Examkin

Gene Mutations and Chromosomal Aberrations

17 soru

Tüm türler17Çoktan seçmeli10Doğru / Yanlış5Eşleştirme2
Soru 1Soru →

In human genetics, sickle-cell anaemia and Down syndrome represent two distinct types of genetic disorders. Which of the following statements correctly differentiates between the underlying genetic causes of these two conditions?

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Cevap: Sickle-cell anaemia is caused by a point mutation in a single gene, whereas Down syndrome is caused by a chromosomal numerical aberration.

Cevap

Sickle-cell anaemia is caused by a point mutation in a single gene, whereas Down syndrome is caused by a chromosomal numerical aberration.
The correct response accurately distinguishes the scope of the genetic error. Sickle-cell anaemia is a classic gene mutation resulting from a single point substitution in the DNA sequence coding for beta-globin. Down syndrome is a chromosomal numerical aberration (aneuploidy/trisomy 21) resulting from the failure of chromosome pair 21 to separate during gamete formation (non-disjunction).

Adım Adım Çözüm

1
Analyze the cause of sickle-cell anaemia
Determined that sickle-cell anaemia is a gene mutation caused by a single nucleotide substitution (GAG to GUG) in the hemoglobin gene.
Gene mutations involve alterations in the molecular sequence of nucleotides within a single gene.
2
Analyze the cause of Down syndrome
Determined that Down syndrome is a chromosomal aberration involving an extra chromosome 21 (trisomy 21) caused by non-disjunction during meiosis.
Chromosomal aberrations involve changes in the number or gross structure of whole chromosomes rather than single nucleotide changes.
3
Compare the classifications
Matched sickle-cell anaemia to gene mutation and Down syndrome to chromosomal numerical aberration.
This establishes the fundamental distinction between point gene mutations and numerical chromosomal abnormalities.

Anahtar Kavram

Distinction between Gene Mutations and Chromosomal Aberrations
Tahmini Süre:1m 0s
Soru 2Soru →

A reciprocal translocation between non-homologous chromosomes alters the arrangement of genes on the affected chromosomes without changing the total count of chromosomes in a diploid cell, whereas meiotic non-disjunction causes aneuploidy by altering the total chromosome number in the resulting gametes.

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Cevap: True

Cevap

The statement is True.
Reciprocal translocation represents a structural chromosomal mutation where segments of non-homologous chromosomes are exchanged without altering the total chromosome count (2n2n). Non-disjunction, on the other hand, is a nuclear division error where chromosomes fail to segregate properly during Anaphase I or II of meiosis, producing gametes with missing or extra whole chromosomes, which leads to numerical aneuploidy (such as Down syndrome, 47,+2147, +21).

Adım Adım Çözüm

1
Analyze the nature of reciprocal translocation
Reciprocal translocation involves non-homologous chromosomes swapping segments. This rearranges gene loci (a structural alteration) but leaves the total count of chromosomes intact.
Structural chromosomal aberrations modify chromosome architecture rather than chromosome complement number.
2
Analyze the mechanism and outcome of meiotic non-disjunction
Non-disjunction occurs when homologous chromosomes fail to segregate in Anaphase I or sister chromatids fail to separate in Anaphase II, yielding abnormal gametes (n+1n+1 or n−1n-1).
Failure of proper spindle separation directly alters gametic chromosome counts, causing numerical aberrations (aneuploidy) upon fertilization.
3
Evaluate the complete comparative statement
The distinction drawn between structural alteration (translocation preserving number) and numerical alteration (non-disjunction causing aneuploidy) is scientifically accurate.
Both biological concepts are correctly defined and contrasted.

Anahtar Kavram

Distinction between structural chromosomal aberrations (e.g., translocation) and numerical chromosomal aberrations (e.g., aneuploidy via non-disjunction)
Soru 3Soru →

Match each type of gene mutation or chromosomal aberration on the left with its corresponding description on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Öğeler

Substitution mutation
Frameshift mutation
Aneuploidy
Polyploidy

Eşleşmeler

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Cevap

Substitution mutation matches replacement of a single nucleotide base; Frameshift mutation matches addition or deletion altering the reading frame; Aneuploidy matches gain or loss of specific individual chromosomes (2n+12n+1 or 2n−12n-1); Polyploidy matches possession of extra complete chromosome sets (3n3n or 4n4n).
Substitution mutation corresponds to swapping a single nucleotide base for another. Frameshift mutation occurs when additions or deletions of nucleotides shift the triplet codon reading sequence. Aneuploidy describes the loss or gain of individual chromosomes, while polyploidy describes organisms having additional full sets of chromosomes.

Adım Adım Çözüm

1
Distinguish between gene mutations and numerical chromosomal aberrations.
Substitution and frameshift are point gene mutations, while aneuploidy and polyploidy involve changes in chromosome number.
Gene mutations affect nucleotide sequences within a gene, whereas numerical aberrations alter chromosome counts.
2
Identify the mechanisms of point mutations.
Substitution replaces a single base, whereas frameshift adds or deletes bases shifting downstream codons.
Codons are read in triplets, so non-multiple-of-three insertions or deletions disrupt all subsequent amino acid translation.
3
Distinguish between aneuploidy and polyploidy.
Aneuploidy affects individual chromosome counts (2n±12n \pm 1), whereas polyploidy involves entire extra sets of chromosomes (3n,4n3n, 4n).
Non-disjunction of a single chromosome pair causes aneuploidy, whereas failure of spindle formation during total cell division leads to polyploidy.

Anahtar Kavram

Gene Mutations vs Chromosomal Numerical Aberrations
Soru 4Soru →

During gametogenesis, non-disjunction of chromosome 21 occurs specifically during Meiosis I. If all resulting gametes are subsequently fertilized by normal haploid gametes, what percentage of the produced zygotes will have trisomy 21?

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Cevap: 50%

Cevap

50% of the resulting zygotes will exhibit trisomy 21.
When non-disjunction of a chromosome pair takes place during Meiosis I, the pair fails to segregate. Consequently, one daughter cell receives both homologous chromosomes while the other receives none. Following normal chromatid separation in Meiosis II, half of the gametes carry an extra chromosome (n+1n+1) and half are deficient by one chromosome (n−1n-1). Fertilization of the n+1n+1 gametes with normal haploid gametes (nn) produces trisomic (2n+12n+1) zygotes. Therefore, 50% of the zygotes will have trisomy 21.

Adım Adım Çözüm

1
Analyze the meiotic stage where non-disjunction occurs
In Meiosis I, non-disjunction means homologous chromosomes fail to separate into distinct daughter cells.
Understanding the exact stage of meiotic failure determines the chromosome distribution in the daughter cells after Meiosis I.
2
Determine the chromosomal content of the resulting four gametes after Meiosis II
Meiosis I yields one cell with an extra chromosome (n+1n+1) and one cell missing that chromosome (n−1n-1). Normal chromatid separation in Meiosis II creates two (n+1)(n+1) gametes and two (n−1)(n-1) gametes.
All four gametes produced are genetically abnormal after a Meiosis I non-disjunction event.
3
Calculate the proportion of trisomic zygotes post-fertilization
Fertilization of the two (n+1)(n+1) gametes by normal haploid (nn) gametes gives (2n+1)(2n+1), which is trisomy 21. Two out of four total zygotes (50%) will be trisomic (and 50% will be monosomic).
Trisomy requires the combination of an (n+1)(n+1) gamete with a normal nn gamete.

Anahtar Kavram

Chromosomal Non-disjunction and Aneuploidy
Soru 5Soru →

A paracentric inversion is a structural chromosomal aberration involving two double-strand breaks on a chromosome followed by a 180-degree rotation of the detached segment including the centromere, thereby altering the chromosome's relative arm length ratio.

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Cevap: False

Cevap

The statement is False. A paracentric inversion occurs within a single arm of a chromosome and does not involve the centromere. A pericentric inversion is the type of inversion that includes the centromere and can change the arm length ratio.
The statement is false because a paracentric inversion is restricted to a single chromosome arm and does not incorporate the centromere. Therefore, it cannot alter the centromere's position or the arm ratio of the chromosome.

Adım Adım Çözüm

1
Define structural chromosomal inversions
Inversions occur when a segment of a chromosome breaks at two points, rotates 180 degrees, and reinserts into the chromosome.
Establishing the general mechanism of inversion aberrations.
2
Distinguish between paracentric and pericentric inversions based on centromere involvement
Para- means 'next to' or 'beside' (confined to one arm, excluding the centromere). Peri- means 'around' (spans across the centromere).
Identifying which structural inversion type contains the centromere.
3
Evaluate the morphological impact on chromosome arm ratios
Because paracentric inversions take place entirely within one arm, the centromere remains intact in its original location, leaving the relative lengths of the short (pp) and long (qq) arms unchanged. Pericentric inversions can change the centromere position relative to the ends, changing arm ratios.
Determining the truth value of the stem's claim.

Anahtar Kavram

Paracentric versus Pericentric Chromosomal Inversions
Tahmini Süre:1m 30s
Soru 6Soru →

During DNA replication, exposure to an alkylating agent causes the insertion of a single extra nucleotide base into the coding region of a functional gene, while an error during spindle fiber assembly in meiosis causes two homologous chromosomes to fail to separate. Which of the following statements correctly distinguishes the molecular nature and scope of these two genetic events?

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Cevap: The nucleotide insertion causes a frameshift gene mutation that alters the reading frame of a single protein, whereas non-disjunction causes a numerical chromosomal aberration altering total chromosome count.

Cevap

The nucleotide insertion causes a frameshift gene mutation that alters the reading frame of a single protein, whereas non-disjunction causes a numerical chromosomal aberration altering total chromosome count.
A single nucleotide addition within a gene alters the codon triplet reading frame (frameshift mutation), affecting only that specific gene product. In contrast, non-disjunction involves the failure of chromosome separation, leading to aneuploidy, which is a numerical chromosomal aberration.

Adım Adım Çözüm

1
Classify the single nucleotide insertion event.
Insertion of a single base into a gene's coding sequence alters the triplet codon reading frame during translation (frameshift gene mutation).
Gene mutations involve chemical or sequence changes within localized nucleotides of a single gene.
2
Classify the homologous chromosome non-separation event during meiosis.
Failure of homologous chromosomes to separate is termed non-disjunction, resulting in gametes with extra or missing whole chromosomes (aneuploidy).
Non-disjunction affects macro-structures (whole chromosomes), making it a numerical chromosomal aberration.
3
Compare the scope and classification of both events.
The insertion is a gene-level mutation affecting one polypeptide, while non-disjunction is a chromosomal aberration affecting total chromosome number.
Gene mutations affect nucleotide sequences, whereas chromosomal aberrations affect gross chromosome structure or number.

Anahtar Kavram

Distinction between Gene Mutations and Chromosomal Aberrations
Tahmini Süre:1m 30s
Soru 7Soru →

Match each type of genetic alteration listed on the left with its precise molecular or cytogenetic mechanism on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Öğeler

Transition mutation
Transversion mutation
Pericentric inversion
Robertsonian translocation

Eşleşmeler

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Cevap

Transition mutation matches replacement of a purine by another purine (or pyrimidine by pyrimidine). Transversion mutation matches substitution of a purine with a pyrimidine (or vice versa). Pericentric inversion matches chromosomal breaks flanking the centromere with 180∘180^\circ inversion containing the centromere. Robertsonian translocation matches centromeric fusion of two acrocentric long arms resulting in loss of short arms and reduced chromosome number.
Transition mutation corresponds to swapping purine-for-purine (A↔GA \leftrightarrow G) or pyrimidine-for-pyrimidine (C↔TC \leftrightarrow T). Transversion mutation corresponds to swapping purines for pyrimidines (A/G↔C/TA/G \leftrightarrow C/T). Pericentric inversion includes the centromere between two break points prior to rotation. Robertsonian translocation specifically joins the qq arms of acrocentric chromosomes near the centromere, shedding the non-essential heterochromatic pp arms.

Adım Adım Çözüm

1
Classify point mutations by chemical base structure alteration
Transition mutations exchange like-for-like ring structures (purine to purine or pyrimidine to pyrimidine), whereas transversion mutations swap single-ring pyrimidines with double-ring purines or vice versa.
This establishes the precise molecular distinction between point substitution categories.
2
Differentiate structural chromosomal inversions
Pericentric inversions involve breaks on both sides of the centromere (including it in the inverted segment), unlike paracentric inversions which occur entirely within one chromosome arm (excluding the centromere).
Including the centromere can alter arm ratios and morphological appearance of the chromosome.
3
Identify special translocation mechanisms involving acrocentric chromosomes
Robertsonian translocation specifically involves breakage near centromeres of acrocentric chromosomes, causing long arms to fuse into a single metacentric or submetacentric chromosome.
This reduces the overall functional chromosome count (2n=452n = 45 in balanced carriers).

Anahtar Kavram

Distinction between point gene mutation mechanisms (transitions vs transversions) and structural/numerical chromosomal aberrations (pericentric inversions vs Robertsonian translocations).
Tahmini Süre:2m 0s
Soru 8Soru →

In a molecular genetics analysis of a patient, a single nucleotide substitution is detected where adenine is replaced by thymine in the sixth codon of the β\beta-globin gene, causing glutamic acid to be replaced by valine. Which of the following correctly classifies this genetic change and the pattern of variation its resulting phenotype displays in human populations?

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Cevap: A gene mutation resulting in discontinuous variation

Cevap

The genetic change is classified as a gene mutation and the resulting phenotype displays discontinuous variation.
The substitution of a single nitrogenous base in a codon alters the amino acid sequence of a specific polypeptide without altering the macroscopic structure or number of chromosomes, defining it as a gene (point) mutation. Because the resulting sickle-cell condition produces distinct, clear-cut phenotypic classes (normal, carrier, affected) without intermediate continuum states, it represents discontinuous variation.

Adım Adım Çözüm

1
Classify the type of genetic modification
Replacing a single nucleotide base (adenine with thymine) within the coding sequence of the β\beta-globin gene constitutes a point mutation (gene mutation), as it affects only the nucleotide sequence of one gene without altering chromosome structure or number.
Gene mutations involve localized alterations in the DNA base sequence of a single gene locus.
2
Determine the resulting pattern of genetic variation
Sickle-cell trait/anemia displays distinct, non-overlapping phenotypic categories (unaffected, sickle-cell trait carrier, or sickle-cell anemia).
Traits governed by single gene loci with clear discrete phenotypic classes exemplify discontinuous variation.

Anahtar Kavram

Gene mutation vs chromosomal aberration and its phenotypic expression as discontinuous variation
Tahmini Süre:2m 0s
Soru 9Soru →

Sickle cell anaemia is a well-known inherited disease caused by a point mutation in the gene encoding the β\beta-globin chain of human haemoglobin. Which of the following alterations at the primary protein structure level directly causes the formation of abnormal haemoglobin S (HbS)?

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Cevap: Replacement of glutamic acid with valine at the sixth position of the β\beta-globin polypeptide chain

Cevap

The formation of abnormal haemoglobin S (HbS) in sickle cell anaemia is caused by the replacement of glutamic acid with valine at the sixth position of the β\beta-globin polypeptide chain.
Sickle cell anaemia is caused by a single nucleotide substitution in the β\beta-globin gene located on chromosome 11, where adenine is substituted by thymine. This alters the mRNA codon from GAG to GUG, causing valine to replace glutamic acid at the 6th position of the β\beta-chain, leading to polymerisation of haemoglobin molecules under low oxygen tension.

Adım Adım Çözüm

1
Identify the genetic nature of sickle cell anaemia
It is a gene (point) mutation caused by a single base pair substitution in the DNA sequence of the β\beta-globin gene.
Understanding whether a trait is caused by a point mutation or structural/numerical chromosomal aberration narrows down the biological mechanisms.
2
Determine the molecular consequence at the protein level
The codon GAG (coding for glutamic acid) is mutated to GUG (coding for valine) at codon position 6 of the β\beta-globin polypeptide.
Replacing a hydrophilic amino acid (glutamic acid) with a hydrophobic amino acid (valine) alters the solubility and structural properties of haemoglobin under low oxygen conditions.

Anahtar Kavram

Gene Point Mutation and Molecular Consequences in Sickle Cell Anaemia
Tahmini Süre:1m 0s
Soru 10Soru →

During a plant breeding experiment, a researcher treats dividing meristematic cells of a diploid crop species (2n=142n = 14) with the chemical mutagen colchicine. Colchicine inhibits microtubule polymerization and prevents the assembly of the mitotic spindle apparatus during cell division, preventing chromatid separation into daughter nuclei. Which of the following correctly describes the chromosome count and the class of mutation present in the resulting cells?

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Cevap: 28 chromosomes, resulting from polyploidy which is a chromosomal aberration

Cevap

28 chromosomes, resulting from polyploidy which is a chromosomal aberration
Colchicine interferes with microtubule assembly, disabling spindle fibers during mitosis. As a result, duplicated sister chromatids are retained within a single cell nucleus instead of being pulled to opposite poles. For an initial diploid cell (2n=142n = 14), this failure of segregation results in tetraploidy (4n=284n = 28). Because this change involves entire sets of chromosomes rather than alterations within single nucleotide chains, it is classified as a numerical chromosomal aberration (polyploidy).

Adım Adım Çözüm

1
Analyze the action of colchicine on dividing cells
Colchicine prevents spindle fiber formation, blocking anaphase separation of sister chromatids.
Microtubules cannot assemble, so duplicated chromosomes remain together in a single nucleus without cytokinesis.
2
Calculate the resulting chromosome number
The diploid number 2n=142n = 14 is doubled to 4n=284n = 28.
Since DNA replication occurred before mitosis and sister chromatids failed to segregate into separate daughter cells, the chromosome set doubles.
3
Classify the type of genetic mutation
Numerical doubling of whole chromosome sets is classified as polyploidy, a major chromosomal aberration.
Gene mutations affect nucleotide sequences within individual genes, whereas numerical alterations of whole chromosome sets are chromosomal aberrations.

Anahtar Kavram

Polyploidy and Numerical Chromosomal Aberrations
Tahmini Süre:1m 30s
Soru 11Soru →

A cytogenetic analysis of a patient reveals a somatic cell chromosomal complement of 4747 chromosomes with an additional sex chromosome (47,XXY47, XXY). Which mechanism is directly responsible for this structural or numerical chromosomal aberration, and what clinical condition does it produce?

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Cevap: Nondisjunction of sex chromosomes during meiotic division, resulting in Klinefelter syndrome

Cevap

Nondisjunction of sex chromosomes during meiotic division, resulting in Klinefelter syndrome
The correct option correctly attributes the karyotype 47,XXY47, XXY to nondisjunction during gametogenesis. When pair 23 fails to disjoin properly during meiosis, an egg carrying XXXX fertilized by a YY sperm (or an XX egg fertilized by an XYXY sperm) produces an individual with 4747 chromosomes (47,XXY47, XXY), which manifests as Klinefelter syndrome.

Adım Adım Çözüm

1
Analyze the given karyotype
The individual has 4747 chromosomes including an extra X chromosome (47,XXY47, XXY).
Normal human somatic cells have 4646 chromosomes (4444 autosomes + 22 sex chromosomes).
2
Identify the genetic mechanism leading to gain of a whole chromosome
The mechanism is meiotic nondisjunction, where chromosomes fail to separate properly during Anaphase I or II.
Gene mutations (substitutions, insertions, deletions) alter nucleotide sequences within a single gene but do not change the total number of chromosomes.
3
Match karyotype 47,XXY47, XXY to its clinical condition
An extra X chromosome in a male (47,XXY47, XXY) causes Klinefelter syndrome.
Trisomy 21 causes Down syndrome, whereas sex-chromosome trisomy 47,XXY47, XXY defines Klinefelter syndrome.

Anahtar Kavram

Chromosomal Aberrations and Nondisjunction
Tahmini Süre:1m 0s
Soru 12Soru →

Down syndrome is a genetic disorder caused by a single nucleotide point mutation in a nuclear gene.

Cevabı ve açıklamayı göster

Cevap: False

Cevap

False
The statement is false because Down syndrome is caused by a chromosomal aberration (trisomy 21 due to non-disjunction), not by a single nucleotide gene mutation.

Adım Adım Çözüm

1
Identify the genetic cause of Down syndrome
Down syndrome is caused by non-disjunction of chromosome 21 during meiosis, leading to trisomy 21 (47 chromosomes in total).
Down syndrome involves an extra chromosome set rather than a change in a single gene sequence.
2
Classify the mutation type
An alteration in total chromosome count is classified as a numerical chromosomal aberration, not a gene (point) mutation.
Gene mutations involve point alterations, insertions, or deletions within a specific gene sequence, whereas chromosomal aberrations affect whole chromosomes or large segments.

Anahtar Kavram

Distinction between gene mutations and chromosomal aberrations
Soru 13Soru →

Which of the following mutagenic agents primarily induces gene mutations by causing adjacent thymine bases on a single DNA strand to covalently bond and form pyrimidine dimers?

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Cevap: Ultraviolet radiation

Cevap

Ultraviolet radiation is the mutagenic agent that causes adjacent thymine bases on a DNA strand to form pyrimidine dimers.
Ultraviolet radiation is absorbed strongly by pyrimidines, promoting covalent linkage between adjacent thymine bases on the same DNA strand. This forms thymine dimers that block normal DNA replication and transcription.

Adım Adım Çözüm

1
Identify the primary mechanism of action described in the stem.
The mechanism is the formation of intrastrand thymine (pyrimidine) dimers due to absorption of radiant energy.
Different physical and chemical mutagens operate through distinct biochemical pathways on nucleic acids.
2
Evaluate physical versus chemical mutagens.
Non-ionizing ultraviolet (UV) light specifically excites pyrimidine rings (thymine and cytosine), leading to covalent dimer formation.
UV light lacks the energy to ionize atoms or break double strands, but has sufficient energy to alter nitrogenous base bonding.

Anahtar Kavram

Physical mutagens and the molecular mechanism of UV-induced pyrimidine dimerization
Soru 14Soru →

Nitrous acid (HNO2\text{HNO}_2) acts as a chemical mutagen by inducing the oxidative deamination of cytosine, converting it into uracil (UU). If a double-stranded DNA molecule containing a newly formed uracil base undergoes two consecutive rounds of DNA replication without prior enzymatic repair, which specific base-pair substitution will ultimately be fixed in the mutant daughter DNA molecule?

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Cevap: A C⋅G→T⋅AC\cdot G \rightarrow T\cdot A transition mutation

Cevap

A C⋅G→T⋅AC\cdot G \rightarrow T\cdot A transition mutation will be fixed in the daughter DNA molecule.
Deamination of cytosine creates uracil. Uracil pairs with adenine during the first replication round. In the subsequent round, adenine pairs with thymine, substituting the original cytosine-guanine pair with a thymine-adenine pair (C⋅G→T⋅AC\cdot G \rightarrow T\cdot A).

Adım Adım Çözüm

1
Identify the primary chemical effect of nitrous acid on cytosine.
Nitrous acid removes an amino group (oxidative deamination) from cytosine, turning it into uracil (UU).
Uracil is a nitrogenous base with hydrogen-bonding properties similar to thymine.
2
Trace the base pairing during the first cycle of DNA replication.
The template strand containing uracil (UU) pairs with adenine (AA) during DNA synthesis, producing a U−AU-A base pair intermediate.
DNA polymerase recognizes uracil as equivalent to thymine during template reading.
3
Trace the base pairing during the second cycle of DNA replication.
The strand with adenine (AA) serves as a template and pairs with thymine (TT), producing a stable A−TA-T (or T−AT-A) double helix.
Standard complementary base pairing pairs adenine with thymine in normal DNA.
4
Compare the original wild-type base pair with the final mutant base pair.
The original C⋅GC\cdot G base pair has been replaced by a T⋅AT\cdot A base pair. Because a pyrimidine-purine pair is replaced by another pyrimidine-purine pair, it is classified as a transition mutation.
Point mutations that exchange a pyrimidine for a pyrimidine (C→TC \rightarrow T) are transitions.

Anahtar Kavram

Chemical Mutagenesis and Base-Pair Transition Mutations
Tahmini Süre:2m 0s
Soru 15Soru →

During meiotic cell division, the failure of homologous chromosomes or sister chromatids to separate properly results in gametes with abnormal numbers of chromosomes. What is the precise biological term for this failure of separation?

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Cevap: Nondisjunction

Cevap

Nondisjunction is the failure of homologous chromosomes or sister chromatids to separate properly during meiotic cell division.
Nondisjunction specifically describes the failure of homologous chromosomes to separate during Anaphase I or sister chromatids to separate during Anaphase II of meiosis, resulting in daughter cells with missing or extra chromosomes.

Adım Adım Çözüm

1
Identify the chromosomal process described in the stem.
The stem describes an error during meiosis where chromosomes or chromatids fail to separate during Anaphase I or Anaphase II.
Accurate biological terminology distinguishes structural chromosome changes from numerical errors caused by separation failures.
2
Differentiate between structural chromosomal aberrations, normal meiotic events, and nuclear division errors.
Translocation is a structural rearrangement, crossing over is normal genetic recombination, and polyploidy involves whole-genome duplication.
Failure of specific chromosome pairs or chromatids to separate during division is termed nondisjunction, which directly causes aneuploidy.

Anahtar Kavram

Nondisjunction and Aneuploidy
Tahmini Süre:1m 0s
Soru 16Soru →

A chromosomal inversion alters the linear sequence of genes on a chromosome without altering the total quantity of genetic material.

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Cevap: True

Cevap

True. Chromosomal inversion rearranges the linear order of genes without changing the total quantity of genetic material.
Chromosomal inversion is a structural aberration in which a segment of a chromosome breaks off, flips 180∘180^\circ, and reinserts itself into the chromosome. Since no chromosomal material is deleted or duplicated, the total amount of genetic material is unchanged, even though gene sequence order is rearranged.

Adım Adım Çözüm

1
Define the structural alteration that occurs during chromosomal inversion.
A chromosome segment breaks at two points, rotates by 180∘180^\circ, and reattaches into the same position on the chromosome.
Understanding the mechanism of inversion is necessary to determine if genetic content is modified in quantity.
2
Evaluate if genetic material is added or removed during this process.
Because no DNA segment is excised permanently or duplicated, the total amount of genetic material remains unchanged.
Distinguishing balanced structural changes (inversion, balanced translocation) from unbalanced ones (deletion, duplication) confirms the truth value of the statement.

Anahtar Kavram

Chromosomal Inversion and Structural Aberrations
Tahmini Süre:1m 0s
Soru 17Soru →

An insertion or deletion of a single nucleotide base within the coding region of a gene alters the reading frame, changing all subsequent triplet codons during protein synthesis.

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Cevap: True

Cevap

The statement is true. Inserting or deleting a single nucleotide base shifts the triplet reading frame during translation, altering every codon following the mutation site.
The statement is correct because adding or removing one nucleotide alters the three-by-three reading frame of codons during protein synthesis, causing a frameshift mutation that alters all downstream amino acids.

Adım Adım Çözüm

1
Examine the nature of the genetic code during translation.
Messenger RNA is read by ribosomes in consecutive, non-overlapping triplets called codons.
Each codon specifies a particular amino acid in the synthesizing polypeptide chain.
2
Analyze the impact of inserting or deleting a single nitrogenous base.
Altering the base count by one shifts the triplet alignment down the entire remaining length of the gene.
Because the ribosome continues reading in groups of three, the reading frame boundary changes at the mutation point.
3
Evaluate the outcome on the protein product.
All downstream codons code for different amino acids, or introduce an early stop codon (nonsense mutation).
This frameshift effect changes the entire primary structure of the protein from that point forward.

Anahtar Kavram

Frameshift Mutation Mechanism
Gene Mutations and Chromosomal Aberrations Alıştırma Soruları — JAMB UTME | Examkin