Mendelian principles MCQ Quiz - Objective Question with Answer for Mendelian principles - Download Free PDF

The correct answer is Two

Concept:

• Complementation analysis is a genetic technique used to determine whether different mutations that produce the same phenotype are located in the same gene or in different genes.
• If two mutants complement each other (produce a wild-type phenotype when combined in a diploid), it indicates that the mutations are in different genes. If they do not complement (still exhibit the mutant phenotype), the mutations are likely in the same gene.
• The goal is to determine the number of genes involved in histidine biosynthesis among the six mutant strains.

Explanation:

• To determine the number of different histidine biosynthesis genes, we analyze the complementation pattern of the diploids formed by pairwise combinations of the mutants (His1–His6).
• A diploid showing complementation (‘+’) indicates that the two haploid mutants have mutations in different genes. Conversely, no complementation (‘-’) indicates that the mutations are in the same gene.

There are two complementation groups, hence two different histidine biosynthesis genes are mutated.

The correct option is : 3

Explanation:

• Mendelian inheritance refers to the patterns of inheritance for traits controlled by single genes with two alleles, one derived from each parent. Mendel's laws include the law of segregation (each individual has two alleles for each gene, which segregate during gamete formation) and the law of independent assortment (genes for different traits can segregate independently during the formation of gametes).
• In co-dominance, both alleles in a gene pair contribute equally and visibly to the organism's phenotype. A classic example is the AB blood type in humans, where both A and B alleles are expressed.
• Sex-linked inheritance involves genes located on the sex chromosomes (X and Y). Traits controlled by these genes are typically inherited differently in males and females. An example is hemophilia, which is carried on the X chromosome.
• In incomplete dominance, the heterozygote's phenotype is intermediate between the phenotypes of the homozygotes. For instance, in snapdragons, crossing red-flowered plants with white-flowered ones yields pink-flowered offspring.

The correct answer is 1/8

Explanation:

Genotype of Parents - EEFfGgHh  X eeFfGGhh

• Gametes from Parent 1 - E, F or f, G or g, H or h 
• Gametes from Parent 2 - e, F or f, G, h

Genotype for individual gene pairs in progeny - Ee,FF/Ff/Ff/ff/ GG/Gg,Hh/hh
Probability of (EeFfGGhh) Genotype of progeny - \(1 \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8} \)

(As genes are not linked individual probability as multiplied by product rule of probability for independent events)

Conclusion-So, the correct answer is 1/8.

The correct answer is P - 25%, Q - 50%, R - 45%, S - 38%

Explanation:

AABB (parent) x aabb (parent) produces F1 hybrids that are all AaBb, assuming complete dominance.

P. Unlinked genes: This means that genes A and B assort independently due to being on different chromosomes or far apart on the same chromosome. The F1 genotype AaBb produced by the initial cross, when test crossed with aabb, would produce offspring in a 1:1:1:1 ratio for the genotypes AaBb, Aabb, aaBb, and aabb. This results in 25% of the progeny being aabb.

Q. Completely linked genes: If genes A and B are completely linked, meaning no crossing over occurs between them, they will be inherited together. The F1 generation (AaBb) can produce only AB and ab gametes (no recombinant gametes because there's no crossing over). Test-crossing these F1 individuals with aabb would result in only two genotypes: AaBb and aabb, each making up 50% of the progeny.

R. 10 map units apart: If the two genes are 10 map units apart, it indicates that crossing over happens between them 10% of the time. This means that 10% of gametes will be recombinant (Ab and aB), and 90% will be the parental types (AB and ab). When the AaBb parent produces 10% recombinant gametes, and considering that we are crossing with aabb, 45% of the offspring will actually be represented by the recombinant and non-partner parental types due to the nature of test crossing, which looks at the probability of obtaining the recessive phenotype (aabb).

S. 24 map units apart: 

• When two genes are 24 map units apart, there is a 24% overall chance of recombination occurring between them in a given meiosis. This means that out of 100 meiotic events, we expect about 24 will result in gametes that have experienced crossing over between these two genes.

Given this, when creating gametes:

• 76% of the gametes will be parental types (AB and ab, since these are the gametes produced without recombination).
• 24% of the gametes will be recombinant types (Ab and aB, produced as a result of recombination).

Half of the recombinant gametes (12 out of 24%) will be aB, and the other half (12 out of 24%) will be Ab.
To get aabb offspring, we need the ab gametes from our F1 individual (which are 38% of all gametes: 76% non-recombinant ab + 12% recombinant ab).

Conclusion:

Therefore, the correct answer is P - 25%, Q - 50%, R - 45%, S - 38%

The correct answer is 63/64

Explanation:

In the given problem, genes A, B, and C each control different phenotypes, and the genotype of the plant is Aa Bb Cc. Each gene has a dominant allele (A, B, C) and a recessive allele (a, b, c).

• For each gene, the probability of showing the recessive phenotype (aa, bb, or cc) in the progeny is 1/4, because each gene follows the Mendelian inheritance pattern of 3:1 ratio (dominant:recessive).
• Thus, the probability of progeny being recessive for all three genes (aa bb cc) is (1/4) x (1/4) x (1/4) = 1/64.
• The probability of progeny showing the dominant phenotype for at least one of the genes is the complement of the probability of being recessive for all three genes.
• Therefore, the probability of progeny showing the dominant phenotype for at least one of the phenotypes controlled by genes A, B, and C is 1 - 1/64 = 63/64.

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 2/3

Explanation-

Autosomal recessive diseases are disorders caused by the inheritance of two copies of a defective gene, one from each parent. Autosomal recessive diseases are caused by mutations in genes located on autosomal chromosomes (chromosomes other than the sex chromosomes). Since they are recessive, a person must inherit two copies of the defective gene (one from each parent) to develop the disease.

• Individuals who inherit two copies of the defective gene (aa) will express the disease phenotype.
• Carriers (Aa) typically do not show symptoms of the disease but can pass the defective gene to their offspring.
• Individuals with two normal copies of the gene (AA) are not affected by the disease and are considered genetically unaffected.

In this pedigree:

• Individual 1 and 2 are unaffected parents.
• Individual 3 is unaffected, which means she must be homozygous/heterozygous (AA or Aa).
• Individual 4 is unaffected but must be a carrier (Aa) because they have an affected child (individual 7).
• Individual 5 is unaffected but must also be a carrier (Aa) because they have an affected child (individual 7).
• Individual 6 is unaffected but has affected parents (individuals 4 and 5). Therefore, individual 6 must be normal (AA) or heterozygote/carrier (Aa).

Thus, the probability that individual 6 is a heterozygote (carrier) is 2 (since they are definitely a carrier) divided by 3 (since they could either be a carrier or homozygous dominant, but we're only interested in the probability of being a carrier). This gives us 2/3.

Conclusion-Therefore, the correct answer is 2/3.

Concept:

• Independent assortment is based on the random separation of homologous pairs of chromosomes during anaphase I of meiosis; it occurs when genes coding for two characters are located on different pairs of chromosomes.
•  When genes assort independently, the multiplication rule of probability can be used to obtain the probability of inheriting more than one trait: a cross including more than one trait can be broken down into simple crosses, and the probabilities of obtaining any combination of traits can be obtained by multiplying the probabilities for each trait

Explanation:

• Consider the query above it is trihybrid cross.
• When genotypes AaBbCc X AaBbCc  are crossed, if we used a Punnett square to determine this probability, we might be working on the solution for months.
• However, we can quickly figure the probability of obtaining this one genotype by breaking this cross into a series of single-locus crosses.
• since we need probability of atleast one dominant trait we need to first calculate the probability of genotype with no dominant allele i.e aabbcc and probability for same is (1/4 X 1/4 X1/4) = 1/64
• Therefore of atleast one dominant allele we have 1-1/64 = 63/64

​

Hence the correct answer is option 3

The correct answer is Option 3 i.e.A_bb and aaB

Key Points

• To solve this problem, we can use Punnett squares to determine the possible genotypes and phenotypes of the offspring.
• Let A represent the dominant allele for disc-shaped fruits and a represent the recessive allele for long fruits.
• Let B represent the dominant allele for a round shape and b represent the recessive allele for a long shape.
• When the plant that produces disc-shaped fruits (AABB) is crossed with another plant that produces long fruits (aabb), all the F1 plants will be heterozygous for both traits (AaBb) because they received one copy of the dominant allele from each parent.
• When the F1 plants are intercrossed, we can use a 16-box Punnett square to determine the possible genotypes and phenotypes of the offspring.
• The F2 offspring can have nine different genotypes and three different phenotypes.
• The genotypes are:

1. AABB: disc-shaped
2. AABb: disc-shaped
3. AaBB: disc-shaped
4. AaBb: disc-shaped
5. aaBB: Spherical fruit
6. aaBb: Spherical fruit
7. Aabb: spherical fruit
8. AAbb: Spherical shaped
9. aabb: long fruit

• The spherical fruit phenotype is produced by genotypes (aaB_ and A_bb)

​Therefore, the correct answer is Option 3.

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 1/8

Explanation:

Genotype of Parents - EEFfGgHh  X eeFfGGhh

• Gametes from Parent 1 - E, F or f, G or g, H or h 
• Gametes from Parent 2 - e, F or f, G, h

Genotype for individual gene pairs in progeny - Ee,FF/Ff/Ff/ff/ GG/Gg,Hh/hh
Probability of (EeFfGGhh) Genotype of progeny - \(1 \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8} \)

(As genes are not linked individual probability as multiplied by product rule of probability for independent events)

Conclusion-So, the correct answer is 1/8.

The correct answer is Option 4 i.e. 1/3.

Explanation

• The trait given is autosomal recessive.
• In this pedigree the man’s (II-3) parents must have both been heterozygotes Aa because they produced affected aa child.
• So among expected progeny of man’s (II-3) parents AA(1/4) Aa (1/2) aa (1/4) : 2/3 probability that the man (II-3) is a carrier of the trait.
• The woman II-4 is affected by the trait, so she must have genotype aa. So expected progeny genotype is Aa or aa.
• So overall probability of the son to have the disease is \(\frac{2}{3} \times \frac{1}{2} = \frac{1}{3} \)

Conclusion:- The calculated risk of occurrence of this trait for III-I is 1/3.

The correct answer is Option 3 i.e. 1/4

Explanation-

• Genotype of Parents - AABbccDd X aaBBccDD
• Gametes from parent 1 - A, B or b, c ,D or d
• Gametes from parent 2- a, B, c, D
• Genotype for individual gene pairs in progeny - Aa,BB/Bb,cc,DD/Dd

Probability of (AaBBccDd) Genotype of progeny -\(1 \times \frac{1}{2} \times 1 \times \frac{1}{2} = \frac{1}{4} \)

(As genes are not linked individual probability as multiplied by product rule of probability for independent events)

Conclusion-So, the correct answer is 1/4.

The correct answer is Option 2 i.e. A and C

Explanation-

• The marker shows complete linkage with the trait (spotted skin color). Whoever has 2 bands, they all are affected. Since 4 has only one band, he is unaffected. However 3 has 2 bands and she is passing the affected allele to 6 and 7 and affecting them. Since having a single allele, individual is getting affected means the trait is dominant.
• Individual 3 and 5 have the marker but are not showing the trait. This is possible if the allele is not 100% penetrant. There is no information given regarding the expressivity (intensity of the trait) in the question.
• The individuals that are affected are heterozygous and are having 2 bands. However, the homozygous recessive individuals (unaffected) have single band.
• So, using band pattern we can differentiate between homozygotes and heterozygotes. So the marker is a co-dominant marker.
• Individual 5 is heterozygote. So probability of him to pass the affected allele to next generation is 0.5.

Conclusion:-So, only Statements A and C are correct.

Concept:

• Classical genetics is the area of genetics that is exclusively concerned with the genetic traits that are passed on through sexual activity.
• Genes, genetic diversity, and heredity are the main topics of study in genetics.
• Heredity is the method by which traits are passed down from one generation to the next.
• Genetics has two subfields: classical genetics and contemporary genetics
•  Traditional genetics does not study DNA and nucleic acids at the molecular level.
• Genotype research is a component of contemporary genetics.
• Additionally, using molecular data, it explains inheritance patterns.

Explanation:

• Classical genetics dates back to the 1860s, when a monk named Gregor Mendel figured out the basic laws of inheritance—mainly through looking at traits in pea plants.
• People in Mendel’s time knew nothing about DNA or the cellular processes that underlie inheritance.
• Mendel was curious about how traits were transferred from one generation to the next, so he set out to understand the principles of heredity in the mid-1860s.
• Peas were a good model system, because he could easily control their fertilization by transferring pollen with a small paintbrush.
• Mendel instead believed that heredity is the result of discrete units of inheritance, and every single unit (or gene) was independent in its actions in an individual’s genome. 
• According to this Mendelian concept, inheritance of a trait depends on the passing-on of these units. 
• For any given trait, an individual inherits one gene from each parent so that the individual has a pairing of two genes. We now understand the alternate forms of these units as ‘alleles’. 
• If the two alleles that form the pair for a trait are identical, then the individual is said to be homozygous and if the two genes are different, then the individual is heterozygous for the trait.
• Based on his pea plant studies, Mendel proposed that traits are always controlled by single genes.
• However, modern studies have revealed that most traits in humans are controlled by multiple genes as well as environmental influences and do not necessarily exhibit a simple Mendelian pattern of inheritance.
• Mendel then theorized that genes can be made up of three possible pairings of heredity units, which he called ‘factors’: AA, Aa, and aa. 
• The big ‘A’ represents the dominant factor and the little ‘a’ represents the recessive factor.  In Mendel’s crosses, the starting plants were homozygous AA or aa, the F1 generation were Aa, and the F2 generation were AA, Aa, or aa. 
• The interaction between these two determines the physical trait that is visible to us.
• Mendel’s Law of Dominance predicts this interaction; it states that when mating occurs between two organisms of different traits, each offspring exhibits the trait of one parent only. 
• The dominant factor is present in an individual, the dominant trait will result. 
• The recessive trait will only result if both factors are recessive.

Hence the correct answer is option 1.