Class 12 Bio Botany Β· Chapter 3

Samacheer Class 12 Bio Botany - Chromosomal Basis of Inheritance

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Sections in this chapter
I. Choose the correct answer 2i. monosomy 20II. Find out the incorrect statement 2IV. Choose the correct statement 1IX. Choose the incorrect statement with reference to Deletion 3V. Find the Odd man out with reference to Allopolyploidy 1VII. Find the Odd man out regarding crossing over 1VIII. Choose the wrongly matched pair 1XI. Assertion (A) & Reason (R) 3XII. Two Marks 22XIII. Three Marks 11XIV. Five Marks 10
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1I. Choose the correct answer2 questions
Q.1An allohexaploidy contains a) Six different genomes b) Six copies of three different genomes c) Two copies of three different genomes d) Six copies of one genomev
Answer:

c) Two copies of three different genomes

Q.2The A and B genes are 10 cM apart on a chromosome. If an AB/ab heterozygote is test crossed to ab/ab, how many of each progeny class would you expect out of 100 total progeny? a) 25 AB, 25 ab, 25 Ab, 25 aB b) 10 AB,10 ab c) 45 AB, 45 ab d) 45 AB, 45 ab, 5 Ab, 5aBv
Answer:

c) 45 AB, 45 ab

2i. monosomy20 questions
Q.4Which of the following sentences are correct? 1. The offspring exhibit only parental combinations due to incomplete linkage 2. The linked genes exhibit some crossing over in complete linkage 3. The separation of two linked genes are possible in incomplete linkage 4. Crossing over is absent in complete linkage a) 1 and 2 b) 2 and 3 c) 3 and 4 d) 1 and 4v
Answer:

c) 3 and 4

Q.5Accurate mapping of genes can be done by three point test cross because increases a) Possibility of single cross over b) Possibility of double cross over c) Possibility of multiple cross over d) Possibility of recombination frequencyv
Answer:

b) Possibility of double cross over

Q.6Due to incomplete linkage in maize, the ratio of parental and recombinants are a) 50:50 b) 7:1:1:7 c) 96.4: 3.6 d) 1:7:7:1v
Answer:

b) 7:1:1:7

Q.7Genes G S L H are located on same chromosome. The recombination percentage is between L and G isT5%, S and L is 50%, H and S are 20%. The correct order of genes is a) GHSL b) SHGL c) SGHL d) HSLGv
Answer:

c) SGHL

Q.8The point mutation sequence for transition, transition, transversion and transversion in DNA are a) A to T, T to A, C to G and G to C b) A to G, C to T, C to G and T to A c) C to G, A to G, T to A and G to A d) G to C, A to T, T to A and C to Gv
Answer:

Point mutations in DNA involve a change in a single nucleotide. Transitions are a type of point mutation where a purine is substituted for another purine (A to G or G to A) or a pyrimidine is substituted for another pyrimidine (C to T or T to C). Transversions are point mutations where a purine is substituted for a pyrimidine or vice versa (e.g., A to T, G to C, C to A, or T to G). The question provides four sequences of mutations and asks to identify the correct one representing two transitions and two transversions. Option (b) lists A to G (transition), C to T (transition), C to G (transversion), and T to A (transversion), which accurately represents two transitions and two transversions in DNA.

Q.9If haploid number in a cell is 18. The double monosomic and trisomic number will be a) 35 and 37 b) 34 and 38 c) 37 and 35 d) 17 and 19v
Answer:

a) 35 and 37

Q.10Changing the codon AGC to AGA represents a) missense mutation b) nonsense mutation c) frameshift mutation d) deletion mutationv
Answer:

a) missense mutation

Q.11Assertion (A): Gamma rays are generally use to induce mutation in wheat varieties. Reason (R): Because they carry lower energy to non-ionize electrons from atom a) A is correct. R is correct explanation of A b) A is correct. R is not correct explanation of A c) A is correct. R is wrong explanation of A d) A and R is wrongv
Answer:

c) A is correct. R is wrong explanation of A

Q.12How many map units separate two alleles A and B if the recombination frequency is 0.09? a) 900 cM b) 90 cM c) 9 cM d) 0.9 cMv
Answer:

c) 9 cM

Q.13When two different genes came from same parent they tend to remain together. i) What is the name of this phenomenon? ii) Draw the cross with suitable example. iii) Write the observed phenotypic ratio.v
Answer:

i) The name of this phenomenon is known as Linkage. Linkage was first reported in the sweet pea plant Lathyrus odoratus by William Bateson and Reginald C. Punnett in 1906. In their experiments, genes for purple flower color and long pollen grain were found to be located close together on the same pair of homologous chromosomes. These genes did not assort independently as predicted by Mendel's law of independent assortment, and this tendency of genes to remain together and be inherited as a unit is called linkage. ii) In a typical linkage cross, if we consider two linked genes A and B on one chromosome and their recessive alleles a and b on the homologous chromosome, a cross between AABB (purple flowers, long pollen) and aabb (red flowers, round pollen) produces F1 offspring that are AaBb. When F1 individuals are testcrossed with aabb, the parental combinations (AB and ab) appear more frequently than the recombinant types (Ab and ab) because the linked genes tend to be inherited together. iii) The observed phenotypic ratio in a testcross of linked genes deviates from the expected 1:1:1:1 ratio. Instead, parental phenotypes appear in greater frequency, such as 7:1:1:7 or similar ratios depending on the distance between the genes, while recombinant phenotypes appear in lower frequency. The exact ratio depends on the map distance between the two genes, with closer genes showing less recombination and more parental-type offspring.

Q.15i) What is the name of this test cross? ii) How will you construct gene mapping from the above given data? iii) Find out the correct order of genes.v
Answer:

i) It is three point test cross – It refers to analysing the inheritance, patterns of three alleles by crossing a triple recessive herterozygote with a triple recessive homozygote.
ii) The relative distance between the three alleles & the order in which they are located can be determined with the help of frequency of recombination between them.
All the loci are linked because all the RF values are considerable less then 50%. In AC loci show highest RF value, they must be farthest apart. There fore the B locus must lie between them. The order of genes should be abc. A genetic map can be drawn.
A final point note that two small map distances. 19.9 m.u and 21.75 m.u is add up to 41.95 m.u which is greater then 40.16 m.u the stance calculated for 1 and g. We must identify the two least number of progenius (totalling 8) in relation to recombination of AC. These two least progenius are double cross over. The two least progenies not only counted once should have count each of them twice because each represents a double recombinant progeny. Hence, We can correct the value adding the number 114 + 125 + 116 + 128 + 5 + 14 + 4 = 500 of the total 1200 this number exactly 41.65% which is identical with the same of two component values.
The test cross parental combination can be rewritten as follows.
Gene order showing double recombinant.

Q.16What is the difference between missense and nonsense mutation?v
Answer:

Missense mutation and nonsense mutation are two types of point mutations that affect the genetic code differently. In a missense mutation, the change in the DNA sequence results in a codon that codes for a different amino acid than the original. The mutation causes the substitution of one amino acid for another in the protein sequence. This may result in a protein with altered function, reduced function, or sometimes no noticeable change depending on the location and nature of the amino acid substitution. For example, in sickle cell anemia, a missense mutation changes the codon for glutamic acid to a codon for valine, resulting in abnormal hemoglobin. In contrast, a nonsense mutation occurs when the change in the DNA sequence converts a codon that normally codes for an amino acid into a stop codon (UAA, UAG, or UGA). This premature termination signal causes translation to stop prematurely, resulting in a truncated or incomplete protein that is usually nonfunctional. Nonsense mutations typically have more severe effects than missense mutations because they produce incomplete proteins. The key difference is that missense mutations result in a change of amino acid in the protein, while nonsense mutations result in premature termination of protein synthesis.

Q.17From the above figure identify the type of mutation and explain it.v
Answer:
  • It is a change in the arrangement of gene loci,
  • Here the duplicated segment is located immediately aftear the normal segment but the gene sepuence order will be reversed – (Paracentric inversion)
Q.18Write the salient features of Sutton and Boveri concept.v
Answer:

Salient features of the chromosomal theory of inheritance:
* Somatic cells of organisms are derived from the zygote by repeated cell division (mitosis). These consist of two identical sets of chromosomes. One set is received from female parent (maternal) and the other from male parent (paternal). These two chromosomes constitute the homologous pair.
* Chromosomes retain their structural uniqueness and individuality throughout the life cycle of an organism.
* Each chromosome carries specific determiners or Mendelian factors which are now termed as genes.
* The behaviour of chromosomes during the gamete formation (meiosis) provides evidence to the fact that genes or factors are located on chromosomes.

Q.19Explain the mechanism of crossing over.v
Answer:

Crossing Over – it is a very significant biological process
It is a precise one with several stages
i) Synapsis:
During zygotene – of prophase. I of meiosis I the homologous chromosomes come and align side by side known as – bivalents.
This pairing – is known as synapsis or syndesis.
Types of synapsis
ii) Tetrad Formation:
Each homologous chromosome of – a bivalent begin to form two identical sister chromatids – held together by a centromere.
Each bivalent has 4 chromatids – (tetrad stage),
iii) Cross Over:
At pachytene stage cross over occur. The points of contact at one or more points between non-sister chromatids is called Chiasmata.
Crossing over is exchange of corresponding segments occur, in the chiasma region.
Synaptonemal Complex (SC)
The highly organised structure of filaments called SC – facilitate chiasma formation.
SC formation & chiasma formation – is absent in Drosophila
Terminalisation:
After crossing over, chiasma starts to moving towards the terminal end of chromatids is known as terminalisation. Complete separation of homologous chromosomes occurs after terminalization.

Q.20Write the steps involved in molecular mechanism of DNA recombination with diagram.v
Answer:

Proposed by Robin Holliday in 1964
Steps:
* Homologous DNA molecules are paired side by side with their duplicated copies of DMAs
* One strand of both DNAs cut in one place by the enzyme endonuclease.
* The cut strands cross and join the homologous strands – Holliday junction.
* Holliday junction migrates away from the original site, a process called branch migration, as a result heteroduplex region is formed.
* DNA strands may cut along through the vertical (V) line or horizontal (H) line.
* The vertical cut will result in heteroduplexes with recombinants.
* The horizontal cut will result in heteroduplex with non recombinants.

Q.21How is Nicotiana exhibit self¬incompatibility? Explain its mechanism.v
Answer:

In Nicotiana self sterility or self-incompatibility is due to multiple alleles.
The pollen from a plant is unable to germinate on its own stigma – and no fertilization.
The gene for self incompatibility can be – β€˜S’ which has allelic series S 1, S 2, S 3, S 4 & S 5.
Cross-fertilizing tobacco – were not always homozygous as S 1 S 1 or S 2 S 2, but heterozygous
Crosses between different S 1 S 2 plants, pollen tube did not develop normally.
But effective – development observed when cross was made with other than S 1 S 2 Eg. S 3 S 4.

Q.22How is sex determined in monoecious plants. Write the genes involved in it.v
Answer:

Zeamays (maize) – monoecious plant
Made & Female flowers are present on the same plant.
* Terminal inflorescence – arise from tassel bear staminate flowers
* Lateral inflorescence – arise from ear or cob bear pistillate flowers.
* Unisexvality in maize – occurs through selective abortion of ear florets and pistils in tassel florets.
* The allele for barren plant (ba)- when homozygous makes the stalk staminate (eliminating silk and ears)
* The allele for tassel seed (ts) – transforms tassel into a pistillate structure (no pollen produced)
* Most of these mutations are shown to be defects in Gibberellins biosynthesis.
* Gibbercilins play an important role in the suppression of stamens in florets on the ears.

Q.23What is gene mapping? Write its uses.v
Answer:

Gene mapping is the diagrammatic representation of the position of genes and the distances between adjacent genes on a chromosome. It is also called a linkage map, and the distances mapped are directly proportional to the frequency of recombination between the genes. Gene mapping is based on the principle that the farther apart two genes are on a chromosome, the greater the chance of recombination occurring between them. Uses of gene mapping include determining gene order and identifying the locus of specific genes, which allows for accurate calculation of distances between genes on a chromosome. Gene maps are useful in predicting the results of dihybrid and trihybrid crosses by showing which genes are linked and which assort independently. Additionally, gene mapping allows geneticists to understand the overall genetic complexity and organization of a particular organism, providing insights into how traits are inherited and how genes are distributed across chromosomes. This information is valuable for understanding inheritance patterns and for applications in breeding and genetic studies.

Q.25Mention the name of man-made cereal. How it is formed?v
Answer:

Tetraploidy: Crosses between diploid wheat and rye.
* Hexaploidy: Crosses between tetraploid wheat Triticum durum (macaroni wheat) and rye.
* Octoploidy: Crosses between hexaploid wheat T. aestivum (bread wheat) and rye. Hexaploidy Triticale hybrid plants
12th Bio Botany Guide Chromosomal Basis of Inheritance Additional Important Questions and Answers
I. Fill in the blanks
1. The scientists who independently rediscovered mendelian works were
De Vries, Correns & Tschermak
2. The worm-shaped cells formed during cell division are called in the earlier period as
Chromosomes
3. Who postulated that the chromosomes of a cell are responsible for transferring heredity
Wilhelm Roux (1883)
4. ……………………… was the first to find out physical mutagen in Drosophila
Muller (1927)
5. ………………………used X-rays for the first time to induce mutation in the fruit fly
H.J. Muller
6. Induced mutations are planted was reported for the first time by
L.J. Stadler
7. Chemical mutagenesis was first reported by
Auerback (1944)
8. Double nullisomy is
2n-2-2
9. Trisomis were first reported by Blakeslee in
Datura Stramonium
10. All possible tetrasomics are available in ……………………… plant
Wheat
11. The kind of Aneuploid are usually lethal are
Nullisomy
12. The alkaloid used to induce polyploidy is
Colchicine
13. Raphano brassicas the sterile hybrid of Radish & Cabbage was produced by
G.D. Karpechenko (1927)
14. The cross between hexaploid wheat Triticum aestivum and rye produced is a
Octoploidy
15. Colchicine is extracted from the root and corms of
Colchicum autumnale
16. Who first reported duplication in drosophila
Bridges (1919)
17. In which types of cells chromosomal aberration is commonly found?
Cancer cells
18. Recombination frequencies are the same for
CIS and trans heterozygotes
19. The map distance between gene A and B is 3 units between B & C is 10 units and between C & A is 7 units – the order of genes in a linkage map constructed on the about would perhaps be
B-A-C
20. The percentage of crossing over will be more if
Linked genes are located apart from each other
21. A point mutation that changes an amino acid coding codon into a stop codon, prematurely terminating synthesis of the encoded protein ………………………
Nonsense mutation
22. Single base change in DNA is known as ………………………
Point mutation
23. Genetic change in a non-sex cell is known as
Somatic mulalion
24. A duplicated DNA sequence next to the original sequence is known as
Tandem duplication
25. A missing sequence of DNA or part of a chromosome
Deletion mutation
26. Mutation that alters the genes reading frame is known as
Frame shift mutation
27. A single base change mutation that alters and amino acid ………………………
Missense
28. A substance that changes, adds, or deletes a DNA base
Multagen
29. The mutation that introduces a section of aminoacids not normally found is known as ………………………
Frame shift mutation
30. A mutation that changes an adenine to guanine is an example of a ………………………
transition
31. A point mutation that has no obvious effect at all on the phenotype is called a ……………………… mutation
silent
32. A point mutation that changes a codon specifying an amino acid into a stop codon is called a ………………………
Non sense mutation
33. Changing the codon AGC to AGA represents ……………………… of a point mutation
missense
34. A point multation that alters a codon so that the encoded aminoacid is substituted with another is called a ………………………mutation
missense
35. A ………………………mutation occurs during the DNA replication that precedes meiosis. while a ……………………… mutation occurs during the DNA replication that preceeds mitosis.
germline, somatic
36. A mutation that introduction of section of aminoacids not normally found is ………………………
Frame shift mutation.
37. A point mutation altering a purine to pyrimidine or vice versa is ………………………
transversion
38. A spontaneous mutation usually originates as an error in ………………………
DNA replication
39. The codon for leucine is CUC. How many different aminoacids could possibly result from a single base substitution
7
40. How may map units separate two alleles if the recombination frequency is o.o7?
7cM
41. In a population of 1000 individuals 360 belong to genotype AA. 480 to Aa and the remaining 160 to aa – Based on the data, the frequency of allela A in the population is
0.7
II. Find out the incorrect statement

3II. Find out the incorrect statement2 questions
Q.43Which one of the following is incorrect regarding chromosomal behaviour during cell division? a) The alleles of a genotype are found in the some locus of a homologous chromosome b) In the S phase of meiotic interphase each chromosome replicates forming two copies of each allele, one on each chromatid. c) The Homologus chromosomes segregate in metaphase I, thereby separating two different alleles. d) In anaphase II of meiosis separation of sister chromatid of homologous chromosomes takes place.v
Answer:

c) The Homologous chromosomes segregate in metaphase I, thereby separating two different alleles. This statement is incorrect because homologous chromosomes do not segregate during metaphase I. During metaphase I, homologous chromosomes are aligned at the metaphase plate, and it is during anaphase I that homologous chromosomes separate and move to opposite poles of the cell. The separation of homologous chromosomes during anaphase I is what actually separates the different alleles carried on these chromosomes. The other statements are all correct: alleles of a genotype are found at the same locus on homologous chromosomes, chromosomes replicate during S phase forming two copies of each allele on sister chromatids, and in anaphase II, sister chromatids of homologous chromosomes separate.

Q.45Which of the following statement is not correct of two genes that show 50% recombination frequency? a) The genes may be on different chromosomes b) The genes are tightly linked c) The genes show independent assortment d) If the genes are present on the same chromosome, they undergo more than one crossover in every meiosis.v
Answer:

b) The genes are tightly linked. This statement is not correct for two genes showing 50% recombination frequency. A recombination frequency of 50% indicates that the genes assort independently, which occurs when genes are on different chromosomes or are very far apart on the same chromosome. Genes that are tightly linked show much lower recombination frequencies, typically less than 50%. The other statements are correct: genes with 50% recombination frequency may be on different chromosomes, they show independent assortment, and if they are on the same chromosome, they must undergo more than one crossover in every meiosis to produce a 50% recombination frequency.

4IV. Choose the correct statement1 questions
Q.48When red eyed female Drosophila is crossed with white eyed male, the FI offsprings would be a) Females are with white eye and males are with red eye. b) Males are with red eye and females are with yellow eye. c) Both males and females are with red eye d) Both males and females are with white eye.v
Answer:

c) Both males and females are with red eye. In Drosophila, the eye color gene is located on the X chromosome. The red eye color is dominant over white eye color. When a red-eyed female (homozygous X^R X^R or heterozygous X^R X^w) is crossed with a white-eyed male (X^w Y), the results depend on the female's genotype. If the female is homozygous red-eyed (X^R X^R), all F1 offspring will have red eyes: females will be X^R X^w (red-eyed) and males will be X^R Y (red-eyed). This is the most common interpretation of the question, resulting in both males and females having red eyes in the F1 generation.

5IX. Choose the incorrect statement with reference to Deletion3 questions
Q.56How can we reverse the sterility of FI hybrid? a) Genetic Engineering b) Protoplasmic fusion c) Induced Mutation d) Induced chromosomal aberrationv
Answer:

d) Induced chromosomal aberration

Q.57If haploid number in a cell is 23. The double monosomic and pentasomy number will be a) 44 and 49 b) 17 and 34 c) 47 and 46 d) 45 and 48v
Answer:

a) 44 and 49

Q.58Genes located close together on the same chromosome and inherited together represented as a) linked genes b) unlinked gene c) syntenic genes d) trans genesv
Answer:

a) linked genes. Linked genes are genes located close together on the same chromosome that tend to be inherited together as a unit. Because they are physically near each other on the chromosome, they do not assort independently during meiosis. The frequency of recombination between linked genes is lower than between unlinked genes, and this recombination frequency is used to determine the distance between genes on a chromosome. Linked genes show a characteristic pattern of inheritance where parental combinations appear more frequently than recombinant combinations in the offspring.

6V. Find the Odd man out with reference to Allopolyploidy1 questions
Q.49a) All organisms which possess two or more basic sets of chromosomes derived from two different specie’s. b) They have four or six copies of its own genome – induced by doubling of the diploid species. c) They can be developed by inter-specific crosses and fertility is restored by chromosome doubling with colchicine treatment. d) They are formed between closely related species only..v
Answer:

Allotriploids and allopolyploids are organisms that possess two or more basic sets of chromosomes derived from two different species. They have four or six copies of their own genome, which is induced by doubling of the diploid species. They can be developed by inter-specific crosses, and fertility is restored by chromosome doubling with colchicine treatment. These polyploids are formed between closely related species only. The odd statement among the given options is (d), as allotriploids and allopolyploids are not restricted to closely related species alone; they can also form between distantly related species, though with lower viability and fertility rates.

7VII. Find the Odd man out regarding crossing over1 questions
Q.51a) It occur in germinal cells during gametogenesis. b) Take place during Pachytene state of prophase I of meiosis.. c) It is directly proportional to the frequency of recombination between them. d) It has universal occurrence has great significance.v
Answer:

Crossing over occurs in germinal cells during gametogenesis, specifically taking place during the pachytene stage of prophase I of meiosis. The frequency of crossing over is directly proportional to the frequency of recombination between two loci on a chromosome. Crossing over has universal occurrence and great significance in generating genetic variation and creating new combinations of alleles. The wrongly matched pair is (d), as crossing over does not have universal occurrence in all organisms and all situations; it is absent in some organisms like males of Drosophila and certain other species.

8VIII. Choose the wrongly matched pair1 questions
Q.54a. If the chromosome has only one centromere it is known as Monocentric b. If the inversion include long and short arm of the chromosome does not include centro mere is known as Paracentric c. If the chromosome has no terminal end – it’s known as Telocentric d. If inversion include centromere it is known as Pericentricv
Answer:

If a chromosome has only one centromere it is known as monocentric. If an inversion includes the long and short arms of the chromosome but does not include the centromere, it is known as paracentric inversion. If a chromosome has no terminal end, it is known as acentric. If an inversion includes the centromere, it is known as pericentric inversion. The incorrect statement is (c), because a chromosome with no terminal end is called acentric, not telocentric. A telocentric chromosome is one where the centromere is located at the terminal end of the chromosome.

9XI. Assertion (A) & Reason (R)3 questions
Q.59Assertion (A): Arabidopsis plant chromosomes have more repeats of TTT nucleotide sequences in the telomeres. Reason (R): Restriction endonuclease enzyme is used in the formation of nucleotide sequence (Telomeres) mui a) (A) is incorrect, (R) is correct b) (A) is correct, (R) is the correct explanation (A) c) (A) is correct, (R) is the incorrect explanation (A) d) (A) and (R) are wrong.v
Answer:

b) (A) is correct, (R) is the correct explanation (A)

Q.60Assertion (A): Linkage and crossing over are two processes that have opposite effects. Reason (R): Linkage keeps particular genes together but crossing over mixes them.v
Answer:

a) If both the Assertion (A) & Reason (R) are true and the reason is a correct explanation of the Assertion..

Q.61Assertion (A): Increase in temperature increases the rate of mutation. Reason (R): While rise in temperature hydrolyses DNA by the restriction endonuclease which degrade Nucleotides.v
Answer:

c) Assertion (A) is true but Reason (R) is false
XII. Two Marks

10XII. Two Marks22 questions
Q.1Define chromosome theory of inheritance.v
Answer:

The chromosome theory of inheritance states that Mendelian factors, which are now known as genes, have specific loci on chromosomes and they carry genetic information from one generation to the next generation. This theory establishes the physical basis of heredity by demonstrating that the behavior of chromosomes during meiosis and fertilization parallels the inheritance patterns observed by Mendel in his experiments.

Q.2State the number of chromosomes of the given organism.v
Answer:

The chromosome numbers of the given organisms are as follows: Ophioglossum has 1262 chromosomes, Arabidopsis has 10 chromosomes, Sugarcane has 80 chromosomes, Rice has 24 chromosomes, Potato has 48 chromosomes, and Maize has 20 chromosomes. Ophioglossum is notable for having the highest chromosome number among these organisms, making it a polyploid species with multiple sets of chromosomes.

Q.3What are Fossil Genes?v
Answer:
  • Some junk DNA is made up of pseudogenes, once working but have lost their ability to make proteins.
  • They are fossilized parts act as evidence for evolution.
Q.4State the works of T.H. Morganv
Answer:
  • His works on Drosophila melanogaster – Sex linkage – helped to confirm chromosome theory of heredity.
  • He received Nobel prize in Physiology of medicine in 1933 fot it.
  • He coined the term crossing over.
Q.5What are co-mutagens ?v
Answer:

Co-mutagens are compounds that do not possess mutagenic properties of their own but enhance or increase the mutagenic effects of known mutagens when present together. These substances act as potentiators or sensitizers that amplify the damage caused by actual mutagens. For example, ascorbic acid increases the damage caused by hydrogen peroxide, a known mutagen, by enhancing its oxidative effects on DNA. Similarly, caffeine increases the toxicity of methotrexate, a chemical mutagen, by interfering with DNA repair mechanisms or by enhancing the mutagen's ability to damage genetic material. Co-mutagens are significant in understanding mutagenesis because they demonstrate that the overall mutagenic effect depends not only on the mutagen itself but also on the presence of other chemical substances in the environment. This has important implications for understanding how multiple factors in an organism's environment can combine to increase the risk of mutations.

Q.6Differentiate between Euploidy & Aneuploidyv
Answer:

Euploidy refers to a condition where the ploidy involves entire sets of chromosomes, meaning the chromosome number is an exact multiple of the haploid number. In euploidy, the balance of genes remains normal even though the total number of chromosomes increases. Examples of euploidy include triploidy (3x), tetraploidy (4x), and polyploidy (∞n). Euploid organisms typically have more regular chromosome behavior during meiosis compared to aneuploids. Aneuploidy, in contrast, refers to a condition where the diploid number is altered by the addition or deletion of one or more individual chromosomes, rather than entire sets. In aneuploidy, the chromosome number is not an exact multiple of the haploid number, resulting in an imbalance of genetic material. Examples of aneuploidy include trisomy (2n+1), tetrasomy (2n+2), monosomy (2n-1), and nullisomy (2n-2). Aneuploid conditions typically cause more severe genetic imbalances and are often lethal or produce significant phenotypic abnormalities because of the gene dosage imbalance.

Q.7Distinguish between Monoploidy & Haploidyv
Answer:

Monoploidy and haploidy are related but distinct concepts in genetics. Monoploidy refers to an organism that contains only one complete set of chromosomes, represented as x. In monoploidy, the chromosome number is referred to as x, which represents a single basic set. For example, in hexaploid wheat where the diploid number (2n) is 72, the haploid number (n) is 36, and the monoploid number (x) is 12, indicating that wheat has a basic chromosome set of 12. Haploidy, on the other hand, refers to the condition where an organism has half the number of somatic chromosomes found in a diploid organism, represented as n. The haploid number is the gametic chromosome number, which is the number of chromosomes present in a gamete. In humans, the haploid number is 23 (n), while in wheat, the haploid number is 36 (n). The key distinction is that monoploidy refers to the basic chromosome set (x) of a species, while haploidy refers to the actual number of chromosomes in a gamete (n), which may be different if the organism is polyploid.

Q.8Independent assortment & Linkage are alternatives of each other – Discussv
Answer:

Independent assortment and linkage are alternative phenomena that describe how genes are inherited. In independent assortment, genes present on different chromosomes assort independently during meiosis, resulting in more new combinations of alleles and fewer parental combinations in the offspring. In contrast, linkage occurs when genes are present on the same chromosome and tend to stay together during inheritance, resulting in more parental combinations and fewer new combinations. These two phenomena are mutually exclusive: genes on different chromosomes follow independent assortment, while genes on the same chromosome exhibit linkage. The degree of linkage depends on the distance between genes on the chromosome; genes that are closer together show stronger linkage, while genes farther apart may show some recombination due to crossing over.

Q.9How does the strength and weakness of linkage depend on linked genes?v
Answer:
  • The strength of linkage increases as the distance between linked genes decreases.
  • The linkage becomes weaker with the increase in the distance between genes.
Q.11Distinguish between tetrad & bivalent Tetrad:v
Answer:
  • During Synapsis homologous chromosomes come together side by side resulting in bivalents
  • As the stage during which each bivalent has 4 chromatids & the stage is known as tetrad stage.
Q.12Define Recombination.v
Answer:

Recombination is the process by which segments of DNA from different sources are broken and recombined to produce new combinations of alleles. During recombination, the DNA strands are broken at specific points, and the segments are exchanged between homologous chromosomes or between different DNA molecules. This process results in the creation of new allelic combinations that differ from the parental combinations. Recombination is a crucial mechanism for generating genetic variation in organisms and is a key feature of sexual reproduction. It occurs primarily during meiosis through the process of crossing over, where non-sister chromatids of homologous chromosomes exchange segments of DNA.

Q.13What is RF (Recombination Frequency)v
Answer:

Recombination frequency (RF) is the frequency with which recombination occurs under certain conditions. It is expressed as a percentage and represents the proportion of recombinant gametes produced relative to the total number of gametes. RF is used to determine the distance between two genes on a chromosome, with one map unit or centimorgan corresponding to one percent recombination frequency.

Q.14A diploid organism is heterozygous for 4 loci. How many types of gametes cars be produced.v
Answer:

When a diploid organism is heterozygous for 4 loci, the number of types of gametes that can be produced is calculated using the formula 2^n, where n represents the number of loci for which the organism is heterozygous. In this case, the organism is heterozygous for 4 loci, so n equals 4. Applying the formula: 2^n = 2^4 = 2 Γ— 2 Γ— 2 Γ— 2 = 16. Therefore, the organism can produce 16 different types of gametes. This calculation assumes that the loci are on different chromosomes or are far enough apart that they assort independently. Each gamete will contain a different combination of alleles from the four loci, and the total number of possible combinations is 16.

Q.15Notes on Colchicine.v
Answer:
  • Alkaloid, extracted from – root and corms of colchicum autumnale
  • In low concentration to the growing lips it induce polyploidy
  • It does not affect the source plant due to the presence of Anticolchicine
Q.16Write down the significance of ploidy.v
Answer:
  • Polyploids – More vigorous & more adaptive
  • Ornamental flowers – (Autotetraploids) larger flowers – longer flowering duration
  • Increase in fresh weight (due to more water content)
  • Aneuploids – help to determine the phenotypic effects (loss or gain of different chromosomes
  • Allopolyploids of angiosperms play a role in an evolution of plants.
Q.17Distinguish between Mendelian disorder & Chromosomal disorder.v
Answer:

Mendelian disorders and chromosomal disorders are two distinct categories of genetic disorders that differ in their causes and inheritance patterns. Mendelian disorders occur due to mutations in a single gene and follow the well-known Mendelian patterns of inheritance, such as autosomal dominant, autosomal recessive, or X-linked inheritance. These disorders are caused by changes in the DNA sequence of a specific gene, which may result in a non-functional or altered protein. Examples of Mendelian disorders include sickle cell anemia, cystic fibrosis, and hemophilia. Chromosomal disorders, in contrast, are produced due to alterations in the number or structure of chromosomes rather than mutations in individual genes. These alterations can involve the addition or deletion of entire chromosomes or large segments of chromosomes, resulting in an imbalance of genetic material. Examples of chromosomal disorders include Down syndrome (trisomy 21), Turner syndrome (monosomy X), and Klinefelter syndrome (XXY). Chromosomal disorders typically affect multiple genes and produce more severe phenotypic effects compared to many Mendelian disorders.

Q.19This is a type of Numerical chromosomal abnormality find it out give a note on it.v
Answer:
  • This numerical chromosomal abnormality is known as double monosomy (2n-l-l)
  • From a diploid set of chromosome if one chromosomes is lost, the condition is known as monosomy (2n-l)
  • If another chromosome is also lost it is known as double monosomy (2n-l-l)
Q.20Bring out the difference between Linkage & Crossingover in inheritancev
Answer:

Linkage and crossing over are two important phenomena in inheritance that are related but distinct in their mechanisms and effects. Linkage is the tendency of genes located close together on the same chromosome to stay together and be inherited as a unit during reproduction. Linkage involves genes on the same chromosome of a homologous pair and reduces the formation of new gene combinations because the linked genes do not assort independently. Linked genes show lower recombination frequencies, with parental combinations appearing more frequently in offspring than recombinant types. Linkage is not very significant in evolution because it limits genetic variation. Crossing over, in contrast, leads to the separation of linked genes through the exchange of segments between non-sister chromatids of homologous chromosomes. During crossing over, homologous chromosomes pair up during meiosis, and segments of DNA are exchanged between the non-sister chromatids, creating new combinations of alleles. Crossing over increases genetic variability by forming new combinations of alleles, leading to the formation of new organisms with novel genetic compositions. This process plays an important role in evolution by generating the genetic diversity necessary for natural selection to act upon, allowing populations to adapt to changing environments over time.

Q.21What is chiasmata?v
Answer:
  • The non – sister chromatids of homologous pair make a contact at one or more points.
  • These points of contact between non-sister chromatids of homologous chromosomes are called chiasmata.
Q.22What is multiple alleles?v
Answer:

Multiple alleles refer to the existence of three or more allelic forms of a gene that occupy the same locus on homologous chromosomes within a population. When any of these multiple allelic forms are present in a given pair of homologous chromosomes in an individual, they are said to constitute multiple alleles. A classic example is the ABO blood group system in humans, where three alleles (IA, IB, and i) determine blood type. Another example is the eye color gene in Drosophila, which has multiple allelic forms. Although an individual organism can carry only two alleles at any given locus, multiple alleles increase genetic variation within a population.

Q.23What is monomorphic?v
Answer:
  • About 94% of all flowering plants have only one type of individual, which produces flowers with male organs (the stamens) and female organs (the carpels).
  • Such plants are termed as sexually monomorphic.
Q.24What is Dimorphic?v
Answer:

Dimorphic refers to organisms, particularly flowering plants, that exhibit two distinct forms or types. Some six percent of flowering plants are dimorphic, meaning they have two separate sexes, with male and female reproductive structures present in different individuals. This sexual dimorphism contrasts with hermaphroditic plants, which have both male and female organs in the same flower or plant.

11XIII. Three Marks11 questions
Q.1Differentiate tetrasomy from tetraploidyv
Answer:

Tetrasomy and tetraploidy are both types of polyploidy, which refers to the condition of having more than two complete sets of chromosomes. Tetrasomy specifically refers to the condition where an individual has two pairs of homologous chromosomes, meaning there are four copies of a particular chromosome in an otherwise diploid set, represented as 2n + 2. This is a type of aneuploidy. Tetraploidy, on the other hand, is a condition where an organism possesses four complete sets of chromosomes in every cell, represented as 4n. Tetraploids can arise spontaneously or be induced by treatments that cause chromosome doubling, such as colchicine treatment. They can be autotetraploids, where all four sets of chromosomes originate from the same species (e.g., grapes, groundnuts, potatoes, coffee), or allotetraploids, which arise from the hybridization of two different species followed by chromosome doubling (e.g., wheat). Therefore, tetrasomy involves the addition of extra copies of one or more specific chromosomes, while tetraploidy involves the duplication of the entire genome.

Q.2Give a tabulation comparing the behaviour of gene & Chromosomev
Answer:

Mendelian factors and chromosomes show parallel behaviour during inheritance, which forms the basis of the chromosomal theory of inheritance. Alleles of a gene occur in pairs in diploid organisms, and similarly, chromosomes also occur in homologous pairs. During gamete formation in meiosis, alleles of a factor separate and segregate into different gametes, just as homologous chromosomes separate during meiosis I, ensuring that each gamete receives only one allele and one chromosome from each pair. Mendelian factors can assort independently during gamete formation when they are located on different chromosomes, and correspondingly, non-homologous chromosomes can also separate independently during meiosis. However, genes located on the same chromosome, called linked genes, do not assort independently because they tend to be inherited together as they are physically present on the same chromosome. This parallel behaviour between the segregation and assortment of genes and the behaviour of chromosomes during meiosis provides strong evidence that genes are located on chromosomes.

Q.3The important aspects about the chromosome behaviour during cell devision.v
Answer:
  • Alleles of a genotype – found in the same locus of a homologous chromosome (A/a)
  • In β€˜S’ – Phase of meiotic interphase – the replication of chromosome occur – (two copies of each allele (AA/aa) one on each chromatid
  • Anaphase II of meiosis, separation of sister chromatids of homologous chromosomes. So each daughter cell (gamete) carries only a single allele of a character (A), (A), (a) and (a)
Q.4Write the differences between coupling and Repulsionv
Answer:

Coupling and repulsion refer to the arrangement of alleles on homologous chromosomes in relation to each other, particularly in the context of linked genes. In the coupling phase (also known as cis configuration), two dominant alleles or two recessive alleles of different genes are located on the same chromosome. These alleles tend to be inherited together into the same gamete. In contrast, the repulsion phase (also known as trans configuration) occurs when a dominant allele of one gene and a recessive allele of another gene are on the same chromosome, while the corresponding recessive and dominant alleles are on the homologous chromosome. In this configuration, the alleles tend to segregate and are inherited into different gametes. Understanding coupling and repulsion is crucial for analyzing linkage and recombination frequencies.

Q.5Define synapsis. What are the types of Synapsisv
Answer:

Synapsis, also known as syndesis, is the process of pairing between homologous chromosomes during the zygotene stage of prophase I of meiosis. This pairing is highly specific and occurs along the entire length of the chromosomes, forming structures called bivalents (or tetrads), where each bivalent consists of two homologous chromosomes, each composed of two sister chromatids. This intimate association is facilitated by the synaptonemal complex. There are three main types of synapsis based on the point of initiation of pairing: 1. Procentric synapsis begins at the centromeres and proceeds towards the telomeres. 2. Proterminal synapsis starts at the telomeres and moves towards the centromeres. 3. Random synapsis can initiate at any point along the chromosome and proceed in either direction.

Q.6Distinguish between sharbati sonora & Castor Aruna.v
Answer:

Sharbati Sonora and Castor Aruna are both important mutant varieties developed through specific breeding techniques. Sharbati Sonora is a high-yielding, high-protein mutant variety of wheat, developed in India in 1967 by Dr. M.D. Swaminathan. It was derived from the Mexican variety Sonora 64 by subjecting its seeds to gamma ray irradiation, a form of mutation breeding. This variety is known for its early maturity and excellent bread-making qualities due to its high gluten content. Castor Aruna, on the other hand, is a mutant variety of castor bean. It was developed by treating castor seeds with thermal neutrons. The key improvement in Aruna is its significantly reduced maturity period, maturing in approximately 120 days compared to the typical 270 days for traditional varieties, while also exhibiting high yield potential.

Q.7How do increase in temperature cause mutation?v
Answer:

An increase in temperature causes mutations by disrupting the stability of DNA structure. Rising temperature breaks the hydrogen bonds that hold the complementary base pairs together in the DNA double helix. This destabilization affects the accuracy of DNA replication and transcription processes. When hydrogen bonds are broken, the DNA strands may separate or undergo incorrect base pairing, leading to errors during replication where incorrect nucleotides may be incorporated. Similarly, transcription errors can occur when the template strand is not properly paired. These errors result in changes to the DNA sequence, which constitute mutations. The altered DNA sequence can lead to the production of defective or non-functional proteins, affecting the organism's phenotype and potentially causing harmful effects.

Q.8Distinguish between the impact of ionizing & non ionizing radiation in causing mutation. Ionizing radiation Non Ionizing radiationv
Answer:

Ionizing radiation and non-ionizing radiation differ significantly in their impact on DNA and their ability to cause mutations. Ionizing radiation, such as X-rays, gamma rays, alpha particles, beta particles, and cosmic rays, possesses short wavelengths and high energy. This high energy allows them to readily ionize atoms and molecules by ejecting electrons, which can lead to direct damage to DNA, including single-strand breaks, double-strand breaks, and complex chromosomal aberrations like breaks in chromosomes and chromatids. Non-ionizing radiation, like ultraviolet (UV) rays, has longer wavelengths and lower energy. Consequently, they have less penetrating power and typically cause damage by exciting electrons rather than ionizing atoms. In DNA, UV radiation primarily leads to the formation of pyrimidine dimers (e.g., thymine dimers), which can distort the DNA helix and interfere with replication and transcription if not repaired. Due to their lower energy and penetration, non-ionizing radiation is often used to treat surface-level biological materials like unicellular microbes, spores, or pollen grains.

Q.9What is significance of ploidy?v
Answer:
  • Many polyploids are more vigorous and more adaptable than diploids.
  • Many ornamental plants are autotetraploids and have large flower and longer flowering duration than diploids.
  • Auto polyploids usually have increase in fresh weight due to more water content.
  • Aneuploids are useful to determine the phenotypic effects of loss or gain of different chromosome.
  • Many angiosperms are allopolyploids and they play a role in an evolution of plants.
Q.10What is chemical mutagens? Give an example?v
Answer:

Chemical mutagens are chemical substances that cause mutations by inducing changes in the DNA structure and function. These are chemical agents capable of altering the genetic material and increasing the mutation rate significantly. A common example is nitrous oxide, which acts as a chemical mutagen by altering the nitrogenous bases of DNA through deamination and other chemical modifications. When nitrous oxide interacts with DNA bases, it causes structural changes that affect the accuracy of DNA replication and transcription processes. These alterations in the bases lead to errors during DNA replication, where incorrect base pairing occurs, resulting in mutations. During transcription, the altered bases may be read incorrectly, leading to the synthesis of mRNA with incorrect codons. Consequently, during translation, these incorrect codons result in the incorporation of wrong amino acids or premature termination, leading to the formation of incomplete and defective polypeptides. Other examples of chemical mutagens include formaldehyde, benzene, and various alkylating agents used in research and industry.

Q.11What is cis configuration (or) coupling?v
Answer:

Cis configuration, also called coupling, refers to a specific arrangement of alleles on homologous chromosomes during linked inheritance. In this configuration, the two dominant alleles of different genes occur together on one homologous chromosome, while the two recessive alleles of the same genes occur together on the other homologous chromosome. For example, if we consider two linked genes A and B, in cis configuration the arrangement would be AB on one chromosome and ab on the homologous chromosome. Due to their physical proximity on the same chromosome, these allele combinations tend to be inherited together into the same gamete during meiosis, unless crossing over occurs between them. This results in the production of parental type gametes in higher frequency compared to recombinant gametes. The closer the genes are to each other on the chromosome, the more frequently they will be inherited together. Cis configuration is contrasted with trans configuration, where dominant and recessive alleles are on opposite chromosomes.

12XIV. Five Marks10 questions
Q.1Whose works supported the chromosomal theory of heredity? Explain.v
Answer:
  • T.H. Morgan works on fruit fly supported the chromosomal theory of inheritance.
  • The alleles for red or white eye colour are present on the X – chromosome but there is no counter part for this gene on the Y chromosome.
  • The genes for yellow body colour and miniature wings are also carried on the X – chromosome.
  • By understanding the sex linked inheritance of these characters it is proved that genes are located on the chromosomes.
  • Thus T.H. Morgan’s works on Drophila came as a support to the chromosomal theory of inheritance.
Q.2Write down the steps in the Holliday’s hybrid DNA model.v
Answer:

The Holliday model describes a mechanism for genetic recombination between homologous DNA molecules. The process begins with the alignment of two homologous DNA molecules side-by-side, each consisting of a double helix. An endonuclease enzyme then nicks one strand of each DNA molecule at corresponding positions. Following the nicking, the strands cross over and anneil to the complementary strand of the other DNA molecule, forming a structure known as a Holliday junction. This junction is a cross-shaped intermediate where strands from both DNA molecules are intertwined. Branch migration then occurs, where the heteroduplex region (a region formed by strands from different DNA molecules) extends along the DNA. Finally, the Holliday junction is resolved by cleavage. Depending on the orientation of the cleavage (either vertically or horizontally across the junction), the outcome can be either recombinant DNA molecules (if cleaved vertically, resulting in exchange of flanking markers) or non-recombinant DNA molecules (if cleaved horizontally, maintaining the original flanking marker combinations but still containing heteroduplex regions).

Q.3Explain sex determination is Silene latifolia (Melandrium album)v
Answer:

Sex determination in the plant Silene latifolia (also known as Melandrium album or white campion) is a complex process influenced by genetic factors, environmental cues, and hormonal signals, as extensively studied by C.E. Allen. The genetic basis involves sex chromosomes, where the Y chromosome plays a crucial role in determining maleness, while the X chromosome specifies femaleness. In Silene latifolia, the sex determination system is XY, similar to many animals. However, the Y chromosome is larger and carries more genes related to male development than the X chromosome. The sex chromosomes in Silene latifolia exhibit distinct regions that are important for sex determination. These regions, often designated by Roman numerals (I, II, III, IV, and V), contribute to the differential expression of genes controlling male and female characteristics. The balance between the genes on the X and Y chromosomes, along with potential environmental influences, ultimately dictates the development of male or female reproductive organs.

Q.4How do Hawaii explain the sex determination in Papayav
Answer:
  • Carica papaya 2n = 36
  • The sex chromosomes look like autosomes
  • Developed from autosomes
  • Y- chromosome carries the genes for male organ
  • X- chromosomes bear the gene for female organ development.
Q.5Explain sex determination in Sphaerocarpos donnelli. It is also known as Bottle liverwort (Bryophyta)v
Answer:
  • gametophyte – haploid with 8 chromosome (n).
  • The sporophyte – diploid & heterogametic
  • Male sfemale gameto phyte – seven autosomes are similar.
  • In female 8th chromosome is X – Larger than the seven autosomes.
  • In male 8th chromosome is Y – Smaller than the autosomes.
  • In sporophyte – contain XY – combinations produces two types of meiospores
  • Meiospore with X – produce – female gemetophyte
Q.7The two loci A/a and D/d are so tightly linked that no recombination is ever observed. If AA dd is crossed to aa DD what phonotypes will be seen in the F2 and in what proportions.v
Answer:

In this scenario, the two loci, A/a and D/d, are so tightly linked that no recombination occurs between them. This means that the alleles present on the parental chromosomes will always be inherited together. The cross is between an individual with the genotype AA dd and an individual with the genotype aa DD. Assuming the genes are on different chromosomes initially or considering their linkage groups, the parental genotypes can be represented as Ad/Ad and aD/aD. When these individuals are crossed, the F1 generation will receive one set of alleles from each parent. Therefore, all F1 individuals will have the genotype Ad/aD. Since no recombination occurs, the only gametes that can be produced by these F1 individuals are the parental types: Ad and aD, each with a probability of 50%. When these F1 individuals self-pollinate (or are crossed with each other), the F2 generation will be produced by combining these gametes. The possible combinations are Ad x Ad, Ad x aD, aD x Ad, and aD x aD. This results in the following genotypes in the F2 generation: Ad/Ad (from Ad x Ad), Ad/aD (from Ad x aD and aD x Ad), and aD/aD (from aD x aD). The proportions will be 1/4 Ad/Ad, 1/2 Ad/aD, and 1/4 aD/aD. If we assume that 'A' is dominant over 'a' and 'D' is dominant over 'd', and that these genes control distinct phenotypes, the resulting phenotypes in the F2 generation will be determined by these genotypes. The genotype Ad/Ad will express the phenotype associated with alleles A and d. The genotype Ad/aD will express the phenotype associated with alleles A and D (since both dominant alleles are present). The genotype aD/aD will express the phenotype associated with alleles a and D. Therefore, the F2 generation will show two phenotypes: one corresponding to the Ad/Ad genotype and another corresponding to the aD/aD genotype, with the Ad/aD genotype exhibiting a third phenotype if A and D are dominant and the genes control different traits. However, if the question implies that the linkage maintains the parental combinations, the F2 phenotypes will directly reflect the parental combinations. Given the genotypes, the F2 phenotypes will be those corresponding to the 'Ad' combination and the 'aD' combination. The proportions will be 1/4 exhibiting the 'Ad' phenotype, 1/2 exhibiting a combined phenotype if both dominant alleles are expressed, and 1/4 exhibiting the 'aD' phenotype. If we consider the parental combinations strictly, the F2 phenotypes will be those derived from Ad/Ad and aD/aD, with Ad/aD potentially showing a different phenotype depending on dominance. Assuming standard dominance, the F2 phenotypes will be: Ad/Ad (phenotype A_d_), Ad/aD (phenotype A_D_), and aD/aD (phenotype aaD_). The proportions are 1/4 A_d_, 1/2 A_D_, and 1/4 aaD_. If the question implies that only the parental combinations are seen, then the phenotypes would be those of 'Ad' and 'aD'. The proportions will be 1/4 Ad/Ad, 1/2 Ad/aD, and 1/4 aD/aD.

Q.9Define point mutation & explain it’s typesv
Answer:

Definition:
Mutation affecting single base or base pair of DNA
Types:
* Indel mutation: (Base pair insertions or. addition. Addition or deletions of nucleotide
pairs.
* Substitution: one base pair is replaced by another
Types – (Two)
* (Purine replaced by Purine)
* Pyrimidine replaced by Pyrimidine
* Transversion purine replaced by pyrimidin or pyridine replaced
Synonymous or silent mutations:
Here change in one codon for an amino acid into another codon for that same amino acid
Missense or Non synonymous mutations
Here the codon for one amino acid is changed in to -a termination or stop codon.
Frameshift mutations.
Additions or deletions of a single base pair of DNA, – changed the reading frame for translation – so there is complete loss of normal protein structure & function.

Q.10Explain how translocation chromosomal aberration is different from crossing over?v
Answer:

Translocation and crossing over are both processes involving the exchange of genetic material, but they differ fundamentally in their nature, occurrence, and consequences. Crossing over is a normal, reciprocal exchange of genetic material between homologous chromosomes that occurs during Prophase I of meiosis. It involves the exchange of equivalent segments between paired homologous chromosomes and is a regular event that occurs in nearly all sexually reproducing organisms. Crossing over produces recombinant gametes with new combinations of alleles, which play a crucial role in generating genetic variation and driving evolution. In contrast, translocation is a chromosomal aberration involving the transfer of a segment of one chromosome to a non-homologous chromosome. It is an abnormal event that does not occur regularly and represents a structural chromosomal abnormality. Translocation can be reciprocal, where segments are exchanged between non-homologous chromosomes, or non-reciprocal, where a segment is simply transferred from one chromosome to another. Unlike crossing over, translocation rarely produces beneficial recombinations and often results in genetic imbalance, potentially causing harmful effects on the organism. Additionally, crossing over maintains the integrity of chromosomes and the balance of genetic material, whereas translocation can lead to loss or duplication of genetic material, causing chromosomal imbalances and genetic disorders.

Q.11Explain structural changes in chromosome with reference to changed to changes in the number of gene lociv
Answer:

There are 2 types
* Deletion
* Duplication
Deletion or Deficiency:
* (loss of a portion of chromosome)
* 2 types
* Terminal deletion (break in any one end
* Intercalary deletion (two breaks & reunion of terminal parts leaving the middle.
* > Unpaired loops some times formed known as deficiency loops (during meiotic prophase)
* > Larger deletions may have lethal effect Duplication or Repeat
* > Same order of genes repeated more than once in the same chromosome.
Eg. Drosophila
Duplication
3 types
* Tandem duplication
* Reverse tandem
* Displaced duplication
i) Tandem duplication
Duplicated segment is located immediately after the normal segment in the same order.
ii) Reverse tandem
Duplicated segment, immediately after the normal segment but gene sequence order will be reversed.
ii) Reverse tandem
Duplicated segment away from the normal segment.
Duplication play a maj or role in evolution.

Q.13Consider two hypothetical recessive auto¬somal genes a and b, where a heterozygote is testcrossed to a double homozygous mutant. Predict the phenotypic ratios under the following conditions: a) a and b are located on separate autosomes. b) a and b are linked on the same autosome but are so far apart that a crossover occurs between them. c) a and b are linked on the same autosome but are so close together that a crossover almost never occurs.v
Answer:

a) If the genes 'a' and 'b' are located on separate autosomes, they will assort independently according to Mendel's law of independent assortment. During the testcross of a heterozygote (AaBb) to a double homozygous recessive individual (aabb), the heterozygote will produce four types of gametes (AB, Ab, aB, ab) in equal proportions (25% each). The double homozygous recessive parent will only produce 'ab' gametes. Therefore, the phenotypic ratio of the offspring will be 1:1:1:1, with four distinct phenotypes corresponding to the combinations AB, Ab, aB, and ab. b) If the genes 'a' and 'b' are linked on the same autosome but are far apart, crossing over will occur frequently between them. This high frequency of crossing over will result in recombination, effectively separating the alleles. The genes will behave as if they are on separate chromosomes, leading to independent assortment. Thus, the phenotypic ratio will be the same as in case (a), i.e., 1:1:1:1. c) If the genes 'a' and 'b' are linked on the same autosome and are very close together, crossing over between them will occur very rarely, almost never. In this scenario, the alleles will tend to be inherited together as they were on the parental chromosomes. The heterozygote parent (assuming it was formed from a cross like AABB x aabb, resulting in an AB/ab genotype in the cis configuration, or AAbb x aaBB, resulting in an Ab/aB genotype in the trans configuration) will primarily produce parental gametes. If the heterozygote is AB/ab (cis), it will produce mostly AB and ab gametes, with very few Ab and aB recombinant gametes. If the heterozygote is Ab/aB (trans), it will produce mostly Ab and aB gametes, with very few AB and ab recombinant gametes. In a testcross with aabb, the progeny phenotypes will therefore be predominantly parental types, with a ratio close to 1:1 (parental types dominating) and a very small proportion of recombinant types.