12th Bio - Botany - Book Back Answers - Chapter 3 - English Medium Guides

 

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Tamil Nadu Board 12th Standard Bio-Botany - Chapter 3: Book Back Answers and Solutions

    This post covers the book back answers and solutions for Chapter 3 – Bio-Botany from the Tamil Nadu State Board 12th Standard Bio-Botany textbook. These detailed answers have been carefully prepared by our expert teachers at KalviTips.com.

    We have explained each answer in a simple, easy-to-understand format, highlighting important points step by step under the relevant subtopics. Students are advised to read and memorize these subtopics thoroughly. Once you understand the main concepts, you’ll be able to connect other related points with real-life examples and confidently present them in your tests and exams.

    By going through this material, you’ll gain a strong understanding of Bio-Botany Chapter 3 along with the corresponding book back questions and answers (PDF format).

Question Types Covered:

• 1 Mark Questions: Choose the correct answer, Fill in the blanks, Identify the correct statement, Match the following 
• 2 Mark Questions: Answer briefly 
• 3, 4, and 5 Mark Questions: Answer in detail

All answers are presented in a clear and student-friendly manner, focusing on key points to help you score full marks.

All the best, Class 12 students! Prepare well and aim for top scores. Thank you!

Unit 3 - Chromosomal Basis of Inheritance

I. Multiple Choice Questions

1. 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 Key:
c) Two copies of three different genomes

2. Match list I with list II

List I	List II
A. A pair of chromosomes extra with diploid	i) monosomy
B. One chromosome extra to the diploid	ii) tetrasomy
C. One chromosome loses from diploid	iii) trisomy
D. Two individual chromosomes lose from diploid	iv) double monosomy

a) A-i, B-iii, C-ii, D-iv
b) A-ii, B-iii, C-iv, D-i
c) A-ii, B-iii, C-i, D-iv
d) A-iii, B-ii, C-i, D-iv
Answer Key:
c) A-ii, B-iii, C-i, D-iv

3. 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 Key:
c) 3 and 4

4. 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 Key:
b) 7:1:1:7 

 
5. 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 Key:
c) C to G, A to G, T to A and G to A

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

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

8. 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 and R are correct
b) A is correct. R is wrong
c) A is wrong. R is correct
d) A and R are wrong
Answer Key:
b) A is correct. R is

II. Short Answer Questions

9. 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.

• The name of this phenomenon is coupling or cis configuration of linkage.
• This was reported in Sweet pea (Lathyrus odoratus)
• Exprimented by Willium Bateson and Reginald C. Punnet in 1906. 

 

• They crossed homozygous purple flowers and long pollen grains with red flowers and round pollen grains.
• All the F1 had purple flower and long pollen grains indicating purple flower long pollen (PL/PL) was dominant over red flower round pollen (pl/pl).
• When they crossed the F1 with double recessive parent (test cross) in results. F2 progenies did not exhibit in 1:1:1:1 ratio as expected with independent assortment.
• A greater number of F2 plants had purple flowers and long pollen or red flowers and round pollen.
• So they concluded that genes tor purple colour and long pollen grain and the genes for red colour and round pollen grain were found close together in the same homologous pair of chromosomes.
• These genes do not allow themselves to be separated.
• So they do not assort independently.
• This type of tendency of genes to stay together during separation of chromosomes is called Linkage.
• The two different genes came from same parent they tend to remain together is called coupling or cis configuration
• The observed phenotypic ratio is 7:1:1:7

 
10. What is the difference between missense and nonsense
mutation?

Missense mutation	Nonsense 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	The mutations where codon for one amino acid is changed into a termination or stop codon is called Nonsense mutation.

 

11. 

 From the above figure identify the type of mutation and explain it.

• The given figure is identify as reverse tandem duplication type of Structural changes in chromosome (Structural chromosomal aberration)
• Reverse tandem duplication
• The duplicated segment is located immediately after the normal segment but the gene sequence order will be reversed.

12. Write the salient features of Sutton and Boveri concept.

• Somatic cells are derived from the zygote by mitosis.
• These consist of two identical sets of chromosomes.
• One set is received from female parent and the other from male parent.
• These two constitute the homologous pair.
• Chromosomes retain their structural uniqueness and individuality throughout the life cycle.
• 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

 
13. Explain the mechanism of crossing over.
The stages of the mechanism of crossing over are 1. Synapsis, 2. Tetrad formation, 3. Cross over and 4. Terminalization.
(i) Synapsis

• The pairing between two homologous chromosomes are called synapsis ar syndesis.
• It is initiated during zygotenc stage of prophase 1 of meiosis 1.
• Homologous chromosomes pairs are aligned side by side called bivalents.

synapsis is of three types,

• Procentric synapsis: Pairing starts from middle of the chromosome.
• Proterminal synapsis: Pairing starts from the telomeres.
• Random synapsis: Pairing may start from anywhere.

(ii) Tetrad Formation

• Each homologous chromosome form two identical sister chromatids.
• It remain held together by a centromere.
• At this stage each bivalent has four chromatids.
• This stage is called tetrad stage.

(iii) Cross Over

• Crossing over occurs in pachytene stage.
• The non-sister chromatids of homologous pair make a contact at one or more points are called Chiasmata.
• At chiasma, cross-shaped or X-shaped structures are formed
• The breaking and re-joining of two chromatids occur in that point.
• This results in reciprocal exchange of equal and corresponding segments between them.

(iv) Terminalisation

• After crossing over, chiasma starts to move towards the terminal end of chromatids
• This is known as terminalisation.
• As a result, complete separation of homologous chromosomes occurs.

   

 

14. How is Nicotiana exhibit self-incompatibility. Explain its mechanism.

• In plants, multiple alleles have been reported in association with self-sterility or self-incompatibility.
• Self-sterility means that the pollen from a plant is unable to germinate on its own stigma.
• This will not be able to bring about fertilization in the ovules of the same plant.
• self-incompatibility or self-sterility observed by East (1925) in Nicotiana.
• The gene for self-incompatibility can be designated as S, which has allelic series S1, S2, S3, S4 and S5.
• The cross-fertilizing tobacco plants were not always homozygous as S1S1 or S2S2, 

• but all plants were heterozygous as S1S2, S3S4, S5S6.
• When crosses were made between different S1S2 plants, the pollen tube did not develop normally.
• But effective pollen tube development was observed when crossing was made with other than S1S2 for example S3S4,
• When crosses were made between seed parents with S1S2 and pollen parents with S2S3, two kinds of pollen tubes were distinguished.
• Pollen grains carrying S2 were not effective, but the pollen grains carrying S3 were capable of fertilization.
• Thus, from the cross S1S2XS354, all the pollens were effective and four kinds of progeny resulted: S1S3, S1S4. S2S3 and S2S4.
• Some combinations are showed in the table

Female parent (Stigma spot)	Male parent (Pollen source)
	S1S2	S2S3	S3S4
S1S2	Self Sterile	S3S2 S3S1	S3S1 S3S2 S4S1 S4S2
S2S3	S1S2 S1S3	Self Sterile	S4S2 S4S3
S3S4	S1S3 S1S4 S2S3 S2S4	S2S3 S2S4	Self Sterile

 
15. How sex is determined in monoecious

• Plants. write their genes involved in it. Zea mays (maize) is an example for monoecious.
• Monoecious means male and female flowers are present on the same plant.
• There are two types of inflorescence.
• 1. The terminal inflorescence bears staminate florets develops from shoot apical meristem called tassel.

Genotype	Dominant / recessive	Modification	Sex
ba/ba ts/ts	Double recessive	Lacks on the stalk, but transformed tassel to pistil	Rudiment ary female
ba/ba ts+/ts+	Recessive & dominant	Lacks silk and have tassel	Male
ba+/ba+ ts+/ts+	Double dominant	Have both tassel and cob	Monoecio us
ba+/ba+ ts/ts	Dominant & recessive	Bears cob and lacks tassel	Normal female

(Sex determination in Maize (Superscript (+)
denotes dominant character)

• 2. The lateral inflorescence develops pistillate florets from axillary bud is called ear or cob.
• Unisexuality in maize occurs through the selective abortion of stamens in ear florets and pistils in tassel florets.
• A substitution of two single gene pairs 'ba' for barren plant and 'ts' for tassel seed makes the difference between monoecious and dioecious (rare) maize plants.
• The allele for barren plant (ba) when homozygous makes the stalk staminate by eliminating silk and ears.
• The allele for tassel seed (ts) transforms tassel into a pistillate structure that produce no pollen.
• The table is the resultant sex expression. based on the combination of these alleles.
• Most of these mutations are shown to be defects in gibberellin biosynthesis.
• Gibberellins play an important role in the suppression of stamens in florets on the ears.

 
16. What is gene mapping? Write its uses.

• The diagrammatic representation of position of genes and related distances between the adjacent genes is called genetic mapping. Also called as linkage map.
• It was first developed by Morgan's student Alfred H Sturtevant in 1913. 
• Uses
• Determine gene order,
• Identify the locus of a gene and
• Calculate the distances between genes.
• It is useful in predicting results of dihybrid and trihybrid crosses.
• To understand the overall genetic complexity of particular organism

 
17. Draw the diagram of different types of aneuploidy (Refer book for clear diagram)         

 

                                                          

18. Mention the name of man-made cereal. How it is formed?

• Triticale is the successful first man made cereal.
• Hexaploidy Triticale hybrid plants demonstrate characteristics of both macaroni wheat and rye
• For example, they combine the high-protein content of wheat with rye's high content of the amino acid lysine, which is low in wheat. 
• It can be explained by chart below 

 

 

III. Long Answer Questions

12th bio botany

IV. Exercise

12thbio botany