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

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 \((2)^3=8\) 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 = \(=\frac{ sco \space in \space region \space I \space (cu-e)+ \space dco}{total \space progeny}\times 100%\)

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

\(\begin {equation} \begin {split} &= \frac{334+26}{1800} \times 100 \% \\&=\frac {360}{1800}\times 100 \% \\&= 20\% \end{split} \end{equation}\)

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\(=\frac{ sco \space in \space region \space II \space (e-se)+ \space dco}{total \space progeny}\times 100%\)

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

\(\begin {equation} \begin {split} &= \frac{190+26}{1800} \times 100 \% \\&=\frac {216}{1800}\times 100 \% \\&= 12\% \end{split} \end{equation}\) 

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 1

Key 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: relative locations of genes on a chromosome

Explanation:

• A genetic linkage map is a map of the relative positions of genetic loci (genes or other hereditary markers) on a chromosome based on the frequencies of recombination between them.
• These maps are constructed by determining the recombination frequency between pairs of genes during crossover events in meiosis. The lower the recombination frequency, the closer the genes are presumed to be on a chromosome.
• Genetic linkage maps do not provide accurate physical distances; instead, they provide estimates of the distances based on genetic recombination data.

Additional Information:

• Genetic linkage maps are useful for identifying the location of genes associated with particular traits or diseases.
• They are different from physical maps of chromosomes, which show the actual physical distances between loci measured in base pairs.
• The creation of genetic linkage maps is an essential tool in genetics and genomics, aiding in the study of genetic inheritance and the function of genes.
• The recombination frequency \((( \theta ))\) is given by the formula:\( [ \theta = \frac{\text{Number of recombinant offspring}}{\text{Total number of offspring}} ]\)
• The recombination frequency is expressed as a percentage and it can be converted into map units, also known as centimorgans (cM). One map unit or centimorgan corresponds to a 1% recombination frequency.

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 mutagenic carcinogens

Explanation:

• The Ames test specifically detects mutagens that are suspected of being carcinogenic by observing whether the tested chemicals cause mutations in the DNA of a bacterium (commonly Salmonella typhimurium).
• If the chemical causes mutations, it suggests that it might also be capable of causing mutations in human DNA, thereby increasing the risk of cancer. This makes the test a valuable tool in chemical safety testing and epidemiological studies related to cancer risk.

Methodology

• Strain Selection: The bacteria used in this test are specially selected strains of Salmonella typhimurium that carry mutations in genes involved in histidine synthesis. These mutations prevent the bacteria from growing on a medium lacking histidine unless a second mutation reverses this inability.
• Exposure to Chemicals: The test substance is added to a culture of the mutated bacteria along with a small amount of histidine. The mixture is plated on a medium that lacks histidine.
• Incubation: During incubation, if the test substance causes mutations that reverse the histidine synthesis block (back-mutation), the bacteria will be able to grow and form colonies.
• Results Interpretation: The number of colonies reflects the mutagenic potential of the substance. A higher number of colonies indicates a higher mutagenic potential.