1A chromosomal aberration in which a segment of a chromosome is lost is called a:
Variation in Chromosome Structure: Deletion
Easy
A.Duplication
B.Translocation
C.Deletion
D.Inversion
Correct Answer: Deletion
Explanation:
A deletion occurs when a portion of a chromosome breaks off and is lost, removing the genes it carried.
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2When a segment of a chromosome is present in two copies within the same chromosome, the aberration is called a:
Variation in Chromosome Structure: Duplication
Easy
A.Duplication
B.Inversion
C.Translocation
D.Deletion
Correct Answer: Duplication
Explanation:
A duplication produces an extra copy of a chromosome segment, increasing the gene dosage for that region.
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3In which chromosomal aberration is a segment reversed end to end within the same chromosome?
Variation in Chromosome Structure: Inversion
Easy
A.Translocation
B.Inversion
C.Duplication
D.Deletion
Correct Answer: Inversion
Explanation:
An inversion occurs when a chromosome segment breaks, flips 180°, and rejoins in the reversed orientation.
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4The transfer of a chromosome segment to a non-homologous chromosome is known as:
Variation in Chromosome Structure: Translocation
Easy
A.Deletion
B.Translocation
C.Duplication
D.Inversion
Correct Answer: Translocation
Explanation:
In a translocation, a segment of one chromosome becomes attached to a different, non-homologous chromosome.
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5The failure of chromosomes to separate properly during cell division is called:
Variation in Chromosome Number: Non-disjunction and Aneuploidy
Easy
A.Non-disjunction
B.Replication
C.Transcription
D.Crossing over
Correct Answer: Non-disjunction
Explanation:
Non-disjunction is the failure of homologous chromosomes or sister chromatids to separate, leading to gametes with abnormal chromosome numbers.
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6A condition in which the chromosome number differs from the normal by one or a few chromosomes is called:
Variation in Chromosome Number: Non-disjunction and Aneuploidy
Easy
A.Euploidy
B.Aneuploidy
C.Polyploidy
D.Haploidy
Correct Answer: Aneuploidy
Explanation:
Aneuploidy is an abnormal number of individual chromosomes, such as one extra or one missing chromosome.
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7Down syndrome is caused by an extra copy of which chromosome?
Trisomies - chromosome 13, 18, 21
Easy
A.Chromosome 13
B.Chromosome 21
C.Chromosome 18
D.Chromosome 23
Correct Answer: Chromosome 21
Explanation:
Down syndrome (trisomy 21) results from three copies of chromosome 21 instead of the usual two.
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8Trisomy 18 is also known as:
Trisomies - chromosome 13, 18, 21
Easy
A.Down syndrome
B.Turner syndrome
C.Edwards syndrome
D.Patau syndrome
Correct Answer: Edwards syndrome
Explanation:
Edwards syndrome is caused by an extra copy of chromosome 18 (trisomy 18).
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9Patau syndrome results from an extra copy of which chromosome?
Trisomies - chromosome 13, 18, 21
Easy
A.Chromosome 18
B.Chromosome X
C.Chromosome 13
D.Chromosome 21
Correct Answer: Chromosome 13
Explanation:
Patau syndrome (trisomy 13) is caused by three copies of chromosome 13.
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10Turner syndrome in humans has which sex chromosome constitution?
Sex linked aneuploidies, Turner, Klinefelter, superfemales
Easy
A.47, XXY
B.45, X (XO)
C.47, XYY
D.47, XXX
Correct Answer: 45, X (XO)
Explanation:
Turner syndrome results from a single X chromosome (45,X), giving affected individuals only 45 chromosomes.
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11Klinefelter syndrome is characterized by which chromosome constitution?
Sex linked aneuploidies, Turner, Klinefelter, superfemales
Easy
A.46, XY
B.47, XXY
C.47, XXX
D.45, X
Correct Answer: 47, XXY
Explanation:
Klinefelter syndrome occurs in males with an extra X chromosome, giving the karyotype 47,XXY.
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12The condition called superfemale (triple X syndrome) has which chromosome constitution?
Sex linked aneuploidies, Turner, Klinefelter, superfemales
Easy
A.45, X
B.47, XYY
C.47, XXX
D.47, XXY
Correct Answer: 47, XXX
Explanation:
Superfemales (triple X) have three X chromosomes, giving the karyotype 47,XXX.
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13Which sex-linked aneuploidy typically results in a phenotypic female who is often sterile and short in stature?
Sex linked aneuploidies, Turner, Klinefelter, superfemales
Easy
A.Klinefelter syndrome
B.Patau syndrome
C.Turner syndrome
D.Edwards syndrome
Correct Answer: Turner syndrome
Explanation:
Turner syndrome (45,X) produces females who are commonly short-statured and sterile.
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14The condition of having more than two complete sets of chromosomes is called:
Polypoidy in plants
Easy
A.Polyploidy
B.Monosomy
C.Aneuploidy
D.Trisomy
Correct Answer: Polyploidy
Explanation:
Polyploidy is the presence of three or more complete chromosome sets, such as triploid or tetraploid organisms.
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15Polyploidy is most commonly observed in:
Polypoidy in plants
Easy
A.Mammals
B.Reptiles
C.Plants
D.Birds
Correct Answer: Plants
Explanation:
Polyploidy is far more common and tolerated in plants, where it often increases size and yield.
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16Compared to plants, polyploidy in animals is generally:
Polypoidy in animals
Easy
A.Rare
B.Very common
C.Universal
D.Always beneficial
Correct Answer: Rare
Explanation:
Polyploidy is rare in animals because it often disrupts sex determination and development.
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17A mutation is best defined as a:
Mutations: Definition
Easy
A.Temporary change in protein shape
B.Change in cell size
C.Loss of a cell organelle
D.Heritable change in the DNA sequence
Correct Answer: Heritable change in the DNA sequence
Explanation:
A mutation is a permanent, heritable change in the nucleotide sequence of an organism's DNA.
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18A mutation involving the change of a single base pair in DNA is called a:
Types of mutation
Easy
A.Frameshift deletion
B.Point mutation
C.Genome duplication
D.Chromosomal mutation
Correct Answer: Point mutation
Explanation:
A point mutation affects a single nucleotide base pair in the DNA sequence.
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19A mutation that changes a codon into a stop codon, prematurely ending translation, is called a:
Molecular basis of mutation
Easy
A.Silent mutation
B.Nonsense mutation
C.Neutral mutation
D.Missense mutation
Correct Answer: Nonsense mutation
Explanation:
A nonsense mutation converts an amino-acid-coding codon into a stop codon, truncating the protein.
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20Which of the following is a physical mutagen that can induce mutations?
radiation and chemically induced mutation
Easy
A.Ultraviolet radiation
B.Oxygen
C.Water
D.Glucose
Correct Answer: Ultraviolet radiation
Explanation:
UV radiation is a physical mutagen that damages DNA, commonly by forming thymine dimers.
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21A chromosome with the gene order produces a mutant with the order . What structural change has occurred?
Variation in Chromosome Structure: Deletion
Medium
A.Paracentric inversion of segment
B.Reciprocal translocation of segment
C.Tandem duplication of segment
D.Interstitial deletion of segment
Correct Answer: Interstitial deletion of segment
Explanation:
Segments and are missing from the interior of the chromosome while the rest of the order is unchanged. Loss of an internal segment is an interstitial deletion.
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22The Bar eye phenotype in Drosophila, which reduces eye facet number, is classically explained by which structural chromosomal change?
Variation in Chromosome Structure: Duplication
Medium
A.A tandem duplication in the region of the X chromosome
B.A terminal deletion of the tip of the X chromosome that removes eye-determining genes
C.A reciprocal translocation between the X and an autosome
D.A pericentric inversion spanning the centromere of the X chromosome
Correct Answer: A tandem duplication in the region of the X chromosome
Explanation:
Bar results from a tandem duplication of the band on the X chromosome. Extra copies of the region alter eye development, a classic example of a position/dosage effect from duplication.
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23During meiosis in a heterozygote carrying a paracentric inversion, a single crossover within the inversion loop typically produces which abnormal products?
Variation in Chromosome Structure: Inversion
Medium
A.A ring chromosome and a linear fragment
B.A dicentric bridge and an acentric fragment
C.Two identical duplicated chromosomes
D.Two monocentric chromosomes with balanced gene content
Correct Answer: A dicentric bridge and an acentric fragment
Explanation:
In paracentric inversions the centromere lies outside the inverted region. A crossover inside the loop yields one dicentric chromatid (forming a bridge at anaphase) and one acentric fragment that is lost.
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24A phenotypically normal person carries a balanced reciprocal translocation. Why are they usually unaffected, yet at risk of having children with abnormalities?
Variation in Chromosome Structure: Translocation
Medium
A.They carry a deletion that is compensated by a duplication only in the germ line
B.They have no net loss or gain of genetic material, but abnormal segregation of translocated chromosomes in meiosis can produce unbalanced gametes
C.They are mosaic for the translocation, expressing it only in reproductive tissues
D.They have extra genetic material that is silenced in somatic cells but reactivated in gametes
Correct Answer: They have no net loss or gain of genetic material, but abnormal segregation of translocated chromosomes in meiosis can produce unbalanced gametes
Explanation:
In a balanced translocation the total gene content is normal, so the carrier is healthy. During meiosis, adjacent segregation of the quadrivalent can create gametes with duplications or deletions, leading to affected offspring.
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25If non-disjunction occurs during meiosis II in one secondary oocyte, what is the expected composition of the resulting gametes from that cell?
Variation in Chromosome Number: Non-disjunction and Aneuploidy
Medium
A.Four gametes all with an extra chromosome
B.Two normal and two gametes
C.Two normal, one , and one gamete
D.Two and two gametes
Correct Answer: Two normal, one , and one gamete
Explanation:
Meiosis I is normal, so sister chromatids fail to separate in only one of the two secondary cells. That cell gives and gametes, while the other divides normally to give two gametes.
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26A newborn presents with a single palmar crease, upslanting palpebral fissures, hypotonia, and intellectual disability. Which trisomy is most consistent with these findings?
Trisomies - chromosome 13, 18, 21
Medium
A.Trisomy X (Triple X syndrome)
B.Trisomy 13 (Patau syndrome)
C.Trisomy 21 (Down syndrome)
D.Trisomy 18 (Edwards syndrome)
Correct Answer: Trisomy 21 (Down syndrome)
Explanation:
A single palmar (simian) crease, upslanting palpebral fissures, hypotonia, and intellectual disability are hallmark features of Down syndrome, caused by an extra copy of chromosome 21.
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27Trisomies 13 and 18 have far lower survival than trisomy 21 primarily because
Trisomies - chromosome 13, 18, 21
Medium
A.Chromosomes 13 and 18 lack centromeres in trisomic cells
B.Chromosome 21 is not expressed in early embryos
C.Chromosomes 13 and 18 are larger and gene-richer than chromosome 21, so their imbalance disrupts more essential developmental pathways
D.Trisomies 13 and 18 always arise from paternal non-disjunction
Correct Answer: Chromosomes 13 and 18 are larger and gene-richer than chromosome 21, so their imbalance disrupts more essential developmental pathways
Explanation:
Chromosome 21 is the smallest autosome with relatively few genes, so its trisomy is more survivable. Chromosomes 13 and 18 carry many more genes, and their dosage imbalance is usually lethal in infancy.
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28A phenotypic female with short stature, webbed neck, and non-functional ovaries is found to have 45 chromosomes. Her karyotype is most likely
Sex linked aneuploidies, Turner, Klinefelter, superfemales
Medium
A.47,XXY
B.46,XX with an X deletion
C.45,X
D.47,XXX
Correct Answer: 45,X
Explanation:
Turner syndrome (45,X) results from the absence of one sex chromosome. Characteristic features include short stature, webbed neck, and gonadal dysgenesis (streak ovaries) with sterility.
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29How many Barr bodies are present in the somatic cells of an individual with Klinefelter syndrome (47,XXY)?
Sex linked aneuploidies, Turner, Klinefelter, superfemales
Medium
A.1
B.3
C.0
D.2
Correct Answer: 1
Explanation:
The number of Barr bodies equals the number of X chromosomes minus one. With two X chromosomes (XXY), Barr body is formed by X-inactivation.
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30A woman with a 47,XXX karyotype ("superfemale" / Triple X) is often only mildly affected because
Sex linked aneuploidies, Turner, Klinefelter, superfemales
Medium
A.All three X chromosomes fuse into a single functional chromosome
B.The additional X chromosomes are largely inactivated as Barr bodies, minimizing gene-dosage imbalance
C.The extra X chromosome is expelled from cells during early cleavage divisions
D.The Y chromosome compensates for the extra X
Correct Answer: The additional X chromosomes are largely inactivated as Barr bodies, minimizing gene-dosage imbalance
Explanation:
X-inactivation silences all but one X per cell, so extra X chromosomes are condensed into Barr bodies. This dosage compensation means 47,XXX individuals are usually only mildly affected.
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31An autotetraploid plant is produced by doubling the chromosome set of a diploid (). If the diploid has , how many chromosomes does the autotetraploid have?
Polypoidy in plants
Medium
A.28
B.14
C.21
D.7
Correct Answer: 28
Explanation:
An autotetraploid has four copies of each chromosome set (). Since the monoploid number is , the autotetraploid has chromosomes.
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32Which agent is most commonly used experimentally to induce polyploidy in plants by preventing spindle fiber formation?
Polypoidy in plants
Medium
A.Colchicine
B.Nitrous acid
C.5-bromouracil
D.Ethidium bromide
Correct Answer: Colchicine
Explanation:
Colchicine binds tubulin and blocks spindle assembly, so chromosomes duplicate but do not segregate. This doubles the chromosome number, making it a standard tool for inducing polyploidy.
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33Allopolyploids differ from autopolyploids in that allopolyploids
Polypoidy in plants
Medium
A.Always have an odd number of chromosome sets
B.Are invariably sterile regardless of chromosome pairing
C.Arise from doubling a single species' genome
D.Contain chromosome sets derived from two or more different species
Correct Answer: Contain chromosome sets derived from two or more different species
Explanation:
Allopolyploids combine genomes from different species (often via hybridization followed by doubling), whereas autopolyploids result from multiplying the chromosome sets of a single species.
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34Polyploidy is far rarer and usually lethal in animals compared with plants mainly because
Polypoidy in animals
Medium
A.Chromosomal sex-determination and dosage balance are disrupted by changes in ploidy in most animals
B.Plant genomes are always smaller than animal genomes
C.Animal cells cannot physically hold extra chromosomes
D.Animals lack spindle fibers during mitosis
Correct Answer: Chromosomal sex-determination and dosage balance are disrupted by changes in ploidy in most animals
Explanation:
Many animals rely on a fine sex-chromosome balance (e.g., X:autosome ratio) for sex determination and development. Polyploidy upsets this balance, so it is usually lethal, unlike in many plants.
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35Which statement best defines a mutation in the genetic sense?
Mutations: Definition
Medium
A.A heritable change in the nucleotide sequence of the genetic material
B.The reshuffling of alleles that occurs during crossing over
C.The movement of chromosomes to opposite poles during anaphase
D.Any temporary change in gene expression in response to the environment
Correct Answer: A heritable change in the nucleotide sequence of the genetic material
Explanation:
A mutation is a permanent, heritable alteration in the DNA sequence. Recombination and reversible expression changes are not mutations because they do not alter the underlying sequence heritably.
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36Insertion of a single nucleotide within the coding region of a gene most likely causes which type of mutation?
Types of mutation
Medium
A.A duplication of the entire gene
B.A silent mutation with no change to the protein
C.A conservative missense mutation
D.A frameshift mutation that alters all downstream codons
Correct Answer: A frameshift mutation that alters all downstream codons
Explanation:
Inserting one nucleotide (not a multiple of three) shifts the reading frame. Every codon after the insertion is changed, usually producing a nonfunctional protein and often a premature stop codon.
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37A point mutation changes the codon (Tyr) to (stop). This is best classified as a
Types of mutation
Medium
A.Missense mutation
B.Nonsense mutation
C.Frameshift mutation
D.Silent mutation
Correct Answer: Nonsense mutation
Explanation:
A base substitution that converts a sense codon into a stop codon is a nonsense mutation. It truncates the polypeptide, typically yielding a shortened, nonfunctional protein.
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38Sickle-cell anemia results from a single base substitution changing glutamic acid to valine in -globin. This is an example of a mutation whose phenotypic effect is
Phenotypic effects
Medium
A.A frameshift producing a completely unrelated protein
B.A missense mutation altering protein structure and function
C.A large deletion removing the entire globin gene
D.A silent mutation with no phenotypic consequence
Correct Answer: A missense mutation altering protein structure and function
Explanation:
The single amino-acid change (Glu → Val) is a missense mutation. It alters hemoglobin's structure, causing it to polymerize and distort red cells into a sickle shape.
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39The base analog 5-bromouracil (5-BU) induces mutations because it
Molecular basis of mutation
Medium
A.Removes amino groups from cytosine to form uracil
B.Resembles thymine but can tautomerize and mispair with guanine, causing transition mutations
C.Inserts between stacked bases and causes frameshift mutations
D.Forms covalent thymine dimers when exposed to UV light
Correct Answer: Resembles thymine but can tautomerize and mispair with guanine, causing transition mutations
Explanation:
5-BU is a thymine analog that is incorporated in place of thymine. Its enol tautomer pairs with guanine, leading to transition mutations during replication.
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40Ultraviolet (UV) light is a mutagen that primarily damages DNA by
radiation and chemically induced mutation
Medium
A.Alkylating guanine at the position
B.Deaminating adenine to hypoxanthine
C.Forming pyrimidine (thymine) dimers between adjacent bases
D.Causing double-strand breaks through ionization
Correct Answer: Forming pyrimidine (thymine) dimers between adjacent bases
Explanation:
UV radiation is non-ionizing and induces covalent bonds between adjacent pyrimidines, forming thymine dimers. These distort the helix and block replication unless repaired.
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41A pericentric inversion heterozygote forms an inversion loop during meiosis I. If a single crossover occurs within the inversion loop, what is the expected outcome for the resulting recombinant gametes?
Variation in Chromosome Structure: Inversion
Hard
A.Two recombinant chromatids, one dicentric and one acentric, with normal arm lengths
B.Four viable balanced gametes with reshuffled but complete gene content
C.Recombinant chromatids identical to parental types with no duplication or deletion
D.Two recombinant chromatids that are both duplicated and deficient, with an altered centromere position
Correct Answer: Two recombinant chromatids that are both duplicated and deficient, with an altered centromere position
Explanation:
In a pericentric inversion (centromere inside the inverted segment), a single crossover within the loop produces recombinant chromatids carrying duplications and deletions, and because the centromere lies within the loop, the recombinants show a changed arm-length ratio. Dicentric/acentric products are characteristic of paracentric inversions, not pericentric.
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42A phenotypically normal individual is a carrier of a balanced reciprocal translocation. During meiosis, the four chromosomes form a cross-shaped quadrivalent. Which segregation pattern yields genetically balanced gametes?
Variation in Chromosome Structure: Translocation
Hard
A.Adjacent-1 segregation
B.Adjacent-2 segregation
C.3:1 segregation
D.Alternate segregation
Correct Answer: Alternate segregation
Explanation:
In a translocation quadrivalent, only alternate segregation (in which the two normal chromosomes go to one pole and the two translocated chromosomes to the other) produces balanced gametes. Adjacent-1, adjacent-2, and 3:1 patterns all yield duplication/deficiency (unbalanced) gametes.
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43A woman with a Robertsonian translocation between chromosomes 14 and 21 [karyotype 45,XX,der(14;21)] is having children. Ignoring lethality and gametes lacking chromosome 21, what is the theoretical risk that a liveborn child has translocation Down syndrome, based on the balanced gamete types?
Trisomies - chromosome 21
Hard
A.Approximately
B.Approximately
C.Approximately
D.Approximately
Correct Answer: Approximately
Explanation:
The carrier produces six gamete types, but three lead to nonviable conceptuses (monosomy 21, monosomy 14, trisomy 14). Of the three viable outcomes—normal, balanced carrier, and translocation trisomy 21—one gives Down syndrome, giving a theoretical risk of about among liveborns (the observed empirical risk is lower).
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44A trisomic (2n+1) plant heterozygous for the extra chromosome (genotype ) is testcrossed to . Assuming the extra chromosome segregates as random chromosome pairs (only balanced gametes function equally) and considering only chromosome segregation, what fraction of the diploid () offspring will express the recessive phenotype?
Variation in Chromosome Number: Non-disjunction and Aneuploidy
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
An trisomic produces gametes by random pairing: among the (single) gametes, the ratio is . Crossed to , the offspring are of the diploid class. Because only half the functional gametes are (haploid), the recessive diploid fraction of total offspring works out to under the standard trisomic segregation model.
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45A patient has the karyotype 47,XXY. According to the Lyon hypothesis, how many Barr bodies are expected, and why does the individual still show a phenotype despite X-inactivation?
Sex linked aneuploidies, Turner, Klinefelter, superfemales
Hard
A.Zero Barr bodies; the Y chromosome blocks X-inactivation entirely
B.Two Barr bodies; the extra X is fully silenced so no phenotypic effect should occur
C.One Barr body; genes in the pseudoautosomal regions and those escaping inactivation remain expressed from both X's
D.One Barr body; the Y chromosome is inactivated instead of the second X
Correct Answer: One Barr body; genes in the pseudoautosomal regions and those escaping inactivation remain expressed from both X's
Explanation:
The number of Barr bodies equals (number of X chromosomes − 1), so 47,XXY gives one Barr body. A phenotype still arises because pseudoautosomal-region genes and other loci that escape inactivation are expressed from both X copies, causing a gene-dosage imbalance.
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46A point mutation changes the DNA triplet from to on the template strand. This substitution involves a purine replaced by a purine (). How is this mutation best classified at the molecular level?
Molecular basis of mutation
Hard
A.Silent mutation
B.Transition
C.Transversion
D.Frameshift
Correct Answer: Transition
Explanation:
A transition is a substitution of one purine for another purine () or one pyrimidine for another pyrimidine (). Since is purine-to-purine, this is a transition. A transversion would interchange a purine and a pyrimidine.
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47The chemical mutagen 5-bromouracil (5-BU) induces mutations primarily because:
radiation and chemically induced mutation
Hard
A.Its enol tautomer mispairs with guanine, causing transitions over successive replications
B.It forms covalent pyrimidine dimers that block replication forks
C.It deaminates cytosine to uracil producing transitions
D.It intercalates between base pairs causing single-base insertions and frameshifts
Correct Answer: Its enol tautomer mispairs with guanine, causing transitions over successive replications
Explanation:
5-BU is a thymine analog. In its common keto form it pairs with adenine, but its enol tautomer mispairs with guanine. Over successive replications this shifts an pair to a pair—a transition. Intercalation describes acridines; dimers describe UV; deamination describes nitrous acid.
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48An allotetraploid was formed by chromosome doubling of a sterile hybrid between species A () and species B (). What is the somatic chromosome number of the fertile allotetraploid?
Polypoidy in plants
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
The sterile hybrid contains one haploid set from each parent: chromosomes. Doubling this hybrid (amphidiploidy) yields chromosomes, restoring a complete pairing partner for every chromosome and restoring fertility.
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49A single base insertion occurs near the 5' end of a coding sequence, but a compensating single base deletion occurs a few codons downstream. What is the most likely consequence?
Types of mutation
Hard
A.Transcription is completely blocked because the promoter is disrupted
B.The entire downstream sequence is scrambled and a nonfunctional protein always results
C.Only the codons between the two events are altered; reading frame is restored downstream, often preserving protein function
D.A single amino acid is substituted with no frameshift at all
Correct Answer: Only the codons between the two events are altered; reading frame is restored downstream, often preserving protein function
Explanation:
An insertion followed by a nearby deletion (a net +1 then −1) restores the original reading frame beyond the deletion. Only the short stretch between the two mutations is mis-read, so the protein is often largely functional—this is the basis of intragenic suppression demonstrated in Crick's experiments.
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50In a Drosophila mapping experiment, a recessive allele becomes phenotypically expressed in a heterozygote that carries a chromosomal deletion on the homolog. What does this phenomenon reveal?
Variation in Chromosome Structure: Deletion
Hard
A.Pseudodominance, indicating the dominant wild-type allele lies within the deleted region
B.A new dominant mutation arising at the deletion breakpoint
C.Position effect variegation due to heterochromatin spreading
D.Incomplete dominance caused by dosage of the deletion
Correct Answer: Pseudodominance, indicating the dominant wild-type allele lies within the deleted region
Explanation:
Pseudodominance occurs when a recessive allele is expressed because its dominant counterpart on the homologous chromosome has been removed by a deletion. This is used to physically map genes to specific deleted regions (deletion mapping).
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51The Bar eye phenotype in Drosophila results from a tandem duplication that can undergo unequal crossing over. When two Bar chromosomes pair and unequal crossing over occurs, one product is Double-Bar and the other is:
Variation in Chromosome Structure: Duplication
Hard
A.Ultrabar with an additional triplication
B.Lethal due to acentric fragment loss
C.A ring chromosome carrying no Bar region
D.Wild-type (reverted to normal eye)
Correct Answer: Wild-type (reverted to normal eye)
Explanation:
Unequal crossing over between two Bar (duplicated) segments produces one chromosome with three copies (Double-Bar) and one chromosome with a single copy—i.e., wild-type revertant. This reciprocal outcome demonstrated the mechanism of unequal crossing over in generating copy-number variation.
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52Patau syndrome (trisomy 13) and Edwards syndrome (trisomy 18) can both arise from meiotic non-disjunction. Which statement best distinguishes the cytogenetic origin contributing to advanced maternal age effects in these autosomal trisomies?
Trisomies - chromosome 13
Hard
A.Most result from paternal meiosis II non-disjunction independent of parental age
B.Most result from maternal meiosis I non-disjunction, linked to reduced recombination in aged oocytes
C.Most arise post-zygotically as somatic mosaicism unrelated to gametogenesis
D.Most originate from Robertsonian translocations regardless of maternal age
Correct Answer: Most result from maternal meiosis I non-disjunction, linked to reduced recombination in aged oocytes
Explanation:
The majority of autosomal trisomies, including trisomy 13, 18, and 21, arise from maternal meiosis I non-disjunction. Aberrant or reduced recombination in oocytes arrested since fetal life makes bivalents prone to mis-segregation, explaining the strong maternal-age correlation.
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53Polyploidy is far rarer and usually less tolerated in animals than in plants. Which explanation is the primary reason?
Polypoidy in animals
Hard
A.Animals lack meiosis and therefore cannot form unreduced gametes
B.Chromosomal sex-determination systems are disrupted by altered X:autosome ratios, causing sterility or lethality
C.Plant cell walls physically prevent chromosome doubling in animals
D.Animal genomes are always smaller, leaving no room for extra chromosomes
Correct Answer: Chromosomal sex-determination systems are disrupted by altered X:autosome ratios, causing sterility or lethality
Explanation:
Most animals have chromosomal (often dosage-based) sex determination. Polyploidy upsets the balance between sex chromosomes and autosomes (e.g., the X:autosome ratio), producing intersexes, sterility, or lethality—so polyploidy is much less viable in animals than in plants, which tolerate it well.
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54A missense mutation replaces a buried hydrophobic valine with a charged glutamate deep in a protein's core, whereas the same amino-acid change on the surface has little effect. This illustrates that the phenotypic severity of a missense mutation depends most on:
Phenotypic effects
Hard
A.Whether the mutation is a transition or a transversion
B.The structural/functional context of the affected residue within the protein
C.Only the number of nucleotides changed in the codon
D.The distance of the codon from the transcription start site
Correct Answer: The structural/functional context of the affected residue within the protein
Explanation:
The impact of a missense change depends on where the residue sits and its role. Burying a charged residue in a hydrophobic core disrupts folding, while a surface substitution is often tolerated. Thus context—not merely the type of nucleotide change—governs phenotypic consequence.
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55A DNA sequence contains the run . During replication, strand slippage in such a short tandem repeat most commonly produces:
Molecular basis of mutation
Hard
A.Deamination of every cytosine in the tract
B.Expansion or contraction of the repeat number (dynamic mutation)
C.A transition at the first base of each repeat unit
D.A pyrimidine dimer spanning two repeats
Correct Answer: Expansion or contraction of the repeat number (dynamic mutation)
Explanation:
Tandem repeats promote replication slippage, where the nascent and template strands mispair by one or more repeat units, adding or removing repeats. This dynamic mutation underlies trinucleotide-repeat expansion diseases such as Huntington's disease.
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56A 45,X (Turner) individual and a 47,XXX (triple-X/superfemale) individual differ strikingly in phenotypic severity, with Turner being far more clinically significant. The best explanation is:
Sex linked aneuploidies, Turner, Klinefelter, superfemales
Hard
A.The Y chromosome in Turner syndrome disrupts development
B.Genes escaping X-inactivation must be present in two doses for normal development, so a single X causes haploinsufficiency
C.Turner individuals have extra autosomes not present in triple-X
D.Triple-X always silences all three X chromosomes completely while Turner cannot inactivate any
Correct Answer: Genes escaping X-inactivation must be present in two doses for normal development, so a single X causes haploinsufficiency
Explanation:
Even though only one X is active, roughly 15% of X-linked genes escape inactivation and require two doses. In 45,X these genes (including pseudoautosomal loci like SHOX) are present in only one copy, causing haploinsufficiency. In 47,XXX the extra X is largely inactivated, giving a milder phenotype.
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57Why do paracentric inversion heterozygotes appear as "crossover suppressors" in genetic mapping, even though crossing over still physically occurs within the loop?
Variation in Chromosome Structure: Inversion
Hard
A.Crossovers are converted into gene conversions with no exchange
B.The inversion prevents synapsis, so no crossovers can initiate
C.Recombination enzymes are physically excluded from inverted regions
D.Recombinant chromatids form dicentric bridges and acentric fragments that are lost, so only parental-type gametes survive
Correct Answer: Recombinant chromatids form dicentric bridges and acentric fragments that are lost, so only parental-type gametes survive
Explanation:
In a paracentric inversion, a single crossover in the loop generates a dicentric chromatid (breaks at anaphase) and an acentric fragment (lost). These recombinant products are non-viable, so recovered gametes are almost all parental type—giving the appearance of suppressed recombination.
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58A child has the karyotype 47,XY,+21 but the parents are chromosomally normal. DNA polymorphism analysis shows the child inherited two different maternal alleles (both maternal homologs) at centromeric loci of chromosome 21. This indicates the non-disjunction occurred at:
Variation in Chromosome Number: Non-disjunction and Aneuploidy
Hard
A.Maternal meiosis I
B.A post-zygotic mitotic division
C.Paternal meiosis I
D.Maternal meiosis II
Correct Answer: Maternal meiosis I
Explanation:
Two different maternal homologs (heterozygosity retained at the centromere) means the homologous chromosomes failed to separate—diagnostic of meiosis I non-disjunction. Meiosis II errors would give two identical (sister) centromeric alleles, i.e., homozygosity at the centromere.
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59Autopolyploids often show reduced fertility compared to their diploid progenitors. In an autotetraploid, the primary cytological cause of this reduced fertility is:
Polypoidy in plants
Hard
A.Absence of any chromosome pairing during prophase I
B.Formation of multivalents (e.g., quadrivalents) leading to unbalanced chromosome segregation
C.Complete failure of DNA replication before meiosis
D.Loss of the centromere function in all homologs
Correct Answer: Formation of multivalents (e.g., quadrivalents) leading to unbalanced chromosome segregation
Explanation:
With four identical homologs, pairing produces multivalents (quadrivalents) or unpaired univalents. These often segregate unevenly at anaphase I, yielding aneuploid, unbalanced gametes and thus reduced fertility—unlike allopolyploids where distinct genomes pair as bivalents.
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60Ionizing radiation (e.g., X-rays) and ultraviolet (UV) light both damage DNA but by different primary mechanisms. Which pairing correctly matches the agent to its characteristic lesion?
Ionizing radiation is highly energetic and causes strand breaks, including double-strand breaks and gross chromosomal rearrangements. UV light (non-ionizing) is absorbed by adjacent pyrimidines, forming cyclobutane pyrimidine dimers (and 6-4 photoproducts) that distort the helix and block replication.
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