Inheritance Biology MCQ Quiz in मल्याळम - Objective Question with Answer for Inheritance Biology - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

The correct answer is 'G-banded human karyotype depicting aneuploidy'. 

Explanation:

A G-banded human karyotype depicting aneuploidy refers to a visual representation of the chromosomes in a human cell, where the chromosomes are stained using a technique called G-banding. In this context, aneuploidy refers to an abnormality in the number of chromosomes, where there is a gain or loss of individual chromosomes.

Key Points 

1. G-banding:
   1. G-banding is a technique used in cytogenetics to stain chromosomes. It involves treating the chromosomes with the enzyme trypsin and then staining them with Giemsa dye.
   2. The result is a distinct banding pattern on each chromosome, allowing for the identification and analysis of individual chromosomes.
2. Human Karyotype:
   1. A karyotype is a visual representation of the complete set of chromosomes in an individual's cell.
   2. In a human karyotype, chromosomes are typically arranged in pairs, with one chromosome from each parent.
3. Aneuploidy:
   1. Aneuploidy is a chromosomal abnormality characterized by an abnormal number of chromosomes in a cell.
   2. Trisomy refers to the presence of an extra chromosome, and monosomy refers to the absence of one chromosome.
4. G-banded Human Karyotype Depicting Aneuploidy:
   1. The term "G-banded human karyotype depicting aneuploidy" specifically refers to a karyotype where the G-banding technique is used to reveal chromosomal abnormalities, such as trisomy or monosomy.

​For example, Down syndrome (Trisomy 21) is a well-known condition associated with aneuploidy. In a G-banded karyotype of an individual with Down syndrome, an extra copy of chromosome 21 would be visually apparent.

• C-banding: This technique specifically stains certain regions of the chromosome, predominantly the centromeric regions, where heterochromatin is concentrated. If the question describes the karyotype as "C-banded" and the image or the chromosomes do not explicitly show this type of staining, this would not be the correct choice.
• G-banding: This is the most common type of chromosome banding used in karyotypes and shows a series of light and dark bands along the chromosomes. These bands are crucial for identifying chromosomal anomalies and structural features. If the karyotype image shows these bands clearly, then describing it as "G-banded" is accurate.
• C-banding highlights centromeric regions and G-banding providing a general banding pattern across the entire chromosome that helps identify structural details and anomalies.

Additional InformationHere's a brief overview of each:

1. Q-banded Normal Human Karyotype:

• Q-banding: Q-banding is a chromosome staining technique similar to G-banding but uses quinacrine mustard stain.
• Normal Human Karyotype: A Q-banded normal human karyotype reveals a distinct banding pattern on chromosomes, facilitating the identification and analysis of individual chromosomes. It doesn't specifically highlight structural details as well as G-banding, but it provides valuable information for cytogenetic analysis.

​2. C-banded Karyotype Depicting Klinefelter Syndrome:

• C-banding: C-banding is a technique that specifically stains constitutive heterochromatin, the condensed regions of chromosomes that are rich in repetitive DNA sequences.
• Klinefelter Syndrome: This genetic condition is characterized by the presence of an extra X chromosome in males (XXY instead of XY). In a C-banded karyotype depicting Klinefelter syndrome, one might observe the extra X chromosome and specific patterns of heterochromatin staining.

​3. G-banded Normal Human Karyotype:

• G-banding: G-banding is a widely used chromosome staining technique that involves treating chromosomes with trypsin and staining with Giemsa dye. It results in a unique banding pattern on each chromosome, allowing for precise identification.
• Normal Human Karyotype: A G-banded normal human karyotype shows the characteristic banding pattern on chromosomes, facilitating the identification of normal chromosome structure and number. It is a standard method for analyzing chromosomal abnormalities, such as aneuploidies, deletions, or translocations.

These techniques are essential tools in cytogenetics, aiding in the visualization and analysis of chromosomes for diagnostic and research purposes. Each method provides unique information about chromosome structure, helping researchers and healthcare professionals understand genetic variations and disorders.

The correct answer is Option 1 i.e. Parental type - 585, Single cross over between y and cv-95, Single cross over between cv and f-275, Double cross over - 45

Explanation-

Double cross over (DCO) = 14% X 32% X Total progeny = 0.14 x 0.32 x 1000 = 44.8 = 45 

Gene distance between 

a) y and cv = (Single cross over of y and cv + Double cross over) / Total progeny

           14% = (Single cross over + 45)/ 1000

• Single cross over + 45 = 14% x 1000 = 0.14 x 1000 = 140
• Single cross over = 140-45 = 95

b) cv and f = (Single cross over of cv and f + Double cross over) / Total progeny

          32% = (Single cross over + 45)/ 1000

• Single cross over + 45 = 32% x 1000 = 0.32 x 1000 = 320
• Single cross over = 320-45 = 275

Parental = Total progeny - Total recombination (Single cross over + Double cross over)

              = 1000-95-275-45

               =585

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

Explanation-

A. The mutation arose in the germline of the father.

• This statement is supported by the fact that the father carries the mutated allele in one of his germline cells, as evidenced by the difference between his two alleles (allele 1 and allele 2). The mutation is then inherited by one of the sons.

B. The mutation arose in the son.

• This statement is not supported by the provided information. The mutation is present in one of the alleles of the son, but it originated in the germline of the father, not in the son himself.
• The mutation was inherited from the father and present in one of his sperm cells that contributed to fertilization.

C. The given DNA sequences are present on the X chromosome.

• This statement cannot be determined from the provided information. The sequences are from a family, but there is no indication of whether they are from the X chromosome or any other chromosome.
• The inheritance pattern and presence of mutations do not inherently specify the chromosome involved.

D. There is a possibility to use RFLP for tracking this variation.

• This statement is correct. Restriction Fragment Length Polymorphism (RFLP) analysis can be used to detect the presence of the mutation by analyzing the restriction fragment patterns generated from the DNA sequences of family members. This technique can help track the inheritance pattern of the mutation within the family.

Conclusion- Therefore, the correct statements are A and D.

The correct answer is Option 2 i.e.The polymorphic DNA bands represents two independent genes.

Explanation-

• The F2 progeny exhibiting a Mendelian phenotypic ratio of 9:3:3:1 suggests the independent assortment of two genes and complete dominance in their allelic relationship.
• Independent assortment occurs when genes located on different chromosomes assort into gametes independently of each other. This is a result of the random alignment and segregation of homologous chromosomes during meiosis.
• In this scenario, there are two genes involved, and they are located on different chromosomes. The assortment of alleles at one gene locus does not influence the assortment of alleles at the other gene locus.
• This means that each gene is inherited independently of the other, and the dominant allele will always be expressed over the recessive allele.

Concept:

• A number of human metabolic diseases are inherited as autosomal recessive traits.

• One of them is Tay-Sachs disease.

• Children with Tay-Sachs disease appear healthy at birth but become listless and weak at about 6 months of age.

• Gradually, their physical and neurological conditions worsen, leading to blindness, deafness, and eventually death at 2 to 3 years of age.

• The disease results from the accumulation of a lipid called GM2 ganglioside in the brain.

• A normal component of brain cells, GM2 ganglioside is usually broken down by an enzyme called hexosaminidase A, but children with Tay-Sachs disease lack this enzyme

Explanation:

• Autosomal recessive traits normally appear with equal frequency in both sexes (unless penetrance differs in males and females), and appear only when a person inherits two alleles for the trait, one from each parent.
• If the trait is uncommon, most parents carrying the allele are heterozygous and unaffected; consequently, the trait appears to skip generations.
• Frequently, a recessive allele may be passed for a number of generations without the trait appearing in a pedigree.
• Whenever both parents are heterozygous, approximately of the offspring are expected to express the trait, but this ratio will not be obvious unless the family is large.
• In the rare event that both parents are affected by an autosomal recessive trait, all the offspring will be affected. When a recessive trait is rare, persons from outside the family are usually homozygous for the normal allele.
• Thus, when an affected person mates with someone outside the family (aa AA), usually none of the children will display the trait, although all will be carriers (i.e., heterozygous).
• A recessive trait is more likely to appear in a pedigree when two people within the same family mate, because there is a greater chance of both parents carrying the same recessive allele.
• Mating between closely related people is called consanguinity.
• Like first cousins if both are heterozygous for the recessive allele; when they mate, of their children are expected to have the recessive trait.

​

hence the correct answer is option 1

The correct answer is Option 1 

Key Points

• Three-point testcross is used to determine the position of three genes in a chromosome of an organism. 
• It is a cross of triple heterozygous with triple homozygous recessive.
• If the genes are sex-linked then the female is the heterozygous strain and the male is the hemizygous strain. 
• For example for all three genes, two allelic for exist, the for each gene in the cross, two different phenotypes will be present in the progeny, therefore, for the three genes there will phenotype  classes will be present.
• Three-point test crosses can be used to determine the order of the genes in the chromosome loci.
• Three-point testcross includes one parent that is heterozygous for all three genes while another parent is homozygous recessive for all the genes, so the phenotype of progeny is determined by the gametes of the triply heterozygous parent.
• During gametes formation, we can see the parental type and recombinant type. 
• Recombinant types have immerged from crossing-over events.
• The possibility of the single crossover is more as compared to double crossing-over events.
• Hence, the recombinant type having less number of progeny is the one where double crossing over took place, it will give us information about the gene that is present in the middle. 

Calculation:

Given: Interference between genes = 0.4

Distance between cu and e = 20cM

Distance between e and se = 12cM

First, we will look at option 1 to calculate the map distance between genes 'cu' and 'e'.

The formula of recombinant frequency = 

We will be substituting sco = 334 and dco = 26 and total progeny = 1800 in the above equation, we get:

Since, 1 map distance is equal to 1% recombinant frequency,  20% recombinant frequency is equal to 20cM.

Now, we will calculate the recombinant frequency of region II, using the data in option 1.

Recombinant frequency

We will be substituting sco = 190 and dco =26 and total progeny = 1800 in the above equation, we get,

Since, 1 map distance is equal to 1% recombinant frequency,  12% recombinant frequency is equal to 12cM.

Hence, the correct answer is Option 1.

The correct answer is Option 1 i.e.short arm of chromosome 10 at G-sub band 1 of band 1Key Points

• In G-banding, trypsin-treated chromosomes are stained, and in R-banding, hot, acidic saline is used to denaturate the chromosomes before Giemsa labelling.
• By denaturing chromosomes in an alkaline solution that is saturated, followed by Giemsa staining, C-banding is a technique that is particularly used to identify heterochromatin.
• When scientists use the G-banding method to visualize chromosomes, they first treat the cells with a chemical stain that produces a distinctive pattern of dark and light bands on the chromosomes.
• These bands are numbered consecutively, starting from the centromere (the central region of the chromosome) and extending outward towards the telomeres (the ends of the chromosome).
• The short arm of a chromosome is labeled "p," while the long arm is labeled "q."
• So, for example, the 10p11 notation indicates that the gene of interest is located on the short arm of chromosome 10.
• Within each band of a chromosome, there are smaller sub-bands that are labeled with letters of the alphabet.
• For example, the first sub-band of the first band is labeled "1," and the second sub-band is labeled "2," and so on.
• So, in the case of the 10p11 notation, the gene of interest is located at the first sub-band of the first band on the short arm of chromosome 10, which is denoted as 10p11.

Therefore, the correct answer is Option 1.

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. 

The correct answer is 0.39

Concept:

• The ABO blood group system is determined by three alleles: I_A, I_B, and i.
• The Hardy-Weinberg equilibrium principle states that allele and genotype frequencies in a population will remain constant from generation to generation in the absence of evolutionary influences such as mutation, selection, or genetic drift.
• The frequencies of alleles are denoted as p (for I_A), q (for I_B), and r (for i), and their sum must equal 1: p + q + r = 1.
• The blood group A can result from two genotypes: I_AI_A (homozygous) and I_Ai (heterozygous).
• Under Hardy-Weinberg equilibrium, the expected genotype frequencies can be calculated using the formula for binomial expansion: (p + q + r)².
• The formula for the frequency of blood group A in the population is:
  Frequency of A = p² + 2pr, where p is the frequency of I_A and r is the frequency of i.

Explanation:

The allele frequencies are given as:

• p = 0.3 (frequency of I_A), q = 0.2 (frequency of I_B), and r = 0.5 (frequency of i).

To find the frequency of blood group A, apply the formula:

• Frequency of A = p² + 2pr.

Substitute the values:

• Frequency of A = (0.3)² + 2(0.3)(0.5)
• = 0.09 (homozygous I_AI_A) + 0.3 (heterozygous IAi)
• = 0.09 + 0.3 = 0.39, or 39% of the population.

Thus, the expected percentage of the population with blood group A is 39%.

The correct option is: 3

Explanation:​

• PKU is an autosomal recessive disorder, meaning it occurs when an individual inherits two defective alleles (one from each parent) of the PAH gene (phenylalanine hydroxylase) on chromosome 12. Heterozygous carriers, who have only one defective allele, are asymptomatic but can pass the mutation to their offspring.

• Phenylketonuria (PKU):
  PKU is a metabolic disorder caused by mutations in the PAH gene, leading to a deficiency in the enzyme phenylalanine hydroxylase. This enzyme is crucial for converting phenylalanine to tyrosine. In its absence, phenylalanine accumulates to toxic levels, causing intellectual disability, developmental delays, and other symptoms if untreated.