Inheritance Biology MCQ Quiz - Objective Question with Answer for Inheritance Biology - Download Free PDF
Last updated on Jun 16, 2025
Latest Inheritance Biology MCQ Objective Questions
Inheritance Biology Question 1:
Given below are a few statements.
A. Codominant molecular markers ___(i)___ be used for identification of heterozygotes.
B. Genome wide association studies (GWAS) ___(ii)___ be performed on germplasm with high genetic diversity.
C. An F2 mapping population ___(iii)___ be used as an immortal population for genetic mapping studies in plants.
D. Bulk segregant analysis (BSA) ___(iv)___ be used for mapping of monogenic qualitative traits.
Which one of the following options represents the correct sequence of terms to fill in the blanks in the above statements so that all the statements are true?
Answer (Detailed Solution Below)
Inheritance Biology Question 1 Detailed Solution
The correct answer is (i) can (ii) can (iii) cannot (iv) can
Explanation:
- Molecular markers and genetic mapping techniques are essential tools in plant and animal genetics. They help researchers identify specific genes or genomic regions associated with traits of interest and analyze genetic diversity.
Statement A: "Codominant molecular markers (i) be used for identification of heterozygotes."
- Codominant molecular markers, such as SSRs (Simple Sequence Repeats) or SNPs (Single Nucleotide Polymorphisms), allow for the identification of heterozygotes because they can distinguish between both alleles at a locus. This is because codominant markers show both alleles simultaneously in a heterozygote, enabling their detection.
Statement B: "Genome-wide association studies (GWAS) (ii) be performed on germplasm with high genetic diversity."
- GWAS requires populations with high genetic diversity because it relies on detecting associations between genetic markers and traits across the genome. High diversity ensures sufficient variation to identify meaningful associations.
Statement C: "An F2 mapping population (iii) be used as an immortal population for genetic mapping studies in plants."
- F2 populations are not considered immortal because they are derived from the selfing of F1 individuals and continue to segregate in subsequent generations.
- Immortal populations, like Recombinant Inbred Lines (RILs) or Doubled Haploid (DH) populations, are fixed and can be propagated indefinitely without further segregation.
Statement D: "Bulk segregant analysis (BSA) (iv) be used for mapping of monogenic qualitative traits."
- BSA is an efficient method for mapping monogenic (single-gene) qualitative traits. It involves pooling DNA from individuals with contrasting phenotypes and identifying markers closely linked to the trait of interest.
Inheritance Biology Question 2:
Based on homology, a protein, CG2024, functions as a homotetramer. The function of each unit within the tetramer is essential for its catalytic activity. CG2024 protein has three domains. Domain 'a' is essential for tetramerization, domain 'b' is essential for catalytic activity and domain 'c' does not contribute to CG2024 function at all. Three mutations, a+, b+ and c+ in the 'a', 'b' and 'c' domains of CG2024, respectively, have been identified. The a+ and b+ disrupt the function of their respective domains.
Based on this information, which one of the following options correctly describes the nature of mutations a+, b+ and c+ (in the same order)?
Answer (Detailed Solution Below)
Inheritance Biology Question 2 Detailed Solution
The correct answer is Dominant, dominant, amorphic
Concept:
- Proteins are functional biomolecules composed of multiple domains, each of which contributes differently to the overall activity or structure of the protein.
- Mutations in specific domains of a protein can alter its function or structure, and their effects are classified as dominant or recessive based on whether the mutation's impact is observed in the presence of a wild-type copy.
- Dominant mutations typically affect the protein's function even when a wild-type copy is present. Recessive mutations require both alleles to be mutated to show a phenotype. Amorphic mutations result in loss of function but may not necessarily affect the organism if the domain is non-essential.
Explanation:
- Domain ‘a’: Essential for tetramerization. The mutation a+ disrupts this domain, affecting the ability to form the tetramer. Since tetramerization is critical for function, the mutation acts as dominant. Even one defective domain prevents proper tetramer formation, irrespective of the presence of wild-type copies.
- Domain ‘b’: Essential for catalytic activity. The mutation b+ disrupts this domain, and catalytic activity is lost. Since catalytic activity requires all units within the tetramer to function, this mutation behaves as dominant, as any defective subunit impacts the entire tetramer's activity.
- Domain ‘c’: Does not contribute to CG2024 function. The mutation c+ in this domain does not affect the protein's overall activity because the domain is non-essential. Thus, this mutation is considered amorphic, resulting in a loss of function for the domain but having no impact on the protein’s overall function.
Inheritance Biology Question 3:
In sesame, the seed coat color is of two types: white or brown. True breeding white- and brown-seeded plants were reciprocally crossed and the results are given below.
Which one of the following types of inheritance explains the depicted transmission of seed color in sesame?
Answer (Detailed Solution Below)
Inheritance Biology Question 3 Detailed Solution
The correct answer is Nuclear gene-mediated maternal effect
Explanation:
- Maternal effect refers to an inheritance pattern where the genotype of the mother determines the phenotype of the offspring, regardless of the offspring's own genotype.
- This occurs because maternal genes influence the environment of the developing embryo, often through substances deposited in the egg during oogenesis.
- In the case of nuclear gene-mediated maternal effect, the genes of the mother (located in the nucleus) dictate the phenotype of the offspring.
- The observed inheritance pattern in sesame seed coat color (white or brown) fits this mechanism, as the phenotype of the offspring is determined by the maternal genotype.
- Nuclear gene-mediated maternal effect inheritance pattern observed in sesame seed coat color. The reciprocal crosses between true-breeding white- and brown-seeded plants show that the offspring's phenotype is determined by the maternal genotype, not the paternal genotype or the offspring's own genotype. This is characteristic of a maternal effect.
Incorrect Options:
- Plastid/mitochondrial gene-mediated maternal/cytoplasmic inheritance: This type of inheritance involves genes located in organelles such as mitochondria or plastids, which are inherited maternally. The phenotype is determined by the cytoplasmic organelle genes. Cytoplasmic inheritance is through the cytoplasm of the egg (organelles), while maternal effect is through factors (mRNA, proteins) deposited in the egg by the mother’s genotype. However, in the case of sesame seed coat color, the inheritance is governed by nuclear genes, not organelle genes, making this option incorrect.
- Polar overdominance: Polar overdominance refers to a rare genetic phenomenon in which the expression of a specific phenotype depends on the combination of alleles from both parents, and one allele dominates in a unique manner.
- Incomplete maternal inheritance: Incomplete maternal inheritance involves both maternal and paternal contributions to the offspring's phenotype, but with maternal influence being predominant.
Inheritance Biology Question 4:
A mutation in a plant gene is female gametophyte lethal. This mutant allele can be normally transmitted through pollen. However, when transmitted through egg, the embryos abort, resulting in inviable seeds irrespective of the male allele. A student harvested the seeds from a heterozygous mutant plant and grew a total of 100 plants. What is the expected number of homozygous plants in this population?
Answer (Detailed Solution Below)
Inheritance Biology Question 4 Detailed Solution
The correct answer is 50
Concept:
- Female gametophyte lethality refers to the mutation in the gene results in defective female gametophytes, causing embryo abortion when the mutant allele is inherited through the egg.
- When a heterozygous plant carries one wild-type allele and one mutant allele, the mutant allele can still be transmitted normally through pollen because it doesn't affect male gametophyte function.
Explanation:
- In a heterozygous plant (Aa), the expected genetic ratio for offspring following Mendelian inheritance is 1:2:1 (AA:Aa:aa).
- The mutant allele (a) causes embryo lethality when inherited through the egg. Thus, homozygous aa offspring are inviable and do not grow into plants.
- Only AA (homozygous wild-type) and Aa (heterozygous) genotypes produce viable seeds and plants. Therefore, the viable genetic ratio becomes 1:2 for AA:Aa.
- Population breakdown: Out of 100 plants, half will be homozygous wild-type (AA), and other halfwill be heterozygous (Aa).
- aa (homozygous mutant): 0 plants (because the embryos abort)
- aa (homozygous mutant): 0 plants (because the embryos abort)
- Calculation:
- Total viable plants = 100.
- Proportion of homozygous wild-type plants (AA) = 1/2 × 100 = 50.
- Proportion of heterozygous plants (Aa) = 1/2 × 100 = 50.
Inheritance Biology Question 5:
Two closely related individuals, suffering from a congenital disease, have several children. A genetic counselor constructs the following pedigree of the family and concluded that this disease is caused by mutations in at least two genes.
Further, based on this information, the following statements were made:
A. At least one of the children will not carry any mutations linked to this condition.
B. The mutations causing this disease are recessive.
C. At least one of the mutant genes is on the Y chromosome.
D. Simultaneous heterozygous mutations in the involved genes can cause the disease.
Which one of the following options is the combination of all correct statements?
Answer (Detailed Solution Below)
Inheritance Biology Question 5 Detailed Solution
The correct answer is A and D
Explanation:
- A congenital disease refers to a condition present from birth, often caused by genetic mutations or abnormalities.
- Inheritance patterns of genetic diseases can vary, such as autosomal dominant, autosomal recessive, X-linked, Y-linked, or polygenic (involving multiple genes).
- The given pedigree shows that the disease is caused by mutations in at least two genes, indicating a polygenic or digenic inheritance pattern.
Statement A: "At least one of the children will not carry any mutations linked to this condition."
- This is correct because if the disease is caused by mutations in at least two genes, it is possible for a child to inherit no mutations from either parent. For example, if both parents are heterozygous for mutations in two different genes, there is a chance that the child could inherit the normal (non-mutant) allele for both genes.
Statement D: "Simultaneous heterozygous mutations in the involved genes can cause the disease."
- This is correct because the pedigree suggests a digenic inheritance model. In such cases, the presence of heterozygous mutations in both genes may interact to cause the disease, fulfilling the condition for digenic inheritance.
Statement B: "The mutations causing this disease are recessive."
- This is incorrect because In digenic inheritance, the interaction between two heterozygous mutations can cause the disease even if neither mutation is strictly recessive.
Statement C: "At least one of the mutant genes is on the Y chromosome."
- This is incorrect because the Y chromosome is only passed from father to son. If the disease were Y-linked, it would only affect males in the family, but the pedigree does not suggest this pattern. Therefore, the mutations are not Y-linked.
Top Inheritance Biology MCQ Objective Questions
The statements below are about possible genetic relatedness between individuals of a monogamous, haplodiploid insect.
A. A female is related to its son by 0.5
B. A female is related to its brother by 0.5
C. A male is related to its mother by 1
D. A male is related to its daughter by 1
Which one of the following options represents the combination of all correct statements?
Answer (Detailed Solution Below)
Inheritance Biology Question 6 Detailed Solution
Download Solution PDFThe correct options are: A, C, and D
Explanation:
A. A female is related to its son by 0.5
- In haplodiploid systems, the mother (diploid) passes one set of her chromosomes to her son (haploid).
- Therefore, each son inherits half of his genes from his mother, making the genetic relatedness 0.5.
- This statement is correct.
B. A female is related to its brother by 0.5
- A diploid female receives half of her chromosomes from each parent. In a haplodiploid system, a male passes all his genes to his offspring.
- A female and her brother share the genes they inherit from their mother; brothers and sisters have a relatedness of 0.25 (due to the haplodiploid genetics), not 0.5.
- This statement is incorrect.
C. A male is related to its mother by 1
- A male (haploid) inherits all his genes from his mother (diploid) as he develops from an unfertilized egg.
- Thus, the male is related to his mother by 1.
- This statement is correct.
D. A male is related to its daughter by 1
- A male (haploid) passes all his genes to his daughter (diploid), who inherits all her chromosomes from her father and experiences recombination from her mother.
- Thus, the daughter carries all her father’s genes, but due to diploidy, the correct coefficient of relatedness is 0.5, not 1.
- In haplo-diploid system, male is related to its daughter by 1
Key Points
Haplodiploid System: In haplodiploid species, males are haploid (having one set of chromosomes) and females are diploid (having two sets of chromosomes).
Genetic Relatedness:
- Mother to Son: In haplodiploid species, a female passes one set of her chromosomes to her haploid son, resulting in a relatedness of 0.5.
- Male’s Relation to Mother: Since males inherit all their genes from their mother, the relatedness is 1.
- Male’s Relation to Daughter: A male passes all his genes to his daughter, making their genetic relatedness 0.5. (The statement D inaccurately states it as 1).
- Females and Brothers: In haplodiploid systems, females and their brothers share 1/4 (or 0.25) of their genes due to the common parent, but B inaccurately indicates 0.5.
Table: Shared gene proportions in haplo-diploid sex-determination system relationships
Sex | Daughter | Son | Mother | Father | Sister | Brother |
Female | 1/2 | 1/2 | 1/2 | 1/2 | 3/4 | 1/4 |
Male | 1 | N/A | 1 | N/A | 1/2 | 1/2 |
A cancer clinic is treating four unrelated patients suffering from chronic myelogenous leukemia. A researcher sequences the Philadelphia chromosome from the leukemic cells of these patients and makes the following statements:
A. The DNA sequence was identical in the translocation breakage and rejoining (TBR) sections in all leukemic cells in all 4 patients.
B. The DNA sequence was identical in all leukemic cells from patient 1 , but every patient had a different TBR sequence.
C. All patients have translocations between long arms of chromosomes 9 and 22.
D. All patients have translocations between long arm of chromosome 9 and short arm of chromosome 22.
Which one of the following options represents a combination of all correct statements?
Answer (Detailed Solution Below)
Inheritance Biology Question 7 Detailed Solution
Download Solution PDFThe correct answer is B and C
Explanation:
Statement A: "The DNA sequence was identical in the translocation breakage and rejoining (TBR) sections in all leukemic cells in all 4 patients."
- This statement is incorrect because the breakpoints and sequences in the Philadelphia chromosome can vary between different patients. Although the Philadelphia chromosome is a consistent feature of chronic myelogenous leukemia (CML), the exact locations where the chromosomes break and rejoin (TBR) can differ among patients.
Statement B: "The DNA sequence was identical in all leukemic cells from patient 1, but every patient had a different TBR sequence."
- This statement is correct. Within a single patient, all leukemic cells will have the same TBR sequence due to the clonal nature of the leukemia. However, the TBR sequences can vary between different patients. This reflects the variability in the breakpoints of the Philadelphia chromosome across different individuals with CML.
Statement C: "All patients have translocations between long arms of chromosomes 9 and 22."
- This statement is correct. The Philadelphia chromosome involves a translocation between the long arms of chromosomes 9 and 22.This is a specific cytogenetic abnormality resulting from a reciprocal translocation between chromosome 9 and chromosome 22.
- The translocation occurs between the long arm (q) of chromosome 9 at position 34 (q34) and the long arm of chromosome 22 at position 11 (q11).
Statement D: "All patients have translocations between long arm of chromosome 9 and short arm of chromosome 22."
- This statement is incorrect. The Philadelphia chromosome involves the long arms of both chromosomes 9 and 22, not the short arm of chromosome 22.
Conclusion
The combination of the correct statements is:
- B: The DNA sequence was identical in all leukemic cells from patient 1, but every patient had a different TBR sequence.
- C: All patients have translocations between long arms of chromosomes 9 and 22.
This combination reflects the consistency of TBR sequences within patients and the common translocation between the long arms of chromosomes 9 and 22.
Brassica juncea has bisexual flowers.
A mutation in the mitochondria leads to cytoplasmic male sterility (CMS). CMS can be restored by a restorer of fertility gene ( Rf ) which is a nuclear gene.
Fertility restoration is a dominant phenotype.
A CMS line is crossed to a homozygous Rf line. The obtained F1 progeny is selfpollinated. What percentage of F2 progeny will be male sterile?
Answer (Detailed Solution Below)
Inheritance Biology Question 8 Detailed Solution
Download Solution PDFThe correct answer is 25.
Explanation:
- CMS line: This line has cytoplasmic male sterility due to mitochondrial mutations, resulting in male-sterile flowers.
- Rf line: This is a homozygous restorer of fertility line with the dominant Rf gene, which restores fertility in the presence of CMS.
Cross Details:
- Cross the CMS line (cc) with the homozygous Rf line (RR):
- The CMS line (cc) cannot produce functional pollen.
- The Rf line (RR) can restore fertility when crossed.
F1 Generation:
- The F1 progeny will be heterozygous for the Rf gene: Rr.
- All F1 plants will be male fertile since they possess the dominant Rf allele (the Rf gene restores fertility).
Self-Pollination of F1:
When the F1 progeny (Rr) is self-pollinated, the genotypic ratio in the F2 generation will be as follows:
- Rr x Rr leads to:
- RR (fertile): 25%
- Rr (fertile): 50%
- rr (male sterile): 25%
F2 Progeny:
- The male sterile phenotype corresponds to the rr genotype.
- The percentage of male sterile plants in the F2 generation is 25%.
Conclusion: Thus, 25% of the F2 progeny will be male sterile.
A cruciform structure of chromosomes during meiosis is a characteristic feature of:
Answer (Detailed Solution Below)
Inheritance Biology Question 9 Detailed Solution
Download Solution PDFThe correct answer is Option 1 i.e.Translocation.
Key Points
- A chromosomal abnormality, or chromosomal aberration refers to any disorder that is characterized by a morphological or numerical change in single or multiple chromosomes, that affects autosomes, sex chromosomes, or both.
Types of chromosomal aberration -
- Deletion:
- Deletion refers to the loss of a segment of a chromosome.
- It can be of two types: terminal and interstitial.
- Terminal deletion involves a single break while interstitial deletion rinvovled two breaks.
- Generally deletion is lethal because it causes genetic imbalance in diploid organisms.
- Duplication:
- Duplication refers to occurence of a segment of DNA of a chromsomes in two or more copies per genome. In this case, the duplicated segments can be located next to one another or they can be dispersed on the same chromosome.
- It is produced as the result of abnormal events during recombination.
- It can be of two types - direct and inverted.
- In direct duplication, the duplicated segments keeps the same orientation with respect to the centromere.
- In inverted duplication, the duplicated segments takes the opposite orientation.
- Inversion:
- Inversion is a type of chromosomal mutation that results in a change in the nucleotide sequence of a gene or a chromosome.
- It starts with two double-stranded breaks within a segment of chromosomes.
Then entire fragment is rotated end-to-end between the fracture lines, and the re-fusion of the fragment. - This causes a change in the order of the genes in the section that is inverted.
- Chromosomal inversion is of two types -
- Paracentric inversion: It does not involve the centromere and only one arm of the chromosome has two breaks.
- Pericentric inversion: It involves a centromere with each arm having one double-stranded break.
- Translocation:
- Translocation refers to exchange of chromosomal segments between two non-homologous chromosomes.
- It can be of two types: reciprocal and non-reciprocal.
- Non-reciprocal translocation involves the transfer of segment in one direction from one chromosome to another.
- Reciprocal translocation involves exchange of the segments of chromosomes between non-homologous chromosomes, this results in the generation of two translocated chromosomes simultaneously.
Explanation:
- Translocation typically affects the product of meiosis.
- In the strains that are homozygous for a reciprocal translocation, meiosis occurs normally, all chromosomes can pair and crossing over does not leads to the production of abnormal chromatids.
- In the strains that are heterozygous for a reciprocal translocation, then all the homologous chromosomes pairs in best way possible this leas to generation of cross-like (cruciform) configuration in meiotic prophase I.
- So, cruciform structure is seen in the event of reciprocal translocation that takes place during meiosis process.
Hence, the correct answer is Option 1.
A colour blind father has a daughter who is also colour blind and has Turner's syndrome. The genotype of the daughter is due to:
Answer (Detailed Solution Below)
Inheritance Biology Question 10 Detailed Solution
Download Solution PDFThe correct answer is Option 3 i.e. Non-disjunction event in the mother
Key Points
- Colour blindness is a genetic condition typically inherited through the X chromosome. It is an X linked recessive disease.
- Since, the daughter is colourblind, we can infer that she received one X chromosome with the colour blindness gene from her father, who is also colourblind.
- Turner syndrome, on the other hand, is a genetic condition that occurs when one of the chromosomes is partially or completely missing in females.
- The presence of Turner syndrome in the daughter is likely due to a non-disjunction event during one of the parent's gamete formation.
- Non-disjunction can happen in either the mother or the father during meiosis, where the chromosomes fail to separate properly, resulting in an abnormal number of chromosomes in the offspring.
- In this case, it would have resulted in the daughter having only one X chromosome instead of the usual two.
Hence, correct answer is option 3.
Additional Information
- Translocation is a different genetic event involving the exchange of genetic material between non-homologous chromosomes, and it is not related to Turner syndrome or color blindness in this scenario.
- Basically it is a type of abnormality where a segment of chromosome breaks off and attaches to another non-homologous chromosome.
- This rearrangement can occur between two different chromosomes or within the same chromosome. There are two main types of translocations :-
- Reciprocal translocation:
- In this type, two non-homologous chromosomes exchange segments with each other.
- A piece of one chromosome breaks off and attaches to another chromosome, and vice versa.
- Robertsonian translocation:
- In this type, the long arms of two acrocentric chromosomes fuse together to form a single chromosome.
- The short arms are usually lost.
- This type of translocation is more common in certain chromosomal disorders, such as Down syndrome, where an extra copy of chromosome 21 is formed due to a Robertsonian translocation.
Following statements were made about imprinting in the human genome.
A. Imprinting control centre (IC) harbors part of the SNRPN gene.
B. Imprinting of genes in an individual cannot be tissue specific.
C. Sperms and eggs exhibit identical pattern of genome methylation, except in the sex chromosomes.
D. At imprinted loci, expression depends on the parental origin.
Select the option with all the correct statements.
Answer (Detailed Solution Below)
Inheritance Biology Question 11 Detailed Solution
Download Solution PDFThe correct answer is Option 1 i.e. A and D
Concept:
- Genomic imprinting is a process of silencing genes through DNA methylation.
- The repressed allele is methylated, while the active allele is unmethylated.
- This stamping process, called methylation, is a chemical reaction that attaches small molecules called methyl groups to certain segments of DNA.
Explanation:
Statement A: Imprinting control centre (IC) harbors part of the SNRPN gene.
- SNRPN gene is located within the Prader-Willi Syndrome critical region on chromosome 15 and is imprinted and expressed from the paternal alleles.
- Consider the explanation above thus this statement is true
Statement B: Imprinting of genes in an individual cannot be tissue specific.
- Tissue specificity of imprinting is widespread, and gender-specific effects are revealed in a small number of genes in muscle with stronger imprinting in males.
- Thus this statement is not true.
Statement C: Sperms and eggs exhibit identical pattern of genome methylation, except in the sex chromosomes.
- Sperm genomes are almost fully methylated (~90% of CpGs) except CGIs
- Whereas oocyte genomes show lower methylation levels (~40% of CpGs), with methylation marks being largely confined to intragenic regions of active genes
- Thus this statement is not true.
Statement D: At imprinted loci, expression depends on the parental origin.
- A gene's expression is governed by its parent of origin due to genomic imprinting.
- Additionally, dispersal can alter the costs and advantages of genomic imprinting.
- Barriers to introgression and inbreeding can change as a result of imprinted genes.
- Genes can exhibit behavior that is dependent on their parent of origin.
- Thus this statement is true.
Hence the correct answer is Option 1: A and D
DNA was isolated from a strain of bacterium with genotype a+ b+ c+ d+ e+ and transformed into a bacterial strain a- b- c- d- e-. The transformants were tested for the presence of the donated genes. The cotransformed genes were found as follows:
a+ and b+; c+ and e+; d+ and c+; b+ and d+;
What is the order of genes on the bacterial chromosome?
Answer (Detailed Solution Below)
Inheritance Biology Question 12 Detailed Solution
Download Solution PDFThe correct answer is Option 4 i.e. a b d c e
Concept:
- The simultaneous introduction of numerous genes into a cell is known as co-transformation, and the genes might be found on the same plasmid or on different plasmids.
- Only genes near each other on a chromosome can undergo transformation; the more frequently two genes are near each other, the more often they will cotransform.
- In contrast, genes sufficiently separated such that they cannot appear together on a piece of foreign DNA will nearly never be co-transformed.
Explanation:
- Consider the situation in this problem here,
- Since we know that only genes close can be co transformed we need to consider the data on that,
- Now, a+ and b+ are getting transformed together whereas, b+ and d+ are also in same scenario so quite as possibility is that they are near so a+, b+ and d+ are near.
- Next, consider that, c+ and e+ whereas d+ and c+ are also in same scenario so quite as possibility is that they are near d+, c+ and e+ are near.
- Keeping the above two situations in consideration the order of genes can be deduced as,
- a+, b+, d+, c+ and e+
Hence the correct answer is Option 4
A researcher used CRISPR-Cas9 system and observed a different type of mutation in two alleles of a target gene in a T0 transgenic plant. These mutations are designated as follows:
Allele 1: addition of a nucleotide
Allele 2: deletion of a nucleotide
The observed mutations can be classified as
Answer (Detailed Solution Below)
Inheritance Biology Question 13 Detailed Solution
Download Solution PDFThe correct answer is biallelic heterozygous mutations.
Concept:
Definitions of Mutation Types:
- Monoallelic mutations: Mutations occurring in only one of the two alleles of a gene in a diploid organism.
- Biallelic mutations: Mutations occurring in both alleles of a gene. This can further be classified into:
- Biallelic homozygous mutations: Both alleles have the same type of mutation (e.g., both have the same addition or deletion).
- Biallelic heterozygous mutations: Each allele has a different mutation (e.g., one allele has an addition and the other has a deletion).
- Chimeric mutations: Refers to a mixture of cells with different genetic compositions, often resulting from techniques like CRISPR-Cas9, but this term usually describes the overall plant rather than specific mutations in alleles.
Explanation:
- Allele 1 has an addition of a nucleotide.
- Allele 2 has a deletion of a nucleotide.
Since there are mutations in both alleles, and the mutations are different (one is an addition and the other is a deletion), this scenario represents a case of biallelic heterozygous mutations.
Conclusion: The correct classification of the observed mutations is biallelic heterozygous mutations. This is because both alleles of the target gene have been mutated, but they have different types of mutations (addition vs. deletion).
Given below are statements on concepts of genetics.
A. The degree to which a particular gene is expressed in a phenotype is called ____________.
B. A heritable change in gene expression that does not result from a change in the nucleotide sequence of the genome is called ________ change.
C. The frequency with which a dominant or homozygous recessive gene is phenotypically expressed within a population is called ____________.
D. An allele that results in the death of organisms that is homozygous for the allele is ___________.
Which one of the following options represents the most appropriate sequence of terms to fill all the blank spaces in the above statements?
Answer (Detailed Solution Below)
Inheritance Biology Question 14 Detailed Solution
Download Solution PDFThe correct answer is A - expressivity, B - epigenetic, C - penetrance, D - recessive lethal.
Explanation:
In genetics, the terms used to describe gene expression, heritable changes, and effects of alleles in phenotypes have precise meanings.
Statement A: "The degree to which a particular gene is expressed in a phenotype is called ____________."
- The degree of expression of a gene in a phenotype refers to expressivity, which describes how much the trait is expressed or how severe the phenotype is. Thus, the correct term for this blank is expressivity.
Statement B: "A heritable change in gene expression that does not result from a change in the nucleotide sequence of the genome is called ________ change."
- A heritable change in gene expression without any change in the nucleotide sequence is called an epigenetic change. Epigenetic changes involve mechanisms such as DNA methylation and histone modification that affect gene expression without altering the DNA sequence.
Statement C: "The frequency with which a dominant or homozygous recessive gene is phenotypically expressed within a population is called ____________."
- The frequency at which a gene is expressed in the phenotype is called penetrance, which refers to the proportion of individuals carrying a particular variant of a gene (allele) that also express an associated trait.
Statement D: "An allele that results in the death of organisms that is homozygous for the allele is ___________."
- An allele that causes death when present in the homozygous state is called a recessive lethal allele.
Key Points
- Expressivity refers to the degree or intensity of a gene's expression in the phenotype.
- Epigenetic change involves heritable modifications in gene expression without changes in the DNA sequence.
- Penetrance is the frequency with which a gene is expressed in the phenotype.
- A recessive lethal allele leads to death when the organism is homozygous for that allele.
Who experimentally demonstrated "Mutations occur randomly"?
Answer (Detailed Solution Below)
Inheritance Biology Question 15 Detailed Solution
Download Solution PDFThe correct answer is Salvador Luria and Max Delbrück
Explanation:
Salvador Luria and Max Delbrück conducted the famous Luria-Delbrück experiment in 1943, which demonstrated that mutations occur randomly in bacteria. This experiment provided critical evidence that genetic mutations are not induced by selective pressures (such as exposure to a virus or antibiotic), but rather they arise randomly and then selection acts on these pre-existing mutations.
Key points of the experiment:
- They worked with bacteriophages (viruses that infect bacteria) and E. coli bacteria.
- The goal was to determine if mutations conferring resistance to the bacteriophage occurred before or in response to the exposure to the virus.
- They found that mutations were already present in the bacterial populations before exposure to the virus, indicating that mutations arise spontaneously and randomly, not in response to environmental pressures.
Alfred Hershey and Martha Chase: Demonstrated that DNA is the genetic material.
Matthew Meselson and Franklin Stahl: Conducted the famous experiment showing the semiconservative replication of DNA.
François Jacob and Jacques Monod: Known for their work on the lac operon and gene regulation in bacteria.