12th Bio Botany
I. Choose the correct answer
1. Accurate 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 frequency
Answer:
b) Possibility of double cross over
2. Due 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) 1:7:7:1
Answer:
b) 7:1:1:7
2. Genes 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) HSLG
Answer:
c) SGHL
3. The 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 G
Answer:
b) A to G, C to T, C to G and T to A
4. If 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 19
Answer:
a) 35 and 37
5. Changing the codon AGC to AGA represents
a) missense mutation
b) nonsense mutation
c) frameshift mutation
d) deletion mutation
Answer:
a) missense mutation
6. Assertion (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 wrong
Answer:
c) A is correct. R is wrong explanation of A
7. How 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 cM
Answer:
c) 9 cM
8. An 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 genome
Answer:
c) Two copies of three different genomes
9. The 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, 5aB
Answer:
c) 45 AB, 45 ab
10. Which 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 4
Answer:
c) 3 and 4
11. When 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.
Answer:
i) The name of this phenomenon is known as Linkage.
This is reported in Sweet pea Lathyrus odoratus by Willium Bateson &
Reginald C Punnet in 1906.
Genes for Purple colour and Long pollen grain were found close together in
the same homologous pair of chromosomes – They do not assort independently
and this condition is known as linkage.
12. If you cross dominant genotype PV/PV male Drosophila with double recessive
female and obtain FI hybrid. Now you cross FI male with double recessive
female,
i) What type of linkage is seen?
ii) Draw the cross with correct genotype.
iii) What is the possible genotype in F2 generation?
13 .What is the difference between missense and nonsense mutation?
Answer:
Mis-sense Mutation:
The mutation where the codon for one amino acid is changed into a codon for
another amino acid is called Missense or non-synonymous mutations.
Non-sense Mutation:
The mutations where the codon for one amino acid is changed into a
termination or stop codon is called Nonsense mutation.
14. From the above figure identify the type of mutation and explain it.
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)
15. Write the salient features of Sutton and Boveri concept.
Answer:
Salient features of the chromosomal theory of inheritance:
1. 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.
2. Chromosomes retain their
structural uniqueness and individuality throughout the life cycle of an
organism.
3. Each chromosome carries specific
determiners or Mendelian factors which are now termed as genes.
4. The behaviour of chromosomes
during the gamete formation (meiosis) provides evidence to the fact that
genes or factors are located on chromosomes.
16. Explain the mechanism of crossing over.
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.
17. Write the steps involved in molecular mechanism of DNA recombination with
diagram.
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.
18. How is Nicotiana exhibit self¬incompatibility? Explain its mechanism.
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 – ‘S’ which has allelic series S1,
S2, S3, S4 & S5.
Cross-fertilizing tobacco – were not always homozygous as S1S1 or S2S2, but
heterozygous
Crosses between different S1S2 plants, pollen tube did not develop
normally.
But effective – development observed when cross was made with other than
S1S2 Eg. S3S4.
19. How is sex determined in monoecious plants. Write the genes involved in
it.
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.
20. What is gene mapping? Write its uses.
Answer:
The diagrammatic representation of the position of genes and related
distances between the adjacent genes is called genetic mapping. It is
directly proportional to the frequency of recombination between them. It is
also called a linkage map.
Uses of genetic mapping:
• It is used to determine gene
order, identify the locus of a gene and calculate the distances between
genes.
• They are useful in predicting the
results of dihybrid and trihybrid crosses.
• It allows the geneticists to
understand the overall genetic complexity of the particular organism.
I. Fill in the blanks
1. The scientists who independently rediscovered mendelian works were
Answer:
De Vries, Correns & Tschermak
2. The worm-shaped cells formed during cell division are called in the
earlier period as
Answer:
Chromosomes
3. Who postulated that the chromosomes of a cell are responsible for
transferring heredity
Answer:
Wilhelm Roux (1883)
4. ……………………… was the first to find out physical mutagen in Drosophila
Answer:
Muller (1927)
5. ………………………used X-rays for the first time to induce mutation in the fruit
fly
Answer:
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
Answer:
Auerback (1944)
8. Double nullisomy is
Answer:
2n-2-2
9. Trisomis were first reported by Blakeslee in
Datura Stramonium
10. All possible tetrasomics are available in ……………………… plant
Answer:
Wheat
11. The kind of Aneuploid are usually lethal are
Answer:
Nullisomy
12. The alkaloid used to induce polyploidy is
Answer:
Colchicine
13. Raphano brassicas the sterile hybrid of Radish & Cabbage was
produced by
Answer:
G.D. Karpechenko (1927)
14. The cross between hexaploid wheat Triticum aestivum and rye produced is
a
Answer:
Octoploidy
15. Colchicine is extracted from the root and corms of
Answer:
Colchicum autumnale
16. Who first reported duplication in drosophila
Answer:
Bridges (1919)
17. In which types of cells chromosomal aberration is commonly found?
Answer:
Cancer cells
18. Recombination frequencies are the same for
Answer:
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
Answer:
B-A-C
20. The percentage of crossing over will be more if
Answer:
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
………………………
Answer:
Nonsense mutation
22. Single base change in DNA is known as ………………………
Answer:
Point mutation
23. Genetic change in a non-sex cell is known as
Answer:
Somatic mulalion
24. A duplicated DNA sequence next to the original sequence is known
as
Answer:
Tandem duplication
25. A missing sequence of DNA or part of a chromosome
Answer:
Deletion mutation
26. Mutation that alters the genes reading frame is known as
Answer:
Frame shift mutation
27. A single base change mutation that alters and amino acid
………………………
Answer:
Missense
28. A spontaneous mutation usually originates as an error in ………………………
Answer:
DNA replication
29. The codon for leucine is CUC. How many different aminoacids could possibly result from a single base substitution
Answer:
7
30. How may map units separate two alleles if the recombination frequency is o.o7?
Answer:
7cM
II. Find out the incorrect statement
31. Which 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.
Answer:
c) The Homologus chromosomes segregate in metaphase I, thereby separating
two different alleles.
32. a) 94% of all flowering plants are sexually monomorphic
b) When three or more allelic forms of a gense occupy the same locus in
given pair of homologous chromosome they are known as Multiple alleles
c) The mutation that result in the change of one
codon of an aminoacid changed into codon of another amino acid is known as
Frame shift mutation.
d) Muller (1928) first time used x-rays to induce mutation in
Drosophila
Answer:
c) The mutation that result in the change of one codon of an aminoacid
changed into codon of another amino acid is known as Frame shift
mutation.
33. Which 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.
Answer:
d) If the genes are present on the same chromosome, they undergo more than
one crossover in every meiosis.
34. a) Selfing of monosomic plants produce nullisomics.
b) In a true diploid both the monoploid and haploid chromosome number are
same.
c) An Auto triploids can be produced artificially by crossing between
haploid and a diploid.
d) An increase in the number of chromosome
sets has been an important factor in the origin of new plant species.
Answer:
c) An Auto triploids can be produced artificially by crossing between
haploid and a diploid.
III. Match the following
Choose the correct statement
1. When 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.
Answer:
c) Both males and females are with red eye
Find the Odd man out with reference to Allopolyploidy
1. a) 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..
Answer:
b) They have four or six copies of its own genome – induced by doubling of
the diploid species.
Find the Odd man out with reference to Altotriploidy
1. a) Thev can be produced artificially bv crossing between autotetraploid
& diploid.
b) They are highly fertile due to large number of gametes.
c) Cultivated triploid bananas are seedless having larger fruits than
diploid.
d) Common doob cross is a natural autotriploid.
Answer:
b) They are highly fertile due to large number of gametes.
Find the Odd man out regarding crossing over
1. a) 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.
Answer:
c) It is directly proportional to the frequency of recombination between
them.
Choose the wrongly matched pair
1.
a. Crossing over term – coined by T.H. Morgan
b. Concept of gene mapping developed by
A.H. Sturtevant
c. Linkage in Lathyrus odoratus is sludied by
Bateson & Punnet
d. Widely accepted Model of DNA recombination was 1st proposed by
Robert Koch
Answer:
d
2. a. The condition in which a pair of chromosomes is lost
Nullisomy
b. The condition in which one chromosome is lost
Monosomy
c The condition in which two chromosome is gained
Tetrasomy
d. The condition in which one chromosome is gained
Diplosomy
Answer:
d
a. 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
Pericentric
Answer:
c
Choose the incorrect statement with reference to Deletion
1. a) Deletions occur due to chemicals, drugs & radiation.
b) On the basis of location of breakage on chromosome it is divided in to
Terminal deletion & inter calary deletion
c) Larger deletions have evolutionary significance.
d) Deletions are recorded in Drosophila & Maize
Answer:
c) Larger deletions have evolutionary significance.
2. How can we reverse the sterility of FI hybrid?
a) Genetic Engineering
b) Protoplasmic fusion
c) Induced Mutation
d) Induced chromosomal aberration
Answer:
d) Induced chromosomal aberration
3. If 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 48
Answer:
a) 44 and 49
4. Genes located close together on the same chromosome and inherited together
represented as
a) linked genes
b) unlinked gene
c) syntenic genes
d) trans genes
Answer:
a) linked genes
Assertion (A) & Reason (R)
1. Assertion (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.
Answer:
b) (A) is correct, (R) is the correct explanation (A)
2. Assertion (A) : Linkage and crossing over are two processes that have
opposite effects. Reason (R) : Linkage keeps particular genes together but
crossing over mixes them.
Answer:
a) If both the Assertion (A) & Reason (R) are true and the reason is a
correct explanation of the Assertion. .
3. Assertion (A) : Increase in temperature increases the rate of
mutation.
Reason (R) : While rise in temperature hydrolyses DNA by the restriction
endonuclease which degrade Nucleotides.
Answer:
c) Assertion (A) is true but Reason (R) is false
Two Marks
1. Define chromosome theory of inheritance.
Answer:
It states the Mendelian factors (genes) have specific locus on chromosomes
& they carry information from one generation to the next
generation.
2. State the number of chromosomes of the given organism.
Answer:
1) Ophioglossum 2) Arabiodopsis 3) Sugarcane 4) Rice 5) Potato 6)
Maize
Answer:
1) -1262;
2) -10;
3) 80;
4) 24;
5) 48;
6) 20
3. What are Fossil Genes?
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.
- 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.
5. What are co-mutagens ?
Answer:
Compounds which are not having own mutagenic properties – but enhance the
effects of known mutagens.
Eg. Ascorbic acid – increase the damage caused by hydrogen peroxide.
Caffeine – increase the toxicity of methotrexate.
6. Differentiate between Euploidy & Aneuploidy
Answer:
Eupoidy :
- Ploidy involving entire sets of chromosomes is known as euploidy
- Triploidy (3x); Tetraploidy (4x); Poly ploidy ( ∞n)
Aneuploidy :
- Here the diploid number is altered either by addition or deletion of one or more chromosomes
- Trisomy; Tetrasomv; Monosomy; Nullisomy (2n+1)(2n+2)(2n-1)(2n-2)
7. Distinguish between Monoploidy & Haploidy
Answer:
Monoploidy
In Monoploidy the chromosome number is referred as x .
Eg.
Hexaploidy wheat
(2n) = 6 x = 72
haplaid = (n) 36
Monoploidy = x = 12
Haploidy :
Half the number of somatic chromosomes is referred as gametic chromosome
number called haploidy (n) Human of haploid = 23 (n) Wheat of haploid = 36
(n)
8. Independent assortment & Linkage are alternatives of each other –
Discuss
Answer:
Independent assortment Linkage
- Genes present in different chromosome they assort independently
- Genes present in same chromosome tend to stay together
- More parental combinations & less new combinations are possible
- More parental combinations are possible
9. How does the strength and weakness of linkage depend on linked genes?
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.
10. Distinguish between crossing over & Reciprocal Translocation.
Crossing over
1. It is legitimate &
natural
2. Occurs between nonsister
chromatids of homologous chromosomes
3. Occurs between nonsister
chromatids of homologous chromosomes
Reciprocal Translocation :
1. It is illegitimate &
chromosomal abnormality
2. Occurs between non sister
chromatids of non homologous chromosomes
3. Also play major rome in formation
of species
11. Distinguish between tetrad & bivalent Tetrad:
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.
12. Define Recombination.
Answer:
In this segments of DNA one broken and recombined to produce new
combination of alleles – known as Recombination.
13. What is RF (Recombination Frequency)
Answer:
The frequency with which recombination occur is a certain condition
14. A diploid organism is heterozygous for 4 loci. How many types of gametes
cars be produced.
Answer:
The formula 2n is applied – Organism hetr oizy gous f or 4 loci = n =
4.
So 2n = 24 = 2 x 2 x 2 x 2 = 16.
The organism produces 16 types of gametes.
15. Notes on Colchicine.
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
16. Write down the significance of ploidy.
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.
17. Distinguish between Mendelian disorder & Chromosomal disorder.
Answer:
Mendelian disorder:
Occur due to mutation of single gene & follow the well known Mendelian
pattern of inheritance.
Eg. Sickle cell anaemia
Chromosomal disorder :
Chromosomal disorders are produced due to alteration in the number of
chromosomes.
Eg. Down syndrome
19. This is a type of Numerical chromosomal abnormality find it out give a note
on it.
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)
21. What is chiasmata?
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.
22. What is multiple alleles?
Answer:
When any of the three or more allelic forms of a gene occupy the same locus
in a given pair of homologous chromosomes, they are said be called multiple
alleles.
23. What is monomorphic?
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.
24. What is Dimorphic?
Answer:
Some 6% of flowering plants which have two separate sexes are called
dimorphic.
Three Marks
Question 1.
Differentiate tetrasomy from tetraploidy
Answer:
Tetrasomy Tetraploidy
Addition of a pair or two individual chromosomes to diploid set is called
Tetrasomy. (2n + 2) Tetraploids have
four eopies of its own genome. They can be induced by doubling the
chromosomes of a diploid species.
(2n + 2 + 2) This condition is known as double tetrasomy
There are two types Auto & Allotetraploidy
Eg: wheat Eg: Grapes, ground nut,
potato & coffee
Question 2.
Give a tabulation comparing the behaviour of gene & Chromosome
Answer:
Medelian Factors Chromosom at
behaviour
1. Alleles of a factor occur in pair Chromosomes occur in pair’s
2. Similar or dissimilar alleles of a factor separate during the gamete
formation The homologous chromosomes
during meiosis.
3. Mendelian factors can assort independently
The paired chromosomes can separate independently
during meiosis, but the linked genes in the same chromosome normaly do not
assort independently.
Question 3.
The important aspects about the chromosome behaviour during cell
devision.
Answer:
• Alleles of a genotype – found in
the same locus of a homologous chromosome (A/a)
• In ‘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)
Question 4.
Write the differences between coupling and Repulsion
Answer:
Coupling
1. The two dominant alleles or
recessive alleles called repulsion or trans configuration
2. It tend to inherit together into
same gametes
Repulsion :
1. If dominant or recessive alleles
are present on occur in the same homologous chromosomes two different but
homologous chromosomes.
2. If they inherit apart in to
different game es are
Question 5.
Define synapsis.
What are the types of Synapsis
Answer:
• During cygotene stage of prophase
I of meiosis I – homologous chromosomes are aligned side by side resulting
in a pair called (bivalents).
• This pairing phenomenon is called
synapsis or syndesis.
Based on the starting poiring of pairing there are 3 types of
synapsis
Procentric Proterminal
Random
Starts from middle Starts from the
telomeres Starts from any where
Question 6.
Distinguish between sharbati sonora & Castor Aruna.
Answer:
Sharbati sonora :
1. Multant variety of Wheat –
approved in 1967
2. Sonora 64 (Mexican variety
subjected to gamma rays to produce sharbati sonora
3. Developed by Dr. M.D.
Swaminathan
4. Early maturing & high protein
content high kneading quality
Castor Aruna :
1. Mutant castor variety
2. Seeds treated with thermal
neutrons
3. Early maturing – (120 days
instead of 270 dyas) & High yielding.
Question 7.
How do increase in temperature cause mutation?
Answer:
Rise temperature breaks the hydrogen bonds between two DNA nucleotides –
& affects the process of replication & transcription.
Question 8.
Distinguish between the impact of ionizing & non ionizing radiation in
causing mutation. Ionizing radiation Non Ionizing radiation
Answer:
Ionizing radiation :
• Short wave length and carry enough
higher energy to ionize electrons from atoms. They breaks the chromosome
& chromatids. Ex. x-rays – gamma rays, alfa rays, beta rays & cosmic
rays.
Non Ionizing radiation :
• Longer wave lengths and carry
lower energy so they have lover penetrating power used to treat unicellular
microbes – spores pollengrains – which have nuclei – near surface membrance.
Eg. UV rays
Question 9.
What is significance of ploidy?
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.
Question 10.
What is chemical mutagens? Give an example?
Answer:
Chemical which include mutation are called mutagens.
Example:
Nitrous oxide alters the nitrogen bases of DNA and disturb the replication
and transcription that leads to the formation of incomplete and defective
polypeptide during translation.
Question 11.
What is cis configuration (or) coupling?
Answer:
The two dominant alleles or recessive alleles occur in the same homologus
chromosomes, tend to inherit together into same gamete are called coupling
(or) cis configuration
Five Marks
1. Whose works supported the chromosomal theory of heredity? Explain.
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.
2. Write down the steps in the Holliday’s hybrid DNA model.
Answer:
- Homologous DNA molearles 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 & join the homologous strands forming the Holliday junction
- Holliday junction – migrates away from the original site, by branch migration – as a result
- heteroduplex region is formed.
- DNA strands may cut along the vertical (V) or horizontal (H) line.
- The vertical cut will result in heteroduplexes with recombinants & the Horizontal with non recombinants.
3. Explain sex determination is Silene latifolia (Melandrium album)
Answer:
- C.E Allen (1917) discovered sex determination in plants.
- Complex precess determined by
1. genes
2. environment
3. hormones
Sex determination silene latifolia – is controlled by 3 distinct regions in
a sex chromosome
- Y – Chromosome – determines maleners
- X – Chromosome – specify femaleness
- X & Y – show different segments (I, II, III, IV, & V)
4. How do Hawaii explain the sex determination in Papaya
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.
5. Explain sex determination in Sphaerocarpos donnelli. It is also known as
Bottle liverwort (Bryophyta)
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
6. Define point mutation & explain it’s types
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)
1. (Purine replaced by Purine)
2. Pyrimidine replaced by
Pyrimidine
3. 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.
7. Explain how translocation chromosomal aberration is different from crossing
over?
Answer:
crossing over Translocation
It is an exchange of genetic material between homologous chromosomes that
occurs during Prophase I of meiosis during gametes formation
It is a genetic abnormality involving the exchange of fragments of genes
between
non-homologous chromosomes
It is a normal event occurring in almost all sexually reproducting
organisms. It is a chromosomal
aberration.
It often produces recombinations which play important role in
evolution. It rarely produces
recominations.
8. Explain structural changes in chromosome with reference to changed to
changes in the number of gene loci
Answer:
There are 2 types
- Deletion
- Duplication
Deletion or Deficiency :
- (loss of a portion of chromosome)
- 2 types
1. Terminal deletion (break in any
one end
2. 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.
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