Inheritance Biology MCQ Quiz - Objective Question with Answer for Inheritance Biology - Download Free PDF

The correct answer is AGBFCDE

Explanation:

• In bacterial chromosomes such as Escherichia coli (E. coli), DNA replication occurs in a bidirectional manner from the origin of replication (OriC).
• In bidirectional replication, two replication forks originate at the 'Ori' and move in opposite directions around the circular chromosome.
• Since both forks move at the same rate, genes closer to the origin (in either direction) will be replicated earlier than genes further away
• The genetic map of E. coli is organized in a circular fashion. Genes are replicated in a specific sequence based on their relative positions on the map and the direction of the replication forks.
• The 'Ori' is located between genes G and A on the circular map.
  • Fork 1 (moving clockwise from Ori): This fork will encounter genes in the order: A, then B, then C, then D, then E.
  • Fork 2 (moving counter-clockwise from Ori): This fork will encounter genes in the order: G, then F.
• Since both forks move at the same rate, genes that are equidistant from the origin, but in opposite directions, will be replicated at roughly the same time.
  • A (from clockwise fork) and G (from counter-clockwise fork) are replicated first.
  • B (from clockwise fork) and F (from counter-clockwise fork) are replicated next.
  • The remaining genes on the clockwise path are C, D, E.

Therefore, the correct order of genes are AGBFCDE

The correct answer is 0.26

Explanation:

Cystic fibrosis is an autosomal recessive disorder. Both parents are carriers (heterozygous), meaning their genotype is Ff (where F is the dominant normal allele and f is the recessive allele for cystic fibrosis).

When two heterozygous parents (Ff x Ff) have children, the possible genotypes and their probabilities are:

• FF (Normal phenotype, homozygous dominant): 1/4
• Ff (Normal phenotype, heterozygous carrier): 1/2
• ff (Cystic fibrosis phenotype, homozygous recessive): 1/4

A "normal child" refers to a child who does not have cystic fibrosis. This includes children with genotypes FF and Ff.

• So, the probability of a normal child (p) = P(FF) + P(Ff) = 1/4 + 1/2 = 3/4.
• The probability of an affected child (q) = P(ff) = 1/4.

To find the probability of having exactly three normal children out of five, we apply binomial probabilty formula:

where,

• P(X=k) is the probability of exactly k successes in n trials
• n is the total number of children (trials) = 5
• k is the number of normal children (successes) = 3.
• p is the probability of a normal child (success) = 3/4.
• q is the probability of an affected child (failure) = 1/4.
• C(n,k) is the binomial coefficient, calculated as 

C(5,3) = 

C(5,3) = 120/ 12 = 10

Calculating p^k and q^(n-k)

• p^k = (3/4)³ = 27/64
• q^(n-k) = (1/4)^(5-3) = (1/4)² = 1/16

Now, substitute these values into the binomial probability formula:

• P(exactly 3 normal children) = 10 x (27/64) x (1/16) = 270/1024 ≈ 0.26 

The correct answer is  27 and 8

Explanation:

• The problem involves a dihybrid cross where the genotypes are AaBBCcDd for both parents. Each gene pair (A/a, B, C/c, D/d) follows independent assortment, and the dominant-recessive relationship applies to each allele.
• We calculate the number of genotypes by considering all possible allele combinations, while the number of phenotypes depends on the dominant traits expressed.
• Number of Genotypes
  • The parent genotypes are AaBBCcDd for both parents. This means the genes involved are A/a, B, C/c, and D/d.
  • For each gene pair:
    • A/a: Crossing Aa × Aa results in three genotypes: AA, Aa, and aa.
    • B: Both parents are BB, so the only possible genotype is BB.
    • C/c: Crossing Cc × Cc results in three genotypes: CC, Cc, and cc.
    • D/d: Crossing Dd × Dd results in three genotypes: DD, Dd, and dd.
  • The total number of genotypes is the product of all possible combinations:
    3 (A/a) × 1 (B) × 3 (C/c) × 3 (D/d) = 27 genotypes.
• Number of Phenotypes
  • The phenotypes depend on the dominant-recessive relationship:
    • A/a: The dominant phenotype (A) is expressed in AA and Aa, while aa expresses the recessive phenotype.
    • B: Since both parents are BB, the phenotype for B is always dominant.
    • C/c: The dominant phenotype (C) is expressed in CC and Cc, while cc expresses the recessive phenotype.
    • D/d: The dominant phenotype (D) is expressed in DD and Dd, while dd expresses the recessive phenotype.
  • The total number of phenotypes is the product of all dominant-recessive combinations:
    2 (A/a) × 1 (B) × 2 (C/c) × 2 (D/d) = 8 phenotypes.

The correct answer is 0.25

Explanation:

• F2 Generation from Self-crossing F1 (TtRr x TtRr):
  • For Plant Height (Monohybrid cross: Tt x Tt):
    • Possible genotypes: TT, Tt, tt
    • Genotypic ratio: 1 TT : 2 Tt : 1 tt
    • Phenotypic ratio: 3 Tall : 1 Dwarf
    • This means for every 4 individuals in F2, 3 are Tall (1 TT, 2 Tt) and 1 is Dwarf (1 tt).
  • For Flower Color (Monohybrid cross: Rr x Rr):
    • Possible genotypes: RR, Rr, rr
    • Genotypic ratio: 1 RR : 2 Rr : 1 rr
    • Phenotypic ratio: 3 Red : 1 White
    • This means for every 4 individuals in F2, 3 have Red flowers (1 RR, 2 Rr) and 1 has White flowers (1 rr).
• The Conditional Probability for the Selected Individual:
  • The question states: "if one tall individual is selected from the F2 population..." This means our sample space for the height trait is restricted only to the tall individuals in F2.
  • Probability of being genotypically homozygous for plant height (TT) among the tall F2 individuals:
    • The tall F2 individuals have genotypes TT and Tt, in a ratio of 1:2.
    • Out of these 3 parts (1 TT + 2 Tt), only 1 part is TT.
    • So, P(TT | Tall) = 
  • Probability of making red flowers (phenotype R_) in F2:
    • The question specifies "make red flowers," which refers to the phenotype (meaning genotypes RR or Rr).
    • From the flower color cross, the probability of an F2 individual having red flowers is This probability is independent of the height selection.
    • So, P(Red Flowers) = 

Since the two traits (height genotype and flower color phenotype) segregate independently, the overall probability is the product of their individual probabilities:

Probability = P(TT | Tall) x P(Red Flowers)

• Probability =  x 
• Probability = 
• Probability = 0.25

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.

The 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

The 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.

The 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:

1. 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.

The correct answer is Option 2 i.e. Maternal uniparental disomy.

Concept:

• Imprinitng refers to the inheritable process where one allele of genes (coming from one of the either of parent) is inactivated naturally while other allele is expressed. 
• It is process of controlling gene  expression without changing the DNA sequence or base pair sequence. 
• Generally the activity is controlled by DNA methylation at a particular bases, for example cytosines. 
• During methylation process, methyl group is attached to the bases. It is generally found in cytosine that follows guanine at CpG dinucleotide.
• This methylation can either leads to activation or inhibition of genes expression in the organism.

Explanation:

• Prader-Willie syndrome (PWS) is a multisystemic complex disorder.
• it occurs at about 1 on 10,000 -30,000 individuals. 
• It is known to affects female and male with equal probability in all races and ethnicity. 
• It is first recognised disorder that is linked to genomic imprinting in humans.
• Following are the three main mechanism for PWS: 

1. Paternal 15q11-q13 deletion - in this cases a proximal deletion of region 15q11-q13  on 15 chromosome results in PWS. 
2. Maternal uniparental disomy - it is second most frequent reason for PWS, it occurs due to error in meiosis. It can be due to following reasons:
   • Heterodisomy- it is result of non-disnjunction of 15 chromosome in meiosis I, homolog that reuslt in one gamtes have two copiers of maternal 15 chromosome. 
   • Isodisomy -  It is results of non-disjunction of chromosome 15 during meiosis II, in this case also gametes have two copies of 15 chromsomes from mother. 
3. imprinting defects - Most of the common causes of the PWS are sporadic,  but in some cases incomplete processing of  the imprint in germ cell mieosis from the father or microdeletion of DNA imprinting reason are cause of PWS. 

Hence, the correct answer is Option 2. 

The 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 - 

1. 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. 
2. 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.
3. 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 - 
   1. Paracentric inversion: It does not involve the centromere and only one arm of the chromosome has two breaks.
   2. Pericentric inversion:  It involves a centromere with each arm having one double-stranded break.
4. 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. 

The 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 :-

1. 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.
2. 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.

The 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

The 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

The correct answer is biallelic heterozygous mutations.

Concept:

Definitions of Mutation Types:

1. Monoallelic mutations: Mutations occurring in only one of the two alleles of a gene in a diploid organism.
2. 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).
3. 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).

The 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.