular gene, as for most genes in diploid organisms, only one normally functional allele is required in order to avoid having a mutant phenotype. C) Allele 2 is abnormal; it makes an abnormal gene product that causes a mutant phenotype even though one dose of the normal gene product made by Allele 1 is present.

MERFF is caused by heteroplasmy for a loss-of-function mutation in a mitochondrial gene. A set of identical twins was reported in Australia where one twin had symptoms of MERFF, and the other did not. Choose all of the statements that could explain this phenomenon. A) One twin had a new mutation in a mtDNA early in development. B) Each twin originally inherited different fractions of mutant mtDNAs. C) The overall fraction of mutant mtDNAs in the twins is the same, but through genetic drift, one twin has a higher fraction in the cell types that matter than the other one does. D) One twin inherited its mtDNA from its father, and the other twin inherited its mtDNA from its mother.

B) Each twin originally inherited different fractions of mutant mtDNAs. C) The overall fraction of mutant mtDNAs in the twins is the same, but through genetic drift, one twin has a higher fraction in the cell types that matter than the other one does. D) One twin inherited its mtDNA from its father, and the other twin inherited its mtDNA from its mother.

Which of the following probabilities is correct regarding the progeny of a mating of an Ss RR individual to an individual who is Ss Rr ? (A - indicates the the second allele is either dominant or recessive.) Alleles of the two genes assort independently. A) Heterozygous both alleles: 50% B) S - RR : 37.5% C) Homozygous recessive: 10% D) ss R - : 15.5%

B) S - RR : 37.5%

A researcher is studying the rII locus of phage T4. Recall that wild-type phage can make plaques on E. coli K, but that single rII− mutant phage cannot. Three rII− strains are obtained: A, B, and C. E. coli strain K is co-infected with two rII− strains simultaneously and the results are recorded: Infection with A and B phage = no plaques formInfection with A and C phage = no plaques formInfection with B and C phage = no plaques form What can be inferred from these results? A) Strains A, B, and C have mutations in three different genes. B) Strains A, B, and C have mutations in the same gene. C) Strains A and B have mutations in the same gene, and strain C has a mutation in a different gene. D) Strains B and C have mutations in the same gene, and strain A has a mutation in a different gene. E) Strains A and C have mutations in the same gene, and strain B has a mutation in a different gene.

B) Strains A, B, and C have mutations in the same gene.

You have a 1500 bp long cDNA in a plasmid vector. You want to sequence the cDNA insert. (You know the sequence of the vector.) The sequence reads you can obtain are only 1000 bp long. How can you get the sequence of the entire cDNA insert? (Check all approaches that will work.) A) Synthesize sequencing primers that will hybridize adjacent to either end of the cDNA insert on different single-stranded circles. In two separate sequencing reactions, determine 1000 bp of sequence from each end of the insert. When turned into double strands virtually, the two sequences will overlap to form a 1500 bp contig. B) Synthesize a sequencing primer that will hybridize adjacent to the insert and use it to obtain 1000 bp of sequence. Use that new sequence information to generate a second primer within the cDNA that will provide a read of the remainder of the 1500 bp sequence. C) Synthesize two primers, one adjacent to each end of the insert. Add them both the same sequencing reaction in order to determine the sequence of the entire insert - 1000 bp from each side. D) Sequence 1000 bp from one end of the insert using a primer adjacent to one side. Compare that sequence to the RefSeq for the organism to find the remainder of the cDNA sequence. E) Make a mini-library of 200 bp overlapping fragments from the 1500 bp insert. Shotgun sequence library clones until you have a 1500 bp contig.

B) Synthesize a sequencing primer that will hybridize adjacent to the insert and use it to obtain 1000 bp of sequence. Use that new sequence information to generate a second primer within the cDNA that will provide a read of the remainder of the 1500 bp sequence. C) Synthesize two primers, one adjacent to each end of the insert. Add them both the same sequencing reaction in order to determine the sequence of the entire insert - 1000 bp from each side. D) Sequence 1000 bp from one end of the insert using a primer adjacent to one side. Compare that sequence to the RefSeq for the organism to find the remainder of the cDNA sequence. E) Make a mini-library of 200 bp overlapping fragments from the 1500 bp insert. Shotgun sequence library clones until you have a 1500 bp contig.

EMS is a chemical that ethylates G residues within DNA and causes mutation because ethylated Gs behave like As. Which of the following mutations could EMS make? Check all that apply. A) CACTGC → TACTGC B) TACTGC → CACTGC C) CACTAC → CACTGC D) CACTGC → CACTAC E)CACTGC → CACGGC

B) TACTGC → CACTGC D) CACTGC → CACTAC

A researcher is studying the rIIA locus of phage T4. Recall that wild-type phage can make plaques on E. coli strains B and K, but that rIIA− phage can use only E. coli B as a host. Three rIIA− strains are obtained: A, B, and C. E. coli strain B is co-infected with two rIIA− strains simultaneously, and the phage lysate is recovered and plated on a lawn of E. coli K. The results are: Infection with A and B phage = no plaques formInfection with A and C phage = plaques formInfection with B and C phage = plaques form What can be inferred from these results? A) The mutation in A and B are in different genes B) The mutations in A and C are in different base pairs, and so are the mutations in B and C. The mutations in A and B may be in the same base pair. C) The mutations in A and C are in the same base pair, and so are the mutations in B and C. The mutations in A and B may be in different base pairs. D) The mutations in A and C are in different genes, and so are the mutations in B and C. E) The mutations in A and B are in the same gene.

B) The mutations in A and C are in different base pairs, and so are the mutations in B and C. The mutations in A and B may be in the same base pair.

In the following pedigree, the indicated trait is caused by what type of allele? A) X-linked Dominant B) Y-linked C) autosomal recessive D) autosomal dominant E) X-linked recessive

B) Y-linked

The ribosome binding experiment illustrated below was used to discover that the codon CUC specifies Leucine, and not Serine. Suppose you wanted to use this experimental protocol to determine whether the codon AGA specifies Arg or Glu? What would you need at the start of the experiment? (Check all that apply.) A) radioactive AGA trinucleotides B) radioactive ribosomes C) radioactive ARG D) radioactive Glu E) synthetic RNAs with the sequence: 5' AGAAGAAGAAGAAGA3' F) AGA trinucleotides G) TCT trinucleotides

B) radioactive ribosomes C) radioactive ARG D) radioactive Glu E) synthetic RNAs with the sequence: 5' AGAAGAAGAAGAAGA3' G) TCT trinucleotides

Two F+ E. coli strains are co-cultured in the same flask. One is Strr Thr+ and the other is Genr Thr−. (Streptomycin and Gentamicin are two different antibiotic drugs.) The culture is then plated on minimal media that is supplemented with Gentamicin only (not threonine) and a few colonies grow. What sort of genetic exchange is most likely occurring? A) conjugation B) transformation C) complementation D) transduction E) electroporation

B) transformation

One strand of DNA has the sequence 5′ GGTCTA 3′. What is the sequence of the other strand? A)5′ ATCTGG 3′ B)5′ TAGACC 3′ C) 3′ TAGACC 5′ D) 5′ CCAGAT 3′ E)3′ GGTCTA 5′

B)5′ TAGACC 3′

Daughter cell produced in meiosis have ___? A)twice the number of chromosomes as daughter cells produced in mitosis. B)half the number of chromosomes as daughter cells produced in mitosis. C)one-fourth the number of chromosomes as daughter cells produced in mitosis. D) four times the number of chromosomes as daughter cells produced in mitosis. E) the same number of chromosomes as daughter cells produced in mitosis.

B)half the number of chromosomes as daughter cells produced in mitosis.

In a standard deck of playing cards, four suits exist (red suits = hearts and diamonds, black suits = spades and clubs). Each suit has 13 cards: Ace (A), 2, 3, 4, 5, 6, 7, 8, 9, 10, and the face cards Jack (J), Queen (Q), and King (K). In a single draw, what is the probability that you will draw a red card? A) 1/4 B) 1/52^13 C) 1/2 D) 1/13 E) 1/13 ^52

C) 1/2

If you roll a pair of dice, what is the probability they they will both turn up numbers over 3? A) 1/3 B) 1/6 C) 1/4 D) 1/12 E) 1/36

C) 1/4

If an individual has 10 gene pairs, how many different gametes can be formed if three of the gene pairs are homozygous and the remaining seven gene pairs are heterozygous? A) 49 B) 1024 C) 128 D) 100 E) 131,072

C) 128

If a trait is controlled by two codominant alleles of one gene, what phenotypic ratio is expected in the offspring of a mating of two heterozygotes? A) 3:1 B) 4:1 C) 1:2:1 D) 2:1

C) 1:2:1

In fruit flies, no crossing-over occurs in male meiosis. One interesting observation that results from this fact is that crosses between F1heterozygotes of the form A b / a B always yield the same ratio of phenotypes in the F2 progeny regardless of whether or not the genes are linked and regardless of the distance between them if they are linked. What is this ratio? A) 9 A_ B_ : 3 aa B_ : 3 A_ bb : 1 aa bb B) 1 A_ B_ : 2 aa B_ : 2 A_ bb : 1 aa bb C) 2 A_ B_ : 1 aa B_ : 1 A_ bb D) 1 A_ B_ : 1 aa B_ : 1 A_ bb : 1 aa bb

C) 2 A_ B_ : 1 aa B_ : 1 A_ bb

Suppose the A and B genes are on the same chromosome but separated by 100 map units. What fraction of the progeny from the cross A B / a b × a b / a b would be A b / a b? A) 10% B) 100% C) 25% D) 50%

C) 25%

You perform Sanger sequencing on a small fragment of the human genome and obtain the following small sequence read: 5' AGGCTTAAGCTTAATCGGGCTAT 3'. In order to determine if this sequence might be within the coding region of a gene, you examine it for open reading frames. How many open reading frames exist that go all the way through this DNA fragment? (Recall that the stop codons are 5' TAA, 5' TAG, and 5' TGA.) A) 2 B) 7 C) 4 D) 8 E) 5

C) 4

To generate a knockin mouse, a scientist introduces into ES cells a construct with a mutant exon and within an adjacent intron, loxP sites flanking a neomycin resistance gene. What event will occur next? A) Transplantation of cells into host blastocyst. B) Recombination between the loxP sites in the construct. C) Homologous recombination between the mutant DNA and the corresponding gene in the mouse genome. D) Recombination between one of the loxP sites in the construct and a loxP site in the mouse genome. E) Crossing of heterozygous mice to a Cre-expressing strain.

C) Homologous recombination between the mutant DNA and the corresponding gene in the mouse genome.

Homologous recombination occurs in a heterozygote in which alleles D and d differ by a single base pair. The D allele has a G (a GC base pair) at one position, whereas the d allele has a C (a CG base pair) at the same position. If branch migration causes heteroduplex formation across this position, what is the expected outcome? A) Mismatch repair will identify abnormal G-G and C-C base pairs and gene conversion will always occur in situations like this one when mismatched bases exist within the heteroduplex region. B) Mismatches will cause the Holliday junction to be unstable and to resolve by the noncrossover pathway. C) Mismatch repair will identify abnormal G-G and C-C base pairs and may convert the D allele to a d allele or the d allele to a D allele. D)Both D and d alleles will remain unchanged because G can base pair with C. E) Mismatch repair will identify an abnormal C-G base pair and will ensure that the cell has two copies of each allele.

C) Mismatch repair will identify abnormal G-G and C-C base pairs and may convert the D allele to a d allele or the d allele to a D allele

The pairwise map distances for four linked genes are as follows: P-Q = 22 m.u., Q-R = 7 m.u., R-S = 9 m.u., Q-S= 2 m.u., P-S = 20 m.u., P-R = 29 m.u. What is the order of these four genes? A) PQRS B) RPSQ C) PSQR D) PQSR

C) PSQR

Two strains of S. cerevisae (yeast) are crossed. One has the genotype A B and the other a b. Which statement correctly describes the tetrads that can be produced by one reciprocal crossover between the A and B genes without gene conversion in the dihybrid? A) Recombination occurs at the two-strand stage to produce a tetrad with A b, A b, a B, and a B chromosomes. B) Recombination occurs at the two-strand stage to produce a tetrad with A B, A b, a B, and a b chromosomes. C) Recombination occurs at the four-strand stage to produce a tetrad with A B, A b, a B, and a b chromosomes. D) Recombination occurs at the four-strand stage to produce a tetrad with A B, A B, a B, and a b chromosomes.

C) Recombination occurs at the four-strand stage to produce a tetrad with A B, A b, a B, and a b chromosomes.

In the following pedigree, the indicated trait is most likely caused by which type of allele? A) Y-linked B) X-linked recessive C) autosomal dominant D) X-linked dominant E) autosomal recessive

C) autosomal dominant

Huntington disease is a rare condition in humans, caused by a dominant allele of a single gene, and that results in a slow but inexorable deterioration of the nervous system. The probability that a person with the Huntington genotype will express the phenotype varies with age. Assume that 50% of those inheriting the HD allele will express the symptoms by age 40. Janelle is a 35-year-old woman whose father has Huntington disease. She currently shows no symptoms. What is the probability that Janelle will show symptoms in five years? A) 1/3 B) 2/9 C) 1/8 D) 1/4 E) 1/2 F) 1/9 G) 2/3

D) 1/4

An AAaa tetraploid plant undergoes self-fertilization. At least one dominant A allele is needed to obtain the dominant phenotype. At what frequency will progeny with the dominant phenotype appear? A) 1/16 B) 1/4 C) 15/16 D) 35/36 E) 1/36

D) 35/36

In a random DNA sequence where AT=GC, how long would you expect the average ORF (a sequence read in a particular frame without encountering a stop codon) to be? Recall that three of the triplet codons are stop signals. A) About 4096 nt long. B) About 60 codons (180 nt) long. C) About 256 nt long. D) About 20 codons (60 nt) long. E) About 61 codons (183 nt) long.

D) About 20 codons (60 nt) long.

Recall that in the Ames Test, potential mutagens are tested for their ability to revert His− auxotrophs to His+. Any particular his- mutation present in a His−auxotrophic bacterial strain would be able to be reverted only by some mutagens and not others. Yet, the purpose of the Ames Test is to determine whether or not a specific substance is a mutagen of any type. How is the Ames Test actually performed to circumvent this problem? A) Both the experimental (with potential mutagen) and control (without potential mutagen) bacteria are plated on minimal medium at the end of the experiment, so that only His+ revertants are detected. B) Separate aliquots of His− auxotrophs are grown in either the presence or absence of the compound. C) Rat liver enzymes are added to both the experimental and control samples. D) Several different His− auxotrophs with different kinds of mutations at the molecular level (transitions, transversion, additions, deletions, etc.) are used in separate experiments to test the same compound for mutagenicity.

D) Several different His− auxotrophs with different kinds of mutations at the molecular level (transitions, transversion, additions, deletions, etc.) are used in separate experiments to test the same compound for mutagenicity.

Courtship behavior in fruit flies is controlled by the fruitless (fru) gene. This behavior consists of males doing a "courtship dance" and orienting that dance specifically toward females; females are passive - they do nothing but run away from the males. The fruprimary transcript is made in both females and males, but it is spliced differently in the two sexes because females make Tra protein, a regulator of fru splicing, and males do not. Only the male form of fru mRNA makes a functional Fru protein, called Fru-M, a brain-specific transcription factor. The results of two experiments were informative regarding the role of Fru-M protein in the brain: (1) Mutant males that lack a functional fru gene court males and females indiscriminately; (2) Transgenic female flies that express Fru-M protein in their brains from a fru-M cDNA do the male courtship dance and orient it specifically towards other females. Check all of the statements below that are valid conclusions based on these experimental results. A) The Fru-F protein causes females to be passive. B) Another gene exists in the fly genome, in addition to fru, that can cause the males to do the courtship dance. C) Another gene exists, in addition to fru, that can cause the males to orient their courtship dance towards females. D) The Fru-M protein is sufficient to cause two normally male-specific fly behaviors to occur in females: the courtship dance and the orientation of the dance towards females. E) The Fru-F protein normally prevents the Fru-M protein from working in the female fly brain. F) Human males make a brain-specific protein homologous to Fru-M that causes them to court females.

D) The Fru-M protein is sufficient to cause two normally male-specific fly behaviors to occur in females: the courtship dance and the orientation of the dance towards females. F) Human males make a brain-specific protein homologous to Fru-M that causes them to court females.

A single DNA microarray can be used to genotype an individual for ~4 million common SNPs. Such microarrays can be purchased from companies that manufacture them for geneticists. What DNA is attached to the microarray? To genotype an individuals for SNP loci, with what would the microarray be probed? A) The microarray contains fluorescently labeled deoxyribonucleotides. The probe is small fragments of the individual's genome; SNP loci in the genome will hybridize with the reverse-complementary nucleotide, thereby hiding the fluor and so that grid square will no longer fluoresce. B) The microarray contains nanopores, nanothreaders, and nanodetectors of current. The probe is an individual's genomic DNA, whose sequence is read by threading the chromosomal DNA strands through the nanopores, thus revealing all of the SNP allele genotypes. C) The microarray contains PCR-generated clusters of fragments of the individual's genome. The probe is ASOs corresponding to each possible SNP locus allele. D) The microarray contains ASOs corresponding to all possible alleles of each of several million SNPs - each ASO in a different grid square. The probe is small fragments of the individual's genome, where each fragment has a fluor attached to it. E) The microarray contains exons of all of the genes in the genome. The probe is small fragments of the individual's genome; exon sequences only will be captured for exome sequencing.

D) The microarray contains ASOs corresponding to all possible alleles of each of several million SNPs - each ASO in a different grid square. The probe is small fragments of the individual's genome, where each fragment has a fluor attached to it.

The so-called hypervariable regions (HV1 and HV2) of the human mitochondrial genome are sometimes used in forensic analysis. They are two noncoding regions of mtDNA, each approximately 300 bp, that show more variation in different people than any other DNA sequences. The DNA within HV1 and HV2 appears to accumulate mutations at ten times the rate of DNA sequences in the nuclear genome. Check all statements that correctly describe situations where it would be advantageous to use mtDNA (as opposed to nuclear DNA) to identify individuals. A) You want to determine which of two brothers is the father of a child. B) You want to determine if two people are first cousins. C) Very little human DNA can be recovered from a crime scene. D) You want to determine if a pair of twins is identical or fraternal. E) You want to determine if the woman you've been told all your life is your mother is really your grandmother.

D) You want to determine if a pair of twins is identical or fraternal. E) You want to determine if the woman you've been told all your life is your mother is really your grandmother.

In chickens, females have two different sex chromosomes (Z and W), while the males have two Z chromosomes. A Z-linked gene controls the pattern of the feathers, with the dominant B allele causing the barred pattern and the b allele causing nonbarred feathers. Which of the following crosses would produce all daughters of one type (barred or nonbarred) and all sons of the other type? A) nonbarred females × barred males B) More than one of the choices is correct. C) barred females × barred males D) barred females × nonbarred males E) nonbarred females × nonbarred males

D) barred females × nonbarred males

Jeff's father has Huntington disease, which is caused by a rare autosomal dominant allele (H ). Two-thirds of people with the dominant Huntington allele show no symptoms by age 50, and Jeff is 50 and has no symptoms. Jeff's wife is is pregnant. What is that chance that Jeff's child will eventually show symptoms of Huntington disease? A) 1/8 B) 2/3 C) 1/6 D) 1/4 E) 1/5 F) 1/24

E) 1/5

Three genes in fruit flies affect a particular trait, and one dominant allele of each gene is necessary to get a wild-type phenotype. What phenotypic ratios would you predict among the progeny if you crossed triply heterozygous flies? A) 27:64 B) 9:23 C) 1:7 D) 37:64 E) 27:37

E) 27:37

To estimate the genome size of a newly discovered organism, you digest one copy of its genomic DNA with the restriction enzyme EcoRI, which has a 6 bp recognition site. The genome is cut into about 100,000 fragments. Which is your best estimate of the size of this organism's genome? A) 60,000,000 bp B) 400,000 bp C) 1,000,000,000 bp D) 40,000,000 bp E) 400,000,000 bp

E) 400,000,000 bp

How many different DNA strands composed of 1000 nucleotides could possibly exist? A) 1000^4 B) 1/1000^4 C) 3 billion D) 1/4 ^1000 E) 4^1000

E) 4^1000

Suppose you used a DNA synthesizer to generate the two short, single-stranded DNA molecules (called oligonucleotides) shown below. You add DNA 1 and DNA 2 to a reaction tube containing DNA polymerase and the four radioactive dNTPs. After DNA polymerase is allowed to work for a few hours, what will be the sequences of the single-stranded DNA molecules that contain radioactive nucleotides? (Assume that in the reaction tube, 10 bp of complementary sequence is sufficient for stable base-pairing.) (DNA 1) 5' CTACTACGGATCGGG 3' (DNA 2) 5' CCAGTCCCGATCCGT 3' A) 5' ACTGGCTACTACGGATCGGG 3' and 5' AGTAGCCAGTCCCGATCCGT 3' B) No DNA molecules will contain radioactive nucleotides because no new bases are added to either strand of DNA. C) 5' CTACTACGGATCGGG 3' and 5' CCAGTCCCGATCCGT 3' D) 5' CTACTACGGATCGGGAAAA 3' and 5' CCAGTCCCGATCCGTAAAAA 3' E) 5' CTACTACGGATCGGGACTGG 3' and 5' CCAGTCCCGATCCGTAGTAG 3'

E) 5' CTACTACGGATCGGGACTGG 3' and 5' CCAGTCCCGATCCGTAGTAG 3'

Assume that the mutation rate for a given gene is 1 × 10−6 mutations per generation. For that gene, how many mutations would be expected if 500 million sperm are examined? A) 5 B) 50 C) 0 D) 1 E) 500

E) 500

The size of the mouse (M. musculus) genome is 2700 megabases (Mb) and it contains about 25,000 genes arranged on 40 chromosomes. If the genes are evenly distributed on chromosomes in the mouse, what is the average number of genes per chromosome? A) 9.3 B) 1 C) 67.5 D) 6250 E) 625

E) 625

Which of the following is a step in translation that requires complementary base pairing? A) Adding the correct deoxyribonucleotide to the growing DNA chain. B) Adding the correct ribonucleotide ot the growing mRNA. C) Binding of RNA polymerase to the promoter. D) Binding of DNA polymerase to the origin of replication. E) Adding the correct amino acid to the growing polypeptide chain.

E) Adding the correct amino acid to the growing polypeptide chain.

Suppose DNA has 6 different bases present at random and at equal frequency (bases X and Y, in addition to AGCT) and X and Y pair with each other. How frequently would a restriction enzyme with a 6-base recognition site be expected to cleave DNA? A) Once every 256 base pairs. B) Once every 65,536 base pairs. C) Once every 625 base pairs. D) Once every 15,625 base pairs. E) Once every 46,656 base pairs.

E) Once every 46,656 base pairs.

A curly-winged fruit fly mates with a pure-breeding fruit fly with normal (straight) wings. The F1 mate with each other to produce an F2 generation that consists of 160 flies with curled wings and 80 with straight wings. What can you infer from this observation? A) Curled wings is a recessive trait. B) All of the hybrid F1 flies had straight wings. C) Wing shape is controlled by two codominant alleles. D) Two interacting genes determine wing shape. E) The dominant curled wing allele is also a recessive lethal.

E) The dominant curled wing allele is also a recessive lethal.

A short sequence read is shown below. The primer used was 5' GGAACGCCCATATCCCGCGG. (A = red; C = green; G = purple; T = black.) How would the data look if you accidentally omitted the deoxyTTP (dTTP) from the sequencing reaction? A) No peaks would exist at all. B) Only the first 3 peaks would be present. C) The red peaks would be missing and the positions of the other peaks would shift so that there would be no spaces. D) The black peaks would be missing and the positions of the other peaks would shift so that there would be no spaces. E) The black peaks would be missing and there would be blank spaces in their place. F) Only the first 4 peaks would be present.

F) Only the first 4 peaks would be present.