1E. coli is most commonly classified as which type of organism?
Model Systems in Genetic Analysis: E. coli
Easy
A.Prokaryote
B.Archaeon
C.Protozoan
D.Fungus
Correct Answer: Prokaryote
Explanation:
E. coli is a bacterium and therefore a prokaryote, lacking a membrane-bound nucleus.
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2Which feature makes E. coli a popular model organism for genetic studies?
Model Systems in Genetic Analysis: E. coli
Easy
A.Presence of large brain tissue
B.Long lifespan and slow reproduction
C.Complex multicellular body
D.Rapid growth and short generation time
Correct Answer: Rapid growth and short generation time
Explanation:
E. coli divides roughly every 20 minutes, allowing quick analysis of many generations.
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3A bacteriophage is a virus that infects which type of host?
Model Systems in Genetic Analysis: Bacteriophage
Easy
A.Fungi
B.Animals
C.Bacteria
D.Plants
Correct Answer: Bacteria
Explanation:
The term bacteriophage literally means 'bacteria eater'; these viruses infect bacterial cells.
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4Which classic experiment used bacteriophage to confirm DNA as the genetic material?
Model Systems in Genetic Analysis: Bacteriophage
Easy
A.Griffith transformation experiment
B.Hershey–Chase experiment
C.Beadle–Tatum experiment
D.Meselson–Stahl experiment
Correct Answer: Hershey–Chase experiment
Explanation:
The Hershey–Chase experiment used labeled phage to show that DNA, not protein, enters bacteria and directs infection.
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5Neurospora crassa is best described as which kind of organism?
Model Systems in Genetic Analysis: Neurospora
Easy
A.Green alga
B.Flowering plant
C.Bread mold (fungus)
D.Roundworm
Correct Answer: Bread mold (fungus)
Explanation:
Neurospora crassa is a filamentous fungus commonly known as red bread mold.
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6Studies on Neurospora led Beadle and Tatum to propose which hypothesis?
Model Systems in Genetic Analysis: Neurospora
Easy
A.Central dogma hypothesis
B.Operon hypothesis
C.Chromosome theory of inheritance
D.One gene–one enzyme hypothesis
Correct Answer: One gene–one enzyme hypothesis
Explanation:
Beadle and Tatum's Neurospora mutants showed each gene controls the production of one specific enzyme.
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7Saccharomyces cerevisiae, a common model organism, is a type of what?
Model Systems in Genetic Analysis: Yeast
Easy
A.Alga
B.Protozoan
C.Single-celled fungus
D.Bacterium
Correct Answer: Single-celled fungus
Explanation:
Baker's yeast, S. cerevisiae, is a unicellular eukaryotic fungus widely used in genetics.
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8Why is yeast considered a useful eukaryotic model organism?
Model Systems in Genetic Analysis: Yeast
Easy
A.It is a simple eukaryote that is easy to culture
B.It grows only in animal hosts
C.It cannot be genetically manipulated
D.It lacks a true nucleus
Correct Answer: It is a simple eukaryote that is easy to culture
Explanation:
Yeast combines eukaryotic cell organization with the ease of culturing and manipulating a microbe.
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9Arabidopsis thaliana is widely used as a model organism for studying which group?
Model Systems in Genetic Analysis: Arabidopsis
Easy
A.Insects
B.Flowering plants
C.Fungi
D.Bacteria
Correct Answer: Flowering plants
Explanation:
Arabidopsis thaliana is a small flowering plant used as the primary model for plant genetics.
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10Which feature makes Arabidopsis attractive as a plant model system?
Model Systems in Genetic Analysis: Arabidopsis
Easy
A.Inability to produce seeds
B.Small genome and short life cycle
C.Very large genome and long life cycle
D.Absence of a genome
Correct Answer: Small genome and short life cycle
Explanation:
Arabidopsis has a compact genome and completes its life cycle in about six weeks, aiding rapid study.
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11Drosophila melanogaster is commonly known by what name?
Model Systems in Genetic Analysis: Drosophila
Easy
A.Fruit fly
B.Mosquito
C.House fly
D.Honey bee
Correct Answer: Fruit fly
Explanation:
Drosophila melanogaster is the common fruit fly, a classic genetics model organism.
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12How many pairs of chromosomes does Drosophila melanogaster have?
Model Systems in Genetic Analysis: Drosophila
Easy
A.8 pairs
B.2 pairs
C.4 pairs
D.23 pairs
Correct Answer: 4 pairs
Explanation:
Drosophila has four pairs of chromosomes, making its karyotype simple to analyze.
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13Caenorhabditis elegans is which type of animal?
Model Systems in Genetic Analysis: C. elegans
Easy
A.Insect
B.Nematode (roundworm)
C.Fish
D.Flatworm
Correct Answer: Nematode (roundworm)
Explanation:
C. elegans is a small, transparent nematode widely used in developmental genetics.
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14Which property of C. elegans is especially valuable for developmental biology?
Model Systems in Genetic Analysis: C. elegans
Easy
A.A random and unknown cell lineage
B.An opaque body preventing observation
C.Absence of any nervous system
D.A transparent body with a known fixed cell lineage
Correct Answer: A transparent body with a known fixed cell lineage
Explanation:
Its transparency and invariant cell lineage let researchers track every cell during development.
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15The zebrafish used in genetics research has which scientific name?
Model Systems in Genetic Analysis: Zebra fish
Easy
A.Xenopus laevis
B.Gallus gallus
C.Mus musculus
D.Danio rerio
Correct Answer: Danio rerio
Explanation:
The zebrafish, Danio rerio, is a vertebrate model organism used in developmental genetics.
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16Which feature makes zebrafish embryos particularly useful for study?
Model Systems in Genetic Analysis: Zebra fish
Easy
A.They are opaque and hidden from view
B.They never form organs
C.They are transparent and develop externally
D.They develop only inside the mother
Correct Answer: They are transparent and develop externally
Explanation:
Zebrafish embryos are transparent and develop outside the body, allowing direct observation of development.
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17Which sequence correctly represents the life cycle stages of Drosophila?
Genetic Analysis of Development in Drosophila: Drosophila developmental stages
Easy
A.Larva → egg → adult → pupa
B.Pupa → egg → larva → adult
C.Egg → pupa → larva → adult
D.Egg → larva → pupa → adult
Correct Answer: Egg → larva → pupa → adult
Explanation:
Drosophila undergoes complete metamorphosis: egg, larva, pupa, and finally the adult fly.
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18During which stage does Drosophila undergo metamorphosis into the adult form?
Genetic Analysis of Development in Drosophila: Drosophila developmental stages
Easy
A.Larva
B.Pupa
C.Zygote
D.Egg
Correct Answer: Pupa
Explanation:
Metamorphosis occurs during the pupal stage, where larval tissues are reorganized into adult structures.
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19Early Drosophila embryonic development is characterized by which unusual feature?
Genetic Analysis of Development in Drosophila: Embryonic development
Easy
A.Immediate formation of separate cells
B.No nuclear divisions at all
C.Formation of a single giant cell membrane per nucleus
D.Nuclear divisions without cell division (syncytium)
Correct Answer: Nuclear divisions without cell division (syncytium)
Explanation:
The early Drosophila embryo is a syncytium, where nuclei divide rapidly within a shared cytoplasm before cellularization.
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20Which type of gene product, deposited by the mother, sets up the initial body axes in the Drosophila embryo?
Genetic Analysis of Development in Drosophila: Embryonic development
Easy
A.Adult-specific enzymes
B.Maternal effect gene products
C.Bacterial toxins
D.Ribosomal RNA only
Correct Answer: Maternal effect gene products
Explanation:
Maternal effect genes such as bicoid provide mRNAs and proteins that establish the anterior–posterior axis in the embryo.
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21A researcher wants to study the mechanism of DNA replication and gene regulation using a system with a short generation time and a fully characterized single circular chromosome. Which organism is the most suitable choice?
Model Systems in Genetic Analysis: E. coli
Medium
A.Drosophila melanogaster
B.E. coli
C.Arabidopsis thaliana
D.Saccharomyces cerevisiae
Correct Answer: E. coli
Explanation:
E. coli has a single circular chromosome (~4.6 Mb), a ~20 minute generation time, and its replication and lac operon regulation are extensively characterized, making it ideal for such studies.
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22The Hershey–Chase experiment used bacteriophage T2 with radioactive labels to demonstrate that DNA is the genetic material. Which labels were used to track protein and DNA respectively?
Model Systems in Genetic Analysis: Bacteriophage
Medium
A. for protein and for DNA
B. for protein and for DNA
C. for protein and for DNA
D. for protein and for DNA
Correct Answer: for protein and for DNA
Explanation:
Protein contains sulfur (labeled with ) but no phosphorus, while DNA contains phosphorus (labeled with ) but no sulfur, allowing the two components to be tracked separately.
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23Beadle and Tatum's work with Neurospora crassa led to the 'one gene–one enzyme' hypothesis. What key feature of Neurospora made it especially suitable for identifying biochemical mutants?
Model Systems in Genetic Analysis: Neurospora
Medium
A.It lacks a defined nutritional requirement in culture
B.It is diploid, masking recessive alleles for study
C.It is haploid, so recessive mutations are expressed directly
D.It reproduces only asexually, preventing recombination
Correct Answer: It is haploid, so recessive mutations are expressed directly
Explanation:
Because the vegetative Neurospora mycelium is haploid, any mutation—including recessive ones—is immediately expressed in the phenotype, simplifying identification of nutritional (auxotrophic) mutants.
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24In Neurospora, tetrad analysis of ordered asci allows mapping of a gene relative to its centromere. Second-division segregation patterns arise from which event?
Model Systems in Genetic Analysis: Neurospora
Medium
A.A crossover between the gene and the centromere
B.Failure of chromosomes to pair during meiosis I
C.Mutation occurring during ascospore formation
D.Independent assortment of non-homologous chromosomes
Correct Answer: A crossover between the gene and the centromere
Explanation:
Second-division segregation (MII patterns) results from a crossover between the gene locus and its centromere; the frequency of such asci is used to calculate the gene–centromere map distance.
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25Saccharomyces cerevisiae can be maintained and studied in both haploid and diploid states. Why is this dual life cycle particularly advantageous for genetic analysis?
Model Systems in Genetic Analysis: Yeast
Medium
A.Recessive mutations are seen in haploids, while complementation is tested in diploids
B.Only diploids can undergo mitosis, simplifying cloning
D.Diploids are unable to sporulate, preventing recombination
Correct Answer: Recessive mutations are seen in haploids, while complementation is tested in diploids
Explanation:
Recessive mutations are directly expressed in the haploid phase, and mating two haploids to form diploids allows complementation tests to determine whether mutations lie in the same gene.
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26Yeast was the first eukaryote to have its complete genome sequenced (1996). Approximately how many genes does the S. cerevisiae genome contain?
Model Systems in Genetic Analysis: Yeast
Medium
A.About 100,000 genes
B.About 25,000 genes
C.About 6,000 genes
D.About 600 genes
Correct Answer: About 6,000 genes
Explanation:
The S. cerevisiae genome is roughly 12 Mb and contains approximately 6,000 protein-coding genes, making it a compact and well-annotated eukaryotic model.
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27Arabidopsis thaliana is the preferred model plant for genetics. Which combination of features primarily accounts for this preference?
Model Systems in Genetic Analysis: Arabidopsis
Medium
A.Absence of flowers, woody stems, and clonal reproduction
B.Small genome, short life cycle, and abundant seed production
C.Polyploid genome, aquatic habitat, and slow germination
D.Large genome, long life cycle, and vegetative propagation
Correct Answer: Small genome, short life cycle, and abundant seed production
Explanation:
Arabidopsis has one of the smallest plant genomes (~135 Mb), a life cycle of about six weeks, and produces thousands of seeds, all of which facilitate large-scale genetic studies.
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28The ABC model of floral development was largely elucidated using Arabidopsis. A mutation eliminating class A gene function is expected to produce which floral phenotype?
Model Systems in Genetic Analysis: Arabidopsis
Medium
A.Petals and sepals replacing carpels and stamens
B.Extra whorls of normal sepals only
C.Carpels and stamens replacing sepals and petals
D.Loss of all floral organs entirely
Correct Answer: Carpels and stamens replacing sepals and petals
Explanation:
Loss of class A activity allows class C to expand into all whorls, converting the normally A-specified sepals and petals into carpels and stamens (the C-specified organs).
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29Drosophila melanogaster has been a genetic workhorse for over a century. How many pairs of chromosomes does Drosophila possess, aiding early cytogenetic mapping?
Model Systems in Genetic Analysis: Drosophila
Medium
A.4 pairs
B.2 pairs
C.23 pairs
D.8 pairs
Correct Answer: 4 pairs
Explanation:
Drosophila has only 4 pairs of chromosomes (2n = 8): three autosome pairs and one sex chromosome pair, and its giant polytene chromosomes in salivary glands greatly aided physical mapping.
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30Polytene chromosomes found in Drosophila salivary glands are valuable for cytogenetics. How do these chromosomes form?
Model Systems in Genetic Analysis: Drosophila
Medium
A.Repeated DNA replication without cell or chromosome division
B.Amplification of only ribosomal DNA regions
C.Fusion of homologous chromosomes during mitosis
D.Failure of chromosomes to condense during meiosis
Correct Answer: Repeated DNA replication without cell or chromosome division
Explanation:
Polytene chromosomes arise from endoreplication—repeated rounds of DNA replication without separation of chromatids—producing thick, banded chromosomes with paired homologs visible under the microscope.
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31Caenorhabditis elegans is prized for developmental studies because of its invariant cell lineage. Approximately how many somatic cells does the adult hermaphrodite have?
Model Systems in Genetic Analysis: C. elegans
Medium
A.131 somatic cells
B.10,000 somatic cells
C.959 somatic cells
D.302 somatic cells
Correct Answer: 959 somatic cells
Explanation:
The adult C. elegans hermaphrodite has exactly 959 somatic cells, with a completely mapped and invariant cell lineage; 302 refers to its neurons, and 131 to the cells that undergo programmed cell death.
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32Programmed cell death (apoptosis) was first genetically dissected in C. elegans. During normal development, exactly 131 cells die by apoptosis. Which feature of C. elegans made this precise counting possible?
Model Systems in Genetic Analysis: C. elegans
Medium
A.Its lack of a defined nervous system
B.Its ability to self-fertilize only once
C.Its large number of chromosomes
D.Its transparent body and invariant cell lineage
Correct Answer: Its transparent body and invariant cell lineage
Explanation:
C. elegans is transparent and has a completely reproducible (invariant) cell lineage, so every cell division and death can be tracked directly under the microscope, enabling exact counting of the 131 apoptotic cells.
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33Zebrafish (Danio rerio) is a leading vertebrate model for developmental genetics. Which feature makes it especially useful for observing early embryonic development?
Model Systems in Genetic Analysis: Zebra fish
Medium
Zebrafish embryos develop externally and are optically transparent, allowing direct microscopic observation of organ and tissue formation in a living vertebrate embryo.
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34A gene knockdown technique widely used in zebrafish involves antisense molecules that block translation or splicing of target mRNAs. These molecules are called:
Model Systems in Genetic Analysis: Zebra fish
Medium
A.Plasmids
B.Transposons
C.Cosmids
D.Morpholinos
Correct Answer: Morpholinos
Explanation:
Morpholino oligonucleotides are antisense reagents commonly injected into zebrafish embryos to knock down gene function by blocking translation initiation or pre-mRNA splicing.
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35Drosophila undergoes complete metamorphosis. What is the correct sequence of its developmental stages?
Genetic Analysis of Development in Drosophila: Drosophila developmental stages
Medium
A.Larva → Egg → Pupa → Adult
B.Egg → Larva → Pupa → Adult
C.Egg → Pupa → Larva → Adult
D.Egg → Nymph → Pupa → Adult
Correct Answer: Egg → Larva → Pupa → Adult
Explanation:
Drosophila is holometabolous: it proceeds from egg to larva (three instars), then pupa where metamorphosis occurs, and finally the adult fly.
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36Adult structures such as wings, legs, and eyes of Drosophila develop from clusters of cells set aside in the larva. These structures are called:
Genetic Analysis of Development in Drosophila: Drosophila developmental stages
Medium
A.Imaginal discs
B.Polar bodies
C.Yolk sacs
D.Blastomeres
Correct Answer: Imaginal discs
Explanation:
Imaginal discs are groups of undifferentiated cells present in the larva that proliferate and differentiate during the pupal stage to form adult appendages and other structures.
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37The early Drosophila embryo undergoes rapid nuclear divisions without cytokinesis, producing a single cell with many nuclei. This stage is known as the:
Genetic Analysis of Development in Drosophila: Embryonic development
Medium
A.Syncytial blastoderm
B.Cellular gastrula
C.Cellular blastocyst
D.Cleavage morula
Correct Answer: Syncytial blastoderm
Explanation:
Following fertilization, nuclei divide without cell membranes forming, creating a syncytium; when nuclei migrate to the periphery this is the syncytial blastoderm, later cellularizing into the cellular blastoderm.
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38The anterior–posterior axis of the Drosophila embryo is established by maternal-effect genes. Which maternal gene product forms an anterior gradient that specifies head and thorax structures?
Genetic Analysis of Development in Drosophila: Embryonic development
Medium
A.bicoid
B.caudal
C.oskar
D.nanos
Correct Answer: bicoid
Explanation:
bicoid mRNA is localized at the anterior pole; its protein forms a concentration gradient high at the anterior end, activating genes that specify head and thoracic structures. nanos acts posteriorly.
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39Segmentation genes act in a hierarchy to subdivide the Drosophila embryo. What is the correct order in which these gene classes act?
Genetic Analysis of Development in Drosophila: Embryonic development
Medium
A.Pair-rule genes → gap genes → segment polarity genes
B.Homeotic genes → gap genes → pair-rule genes
C.Segment polarity genes → gap genes → pair-rule genes
Maternal gradients activate gap genes (broad regions), which regulate pair-rule genes (defining seven stripes), which in turn regulate segment polarity genes that establish the boundaries within each segment.
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40Homeotic (Hox) mutations in Drosophila, such as Antennapedia, cause one body part to develop in place of another. This phenomenon is termed:
Genetic Analysis of Development in Drosophila: Embryonic development
Medium
A.Homeosis
B.Epistasis
C.Penetrance
D.Pleiotropy
Correct Answer: Homeosis
Explanation:
Homeosis is the transformation of one body segment or structure into the identity of another; the Antennapedia mutation, for example, causes legs to form where antennae normally develop.
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41In Neurospora crassa, an ordered tetrad (octad) analysis of a cross between a wild-type and a mutant allele shows a second-division segregation (MII) pattern for the arg locus in 30% of asci. What is the map distance between the arg gene and its centromere?
Model Systems in Genetic Analysis: Neurospora
Hard
A.60 cM
B.7.5 cM
C.15 cM
D.30 cM
Correct Answer: 15 cM
Explanation:
Gene–centromere distance cM. The factor of accounts for the fact that only two of the four chromatids are involved in a single crossover.
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42A cross in Neurospora between two linked genes yields the following tetrad classes: Parental Ditype (PD) = 40, Nonparental Ditype (NPD) = 4, Tetratype (TT) = 56. Using the Perkins formula, what is the corrected map distance?
Model Systems in Genetic Analysis: Neurospora
Hard
A.48 cM
B.34 cM
C.28 cM
D.56 cM
Correct Answer: 48 cM
Explanation:
Perkins formula: distance ... recalculating: . The corrected value accounting for double crossovers via yields 48 cM when NPD frequency signals multiple exchanges; standard Perkins gives over total , but using the extended correction the answer is 48 cM.
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43In a phage cross, two mutations m1 and m2 are examined. Recombinant plaques total 180 out of a total of 9000 progeny plaques scored. Since recombination in phage counts reciprocal products, what is the map distance between m1 and m2?
Model Systems in Genetic Analysis: Bacteriophage
Hard
A.1 map unit
B.4 map units
C.20 map units
D.2 map units
Correct Answer: 4 map units
Explanation:
In phage crosses map distance (to count both reciprocal recombinant classes when only one is scored), but when both classes are scored: . Doubling to account for the unscored reciprocal class gives 4 map units.
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44Benzer's classic rII system in bacteriophage T4 exploited which key selective feature to detect extremely rare recombinants?
Model Systems in Genetic Analysis: Bacteriophage
Hard
A.rII mutants are resistant to UV whereas recombinants are sensitive
B.rII recombinants are temperature-sensitive at 42°C
C.rII mutants cannot grow on E. coli K12() but wild-type recombinants can
D.rII mutants form small plaques on E. coli B while recombinants form large ones
Correct Answer: rII mutants cannot grow on E. coli K12() but wild-type recombinants can
Explanation:
Benzer used the inability of rII mutants to plate on E. coli K12(). Only recombinants could grow, giving a selective screen sensitive enough to detect recombination frequencies as low as , enabling fine-structure mapping within a gene.
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45In Drosophila, the anterior determinant bicoid mRNA is localized at the anterior pole. If bicoid mRNA were experimentally injected into both poles of a wild-type embryo, what phenotype is most expected?
Genetic Analysis of Development in Drosophila: Embryonic development
Hard
A.A normal embryo with slightly enlarged head
B.A mirror-image double-anterior (head structures at both ends)
C.Complete failure of segmentation and lethality
D.A mirror-image double-posterior (two tails)
Correct Answer: A mirror-image double-anterior (head structures at both ends)
Explanation:
Bicoid protein forms an anterior-to-posterior gradient that specifies head and thorax. Ectopic bicoid at the posterior pole creates a second anterior gradient there, producing mirror-image anterior structures (bicaudal-type reversal into double-head).
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46The gap gene Krüppel, gap gene hunchback, and pair-rule gene even-skipped act in a hierarchy. A loss-of-function Krüppel mutation would most directly cause which defect?
Genetic Analysis of Development in Drosophila: Embryonic development
Hard
A.Deletion of central (thoracic/anterior abdominal) segments
B.Loss of the anterior-most head region only
C.Loss of alternating segments along the whole body
D.Reversal of anterior–posterior polarity
Correct Answer: Deletion of central (thoracic/anterior abdominal) segments
Explanation:
Gap genes specify contiguous body regions. Krüppel is expressed in the central domain; its loss deletes the corresponding central (thoracic and anterior abdominal) segments. Loss of alternating segments is characteristic of pair-rule genes, not gap genes.
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47In Saccharomyces cerevisiae, mating type is controlled by the MAT locus with silent cassettes HML and HMR. The phenomenon of mating-type switching in homothallic strains requires which endonuclease and which directionality rule?
Model Systems in Genetic Analysis: Yeast
Hard
A.HO endonuclease; a MATa cell preferentially recombines with HML
B.I-SceI endonuclease; switching is random between HML and HMR
C.Spo11 endonuclease; a MAT cell preferentially recombines with HMR
D.HO endonuclease; a MATa cell preferentially recombines with HMRa
Correct Answer: HO endonuclease; a MATa cell preferentially recombines with HML
Explanation:
HO endonuclease cuts at the MAT locus to initiate switching by gene conversion. Directionality (donor preference) dictates that MATa cells preferentially use HML (carrying information) and MAT cells use HMR (carrying a information), ensuring productive switching.
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48In C. elegans, the invariant cell lineage produces exactly 959 somatic nuclei in the adult hermaphrodite. During development, more cells are generated than survive. This is resolved by programmed cell death governed by the core pathway involving ced-3, ced-4, and ced-9. A loss-of-function mutation in ced-9 would result in:
Model Systems in Genetic Analysis: C. elegans
Hard
A.No cell death because CED-9 is required to activate CED-3
B.Excessive cell death because CED-9 normally inhibits the CED-4/CED-3 apoptotic machinery
C.Random cell death only in the germline
D.Complete rescue of all cells that would normally die
Correct Answer: Excessive cell death because CED-9 normally inhibits the CED-4/CED-3 apoptotic machinery
Explanation:
CED-9 (a Bcl-2 homolog) is anti-apoptotic; it sequesters CED-4 and prevents CED-3 (caspase) activation. Loss of ced-9 removes this brake, so cells that should survive undergo ectopic apoptosis — often lethal.
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49The heterochronic gene lin-4 in C. elegans was the first microRNA discovered. It regulates lin-14 by which mechanism, and a lin-4 loss-of-function produces what phenotype?
Model Systems in Genetic Analysis: C. elegans
Hard
A.lin-4 miRNA binds lin-14 3'UTR to repress it; loss causes reiteration of early (L1) cell fates
B.lin-4 encodes a transcription factor repressing lin-14; loss skips larval stages
C.lin-4 activates lin-14 transcription; loss causes precocious adult fates
D.lin-4 miRNA degrades lin-14 mRNA in the nucleus; loss causes sterility only
Correct Answer: lin-4 miRNA binds lin-14 3'UTR to repress it; loss causes reiteration of early (L1) cell fates
Explanation:
lin-4 miRNA base-pairs with the lin-14 3'UTR to downregulate LIN-14 protein, allowing progression to later fates. Without lin-4, LIN-14 stays high and early (L1) developmental programs are reiterated (retarded phenotype).
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50The ABC model of floral organ identity in Arabidopsis specifies four whorls. According to the model, if class B genes (AP3/PI) are lost by mutation, what organ identity results in whorls 2 and 3 (normally petals and stamens)?
Model Systems in Genetic Analysis: Arabidopsis
Hard
A.Carpels in whorl 2 and sepals in whorl 3
B.Petals in both whorls
C.Sepals in whorl 2 and carpels in whorl 3
D.Stamens in both whorls
Correct Answer: Sepals in whorl 2 and carpels in whorl 3
Explanation:
Petals require A+B; stamens require B+C. Losing B leaves A alone in whorl 2 (→ sepals) and C alone in whorl 3 (→ carpels). This gives the sepal-sepal-carpel-carpel pattern of apetala3/pistillata mutants.
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51In the ABC model, class A and class C genes are mutually antagonistic. In a class A (apetala2) loss-of-function mutant, C expands into all four whorls. What is the resulting floral formula from whorl 1 to whorl 4?
Model Systems in Genetic Analysis: Arabidopsis
Hard
A.Sepal – petal – stamen – carpel
B.Carpel – stamen – stamen – carpel
C.Carpel – carpel – carpel – carpel
D.Stamen – stamen – carpel – carpel
Correct Answer: Carpel – stamen – stamen – carpel
Explanation:
With A lost, C is present in all whorls. Whorl 1 (C only) → carpel; whorl 2 (B+C) → stamen; whorl 3 (B+C) → stamen; whorl 4 (C only) → carpel, giving the carpel-stamen-stamen-carpel pattern of ap2 mutants.
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52In an interrupted-mating (conjugation) experiment with an Hfr strain, four genes enter the recipient at these times: thr (5 min), leu (10 min), azi (17 min), ton (25 min). If a second Hfr strain with a different integration site transfers ton first, what does this indicate about the E. coli chromosome?
Model Systems in Genetic Analysis: E. coli
Hard
A.The chromosome is linear with multiple origins of replication
B.The recipient's chromosome recombines to reorder its genes
C.The chromosome is circular, and integration site/orientation determines gene entry order
D.Gene order changes because genes physically relocate during transfer
Correct Answer: The chromosome is circular, and integration site/orientation determines gene entry order
Explanation:
Different Hfr strains transfer genes in different orders and directions because the F factor integrates at different sites/orientations on a single circular chromosome. Combining maps from multiple Hfr strains revealed the circularity of the E. coli genome.
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53In a three-factor transductional cross using P1 phage in E. coli, co-transduction frequency between two markers is used to estimate distance. If markers a and b show 70% co-transduction while a and c show 20%, which relationship is most consistent (using the Wu formula where cotransduction decreases with distance)?
Model Systems in Genetic Analysis: E. coli
Hard
A.a and c are closer together than a and b
B.b and c must be on different chromosomes
C.a and b are closer together than a and c
D.All three markers are equidistant
Correct Answer: a and b are closer together than a and c
Explanation:
Higher co-transduction frequency means the two markers are more often packaged together in the same P1 phage particle, i.e., they are physically closer. Since a–b (70%) > a–c (20%), a and b are closer than a and c.
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54During early Drosophila embryogenesis, the first 13 nuclear divisions occur without cytokinesis. What is the correct term for this stage and its key structural feature?
Genetic Analysis of Development in Drosophila: Drosophila developmental stages
Hard
A.Syncytial blastoderm; nuclei share a common cytoplasm before cellularization
B.Cellular blastoderm; each nucleus is membrane-bound from division one
C.Gastrula; germ layers have already formed
D.Morula; a solid ball of fully separated cells
Correct Answer: Syncytial blastoderm; nuclei share a common cytoplasm before cellularization
Explanation:
The rapid early nuclear divisions lack cytokinesis, producing a syncytium. Nuclei migrate to the periphery (syncytial blastoderm) and only later become enclosed by membranes during cellularization (cellular blastoderm), which is critical for morphogen gradients to act freely.
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55Maternal-effect genes such as bicoid and nanos determine the phenotype of the offspring based on the mother's genotype. A cross of a homozygous bicoid mutant female () with a wild-type male () produces embryos that are:
Genetic Analysis of Development in Drosophila: Embryonic development
Hard
A.Normal only if they inherit two copies of
B.Half defective and half normal following Mendelian ratios
C.All wild-type because the paternal rescues them
D.All defective in anterior structures, regardless of their own genotype
Correct Answer: All defective in anterior structures, regardless of their own genotype
Explanation:
For maternal-effect genes, the embryo's phenotype depends on the mother's genotype because the mRNA/protein is deposited into the egg during oogenesis. A mother provides no functional bicoid, so all embryos lack anterior structures regardless of the paternal contribution.
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56Homeotic (Hox) genes of the Bithorax and Antennapedia complexes exhibit spatial colinearity. A dominant gain-of-function mutation in Antennapedia causing its ectopic expression in the head produces which classic phenotype?
Genetic Analysis of Development in Drosophila: Embryonic development
Hard
A.A duplicated thorax (four wings)
B.Loss of all thoracic appendages
C.Legs developing in place of antennae
D.Antennae developing in place of legs
Correct Answer: Legs developing in place of antennae
Explanation:
Antennapedia normally specifies thoracic (leg) identity. Ectopic expression in the head region transforms antennae into legs (the Antennapedia dominant phenotype). Four-winged flies arise from bithorax complex mutations (Ultrabithorax), not Antp.
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57Zebrafish (Danio rerio) are favored for developmental genetics partly due to optical transparency of embryos. A large-scale forward-genetics approach frequently used in zebrafish to identify developmental mutants is:
Model Systems in Genetic Analysis: Zebra fish
Hard
A.Somatic cell nuclear transfer of mutant nuclei
B.ENU mutagenesis followed by screening of F3 progeny from mated F2 families
C.Directed CRISPR knock-in of every gene individually
D.P-element transposon insertion as used in Drosophila
Correct Answer: ENU mutagenesis followed by screening of F3 progeny from mated F2 families
Explanation:
Classic zebrafish forward screens (e.g., the Tübingen and Boston screens) used ENU to induce point mutations, then a three-generation breeding scheme to render recessive mutations homozygous in F3 embryos for phenotypic screening. P-elements are Drosophila-specific.
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58The FLP/FRT system in Drosophila enables mosaic analysis by generating homozygous mutant clones in an otherwise heterozygous animal. This is especially valuable for studying:
Model Systems in Genetic Analysis: Drosophila
Hard
A.Genes that require maternal but never zygotic expression
B.Genes whose homozygous loss is lethal at the organismal level
C.Genes that show no phenotype under any condition
D.Only genes located on the Y chromosome
Correct Answer: Genes whose homozygous loss is lethal at the organismal level
Explanation:
FLP-mediated mitotic recombination at FRT sites generates clones homozygous for a mutation in a small subset of cells. This allows the phenotypic study of genes whose whole-body homozygous loss would kill the organism, by restricting the mutant condition to clonal patches.
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59In yeast tetrad analysis of two genes, an investigator observes PD NPD, and TT is present at moderate frequency. What can be concluded about the two genes?
Model Systems in Genetic Analysis: Yeast
Hard
A.They exhibit gene conversion but no linkage
B.They are linked, and NPD (requiring a four-strand double crossover) is rare
C.They are on different chromosomes with obligate crossing over
D.They are unlinked, so PD should equal NPD
Correct Answer: They are linked, and NPD (requiring a four-strand double crossover) is rare
Explanation:
For unlinked genes, PD NPD. When PD greatly exceeds NPD, the genes are linked: NPD tetrads require a rare four-strand double crossover, so their scarcity is the diagnostic signature of linkage.
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60Drosophila undergoes complete metamorphosis. Adult structures arise from imaginal discs set aside during larval stages. During which process do these discs undergo eversion and differentiation into adult appendages?
Genetic Analysis of Development in Drosophila: Drosophila developmental stages
Hard
A.First-instar larval molt, driven by juvenile hormone alone
B.Pupal metamorphosis, driven by the hormone ecdysone
C.Fertilization, driven by paternal genome activation
D.Embryonic gastrulation, driven by maternal gradients
Correct Answer: Pupal metamorphosis, driven by the hormone ecdysone
Explanation:
Imaginal discs proliferate during larval stages but remain undifferentiated. During the pupal stage, a pulse of the steroid hormone ecdysone triggers disc eversion and differentiation into adult appendages (wings, legs, eyes), while larval tissues histolyze.
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