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

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

Concept:

• The base analogue of adenine called 2-aminopurine (2-AP) mispairs with cytosine and results in base-pair substitutions of the transition type.
• The adenine base analog 2-aminopurine (2AP) is a potent base substitution mutagen in prokaryotes because of increased ability to form a mutagenic base pair with an incoming dCTP.

​Explanation:

• Research studies confirmed that 2-AP induces both A:T leads to G:C and G:C leads to A:T transitions, with the former occurring more frequently than the latter.
• These compounds result in DNA lesions when they are integrated into DNA.
• These lesions will result in mutations if the mismatch correction system does not rectify the mismatches.
• Normally, these lesions do not activate the SOS functions and act as miscoding lesions.

​​hence the correct answer is option 2

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.

The correct answer is Option 3

Explanation:

• A silent mutation is a type of mutation where a change in the DNA sequence does not affect the amino acid sequence of the protein produced from that gene. This type of mutation usually occurs in the third base of a codon, where changes due to the degeneracy of the genetic code do not alter the amino acid being coded.
• In the case of a silent mutation, the genetic sequence (genotype) has changed due to the mutation; however, because the mutation does not alter the protein (due to the redundancy in the genetic code), the observable traits or characteristics (phenotype) of the organism remain unchanged compared to the wild type.

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

Explanation:

• Frameshift mutations are particularly impactful on the function of proteins because they shift the "reading frame" of the genetic code. The genetic code is read in triplets (codons), each coding for a specific amino acid. A frameshift mutation alters how these triplets are organized and interpreted by the cellular machinery responsible for protein synthesis.
• The correct answer involves scenarios B and D, where the removal of a single nucleotide at the beginning and the deletion of four consecutive nucleotides in the middle of the coding sequence both lead to frameshift mutations.
• Frameshift mutations significantly alter the downstream amino acid sequence, potentially changing the entire structure and function of the encoded protein.
• This type of mutation can have a profound effect on protein function, often resulting in nonfunctional proteins or significantly altered biological properties.
• Such mutations are more likely to result in major changes or loss of function in the resulting protein compared to in-frame deletions or changes near the end of the coding sequence, which may have more limited effects on the protein's overall structure and function.
• Frameshift mutations early in the genetic sequence tend to have more dramatic effects because they affect a larger portion of the protein.