Important Questions

Punjab State Board PSEB 12th Class Biology Important Questions Chapter 6 Molecular Basis of Inheritance Important Questions and Answers.

PSEB 12th Class Biology Important Questions Chapter 6 Molecular Basis of Inheritance

Question 1.
Name the specific components and the linkage between them that form deoxyadenosine.
Answer:
Nitrogenous base (Adenine) and pentose sugar and N-glycosidic linkage.

Question 2.
Why is RNA more reactive in comparison to DNA?
Answer:
RNA is more reactive because:

• It is single stranded.
• Every nucleotide has an additional OH^– group present at position 2 in the ribose.
• Mutates faster as compared to DNA.

Question 3.
In an experiment, DNA is treated with a compound which tends to place itself amongst the stacks of nitrogenous base pairs. As a result of this, the distance between two consecutive base increases from 0.34 nm to 0.44 nm. Calculate the length of DNA double helix, which has 2 × 10⁹ bp in saturating the presence of this compound. [NCERT Exemplar]
Answer:
The new length of DNA helix would be
= 2 × 10^-9 × 0.44 × 10^-9bp.

Question 4.
In a nucleus, the number of RNA nucleoside triphosphates is 10 times more than the number of DNA nucleoside triphosphates,
still only DNA nucleotides are added during the DNA replication, and not the RNA nucleotides. Why?
[NCERT Exemplar]
Answer:
DNA polymerase is highly specific to recognise only deoxyribonucleoside r triphosphates. Therefore, it cannot hold RNA nucleotides.

Question 5.
Name the enzyme involved in the continuous replication of DNA strand. Mention the polarity of the template strand.
Answer:
DNA polymerase is involved in continuous replication of DNA strand. The polarity of template strand is 3′ → 5′.

Question 6.
What is a cistron?
Answer:
Cistron is a segment of DNA coding for a polypeptide chain.

Question 7.
Name the transcriptionally active region of chromatin in a nucleus.
Answer:
Euchromatin or Exon.

Question 8.
Write the function of RNA polymerase n.
Answer:
RNA polymerase II transcribes precursor of mRNA or hnRNA.

Question 9.
Mention the two additional processings which /mRNA needs to
undergo after splicing so as to become functional.
Answer:
Capping and tailing.

Question 10.
Give an example of a codon having dual function.
Answer:
AUG has dual function. It acts as initiation codon and also codes for methionine.

Question 11.
Sometimes cattle or even human beings give birth to their young ones that have extremely different sets of organs like limbs/position of eye(s), etc. Why? [NCERT Exemplar]
Answer:
This is due to a disturbance in coordinated regulation of expression of sets of genes associated with organ development or due to mutations.

Question 12.
How does a degenerate code differ from an unambiguous one?
Answer:
Degenerate code means that one amino acid can be coded by more than one codon. Unambiguous code means that one codon codes for only one amino acid.

Question 13.
What is aminoacylation? State its significance.
Answer:
Aminoacylation of t-RNA involves activation of amino acids by ATP which gets linked to OH’ present at the 3′ end of specific t-RNA. The process is also called charging of t-RNA.

Question 14.
Mention how c|oes DNA polymorphism arise in a population?
Answer:
DNA polymorphism in a population arise due to presence of inheritable mutations at high frequency.

Question 15.
How is repetitive/satellite DNA separated from bulk genomic DNA for various genetic experiments?
Answer:
By using density gradient centrifugation, where satellite DNA forms small peaks.

Short answer type questions

Question 1.
Recall the experiments done by Frederick Griffith, Avery, MacLeod and McCarty, where DNA was speculated to be the genetic material. If RNA, instead of DNA was the genetic material, would the heat killed strain of Pneumococcus have transformed the 12-strain into virulent strain? Explain.
[NCERT Exemplar]
Answer:
RNA is more labile and prone to degradation, owing to the presence of 2’OH group in its ribose. Hence, heat-killed S-strain may not have retained its ability to transform the R-strain into virulent form if RNA was its genetic material.

Question 2.
Name a few enzymes involved in DNA replication other than DNA polymerase and ligase. Name the key functions for each of them. [NCERT Exemplar]
Answer:
The enzymes involved in DNA replication other than DNA polymerase and ligase are listed below with their functions:\

• Helicase – Opens the helix
• Topoisomerases – Removes the tension caused due to unwinding
• DNA ligase – Joins the cut DNA strands

Question 3.
Following are the features of genetic codes. What does each one indicate?
Stop codon; Unambiguous codon; Degenerate codon; Universal codon.
Answer:

• Stop codon is a codon, which does not code for any amino acid and here the polypeptide chain is released e.g., 3 codons – UAA, UAG, UGA.
• Unambiguous codon : One codon codes for only one specific amino acid.
• Degenerate codon : Here, one amino acid is coded by more than one codon e.g., amino acid glycine is coded by four codons (GGU, GGC, GGA, GGG).
• Universal codon means a codon specifies the same amino acid in all
  the organisms even in a virus. ,

Question 4.
A single base mutation in a gene may not ‘always’ result in loss or gain of function. Do you think the statement is correct? Defend your answer. [NCERT Exemplar]
Answer:
The statement is correct because of degeneracy of codons, mutations at third base of codon, usually does not result into any change in phenotype. This is called silent mutations but at other times it can lead to loss or formation of malformed protein changing the phenotype.

Question 5.
(a) Name the scientist who called tRNA an adapter molecule.
(b) Draw a clover leaf structure of fRNA showing the following:
(i) Tyrosine attached to its amino acid site.
(ii) Anticodon for this amino acid in its correct site (codon for tyrosine is UCA).
(c) What does the actual structure of JRNA look like?
Answer:
(a) Francis Crick
(b)
(c) The actual structure of t-RNA looks like inverted L.

Question 6.
A low level of expression of lac operon occurs at all the time. Can you explain the logic behind this phenomena. [NCERT Exemplar]
Answer:
In the complete absence of expression of lac operon, permease will not be synthesised which is essential for transport of lactose from medium into the cells. And if lactose cannot be transported into the cell, then it cannot act as inducer. Hence, cannot relieve the lac operon from its repressed state. Therefore, lac operon is always expressed.

Question 7.
Would it be appropriate to use DNA probes such as VNTR in DNA fingerprinting of a bacteriophage? [NCERT Exemplar]
Answer:
Bacteriophage does not have repetitive sequences such as VNTRs in its genome, as its genome is very small (i.e., 5386bp) and have all the coding sequence. Therefore, DNA fingerprinting is not done for phages.

Long answer type questions

Question 1.
Describe Meselson and Stahl’s experiment that was carried in 1958 on E.coil. Write the conclusion they arrived at after the experiment.
Matthew Meselson and Franklin Stahl (1958) grew E. coli in a medium containing ¹⁵NH₄Cl (¹⁵N is the heavy isotope of nitrogen) as
the only nitrogen source for many generations. The result was that ¹⁵N was incorporated into newly synthesised DNA (as well as other nitrogen containing compounds). This heavy DNA molecule could be distinguished from the normal DNA by centrifugation in a cesium chloride (CsCl) density gradient.

They transferred the cells into a medium with normal ¹⁴NH₄Cl and took samples at various definite time intervals as the cells multiplied, and extracted the DNA that remained as double-stranded helices. The various samples were separated independently as CsCl gradients to measure the densities of DNA.

Thus, the DNA that was extracted from the culture one generation after the transfer from ¹⁵N to ¹⁴N medium (that is after 20 minutes; E.coli divides in 20 minutes) had a hybrid or intermediate density. DNA extracted from the culture after another generation (that is after 40 minutes, II generation) was composed of equal amounts of hybrid DNA and of light DNA.

Matthew Meselson and Franklin Stahl’s experiment demonstrated that DNA replication is semi-conservative.

Generation I Generation II

Question 2.
(a) Describe the series of experiments of F. Griffith. Comment on the significance of the results obtained.
(b) State the contribution of MacLeod, McCarty and Avery.
Answer:
(a) Frederick Griffith (1928), a British doctor, was studying the pathogenicity of different strains of Streptococcus pneumoniae. It has two strains- (i) virulent cause pneumonia, has S-type of bacteria, covered by sheath of mucilage, (ii) non-virulent do not produce the disease, has R-type of bacteria, devoid of sheath of mucilage.

Griffith found that on injecting live R-type bacteria did not produce the disease while live S-type caused pneumonia and the death in mice. However, when heat-killed S-type injected, they did not produce the disease. Finally, Griffith injected a combination of live-R-type and heat-killed S-type bacteria into mice. Some mice survived while others developed the disease of pneumonia and died. Autopsy of dead mice showed that they possessed both the type of bacteria (virulent-S-type and non-virulent-R-type) in living form through the mice that had been injected with dead virulent (S-type) and living non-virulent (R-type) bacteria.

From the above experiment, Frederick Griffith concluded that the occurrence of living S-type virulent bacteria is possible only by their formation from R-type non-virulent bacteria which pick-up the trait of virulence from dead bacteria. This phenomenon is called Griffith effect or transformation. Griffith proposed that the transforming principle is a chemical substance released by heat-killed S-type, which changed the S-type into S-bacteria. It was a permanent change as the new S-type formed only S-type progeny.

(b) Avery, MacLeod, McCarty discovered that DNA from the heat-killed S-strain caused the living R-strain bacteria to become transformed into living S-type. They found proteases and RNAases did not affect transformation while DNAases inhibit transformation. They concluded that DNA is the hereditary material.

Punjab State Board PSEB 12th Class Biology Important Questions Chapter 5 Principles of Inheritance and Variation Important Questions and Answers.

PSEB 12th Class Biology Important Questions Chapter 5 Principles of Inheritance and Variation

Question 1.
Mention any two contrasting traits with respect to seeds in pea plant that were studied by Mendel.
Answer:
Round/Wrinkled, Yellow/Green.

Question 2.
Name the type of cross that would help to find the genotype of a pea plant bearing violet flowers.
Answer:
Test Cross.

Question 3.
State a difference between a gene and an allele.
Answer:
Gene is a unit of heredity passed from’one generation to next generation and determine the expression of any morphological or physiological inheritable character of an organism.

Alleles are alternative form of a gene, occupying the same position on homologous chromosomes and affecting the alternative forms (contrasting traits) of the same character.

Question 4.
A garden pea plant (A) produced inflated yellow pod, and another plant (B) of the same species produced constricted green pods. Identify the dominant traits.
Answer:
Inflated green pod is the dominant trait.

Question 5.
Mention the type of allele that expresses itself only in homozygous state in an organism.
Answer:
Recessive allele.

Question 6.
A garden pea plant produced axial white flowers. Another of the same species produced terminal violet flowers. Identify the dominant traits.
Answer:
Axial, violet flower.

Question 7.
What are ‘true breeding lines’ that are used to study inheritance pattern of traits in plants?
Answer:
True breeding lines are plants which have undergone continuous self-pollination for several generations. These are homozygous for traits.

Question 8.
Name the stage of cell division where segregation of an independent pair of chromosomes occurs.
Answer:
Anaphase-I of Meiosis-I.

Question 9.
A male honeybee has 16 chromosomes whereas its female has 32 chromosomest Give one reason.
Answer:
A male honeybee with 16 chromosomes develops parthenogenetically from an unfertilised egg and is haploid (n) whereas the female honeybee with 32 chromosomes develops from a fertilised egg, the zygote and is diploid (2n).

Question 10.
Why is it that the father never passes on the gene for haemophilia to his sons? Explain.
Answer:
Haemophilia is a sex-linked recessive disease and the defective gene is present on X chromosome only and not on Y chromosome. Father never passes X chromosome to the son as father only contributes Y chromosome to the son.

Question 11.
Write the chromosomal defect in individuals affected with Klinefelter’s syndrome.
Answer:
Klinefelter syndrome is caused due to the presence of an additional X-chromosome in the genotype of an individual i.e., 44 + XXY.

Question 12.
State the chromosomal defect in individuals with Turner’s syndrome.
Answer:
In Turner’s syndrome, the karyotype of the individual is 44 + XO. The X-chromosome is missing. It is due to the non-disjunction of X-chromosomes during oogenesis/spermatogenesis.

Short answer type questions

Question 1.
Mendel published his work on inheritance of characters in 1865, but it remained unrecognised till 1900. Give three reasons for the delay in accepting his work.
Answer:

• The communication was not easy in those days and his work could not be widely publicised.
• His concept of genes as stable and discrete units that controlled the expression of traits and of the pair of alleles which did not ‘blend’ each other was not accepted by contemporaries as an explanation for the apparently continuous variation seen in nature.
• Mendel’s approach of using mathematics to explain biological phenomena was totally new and unacceptable to many of the biologists of his time.
• Though Mendel’s work suggested that factors (genes) were discrete units, he could not provide any physical proof for the existence of factors what they were made of.

Question 2.
The pedigree chart given below shows a particular trait which is absent in parents but present in the next generation irrespective of sexes.
Draw your conclusion on the basis of the pedigree. [NCERT Exemplar]

Answer:
The pedigree chart shows that the trait is autosomal linked and recessive in nature. But, the parents are carriers (i.e. heterozygous) hence, among the offsprings only few show the trait irrespective of sex. The other offsprings are either normal or carrier.

Question 3.
What is Down’s syndrome? Give its symptoms and cause. Why is it that the chances of having a child with Down’s syndrome increases if the age of the mother exceeds forty years?
[NCERT Exemplar]
Answer:
Down’s syndrome is a human genetic disorder caused due to trisomy of chromosome 21. Such individuals are anploid and have 47 chromosomes (2n + 1). The symptoms include mental retardation, growth abnormalities, constantly open mouth, dwarfness, etc. The reason for the disorder is the non-disjunction (failure to separate) of homologous chromosome of pair 21 during meiotic division in the ovum.

The chance of having a child with Down’s syndrome increase with the age of the mother (40+) because ova are present in females since their birth and therefore older cells are more prone to chromosomal non-disjunction because of various physicochemical exposures during the mother’s life-time.

Question 4.
Differentiate between male and female heterogamety.
Answer:
In male heterogamety, the male is heteromorphic and have XY or XO type of sex chromosomes and produce two types of sperms, 50% with X- chromosome and 50% with Y-chromosome or none. The sex of the offspring depends upon the type of sperm, which fuses with egg e.g., mammals, Drosophila, grasshopper.

In female heterogamety, the female is heteromorphic and heterogametic and have ZW or ZO type of sex chromosomes and produce two types of eggs. The sex of the offspring depends upon the type of egg, which is fertilised, e.g., bird and some reptiles, butterflies and moths.

Question 5.
Explain mechanism of sex-determination in birds.
Answer:
Sex-determination in birds is opposite to human beings. Here the females contain heteromorphic sex chromosomes (AA + ZW) while the males have homomorphic (AA + ZZ) chromosomes. Thus, there is female heterogamety. The females are heterogametic and produce two types of eggs (A + Z) and (A + W). The male gametes are of one type (A + Z).

Question 6.
Why are human females rarely haemophilic? Explain. How do haemophilic patients suffer?
Answer:
Haemophilia is a sex-linked recessive disorder. The females have XX chromosomes and the males have XY chromosomes. If one of the two X chromosomes is normal, she remains a carrier and not diseased. Females will haemophilic only when both the X chromosomes carry the haemophilia gene, and this is possible only when the mother is a carrier and father is haemophilic. Haemophilic patients suffer from non-stop bleeding and nd clotting.

Question 7.
How do genes and chromosomes share similarity from the point of view of genetical studies? [NCERT Exemplar]
Answer:
By 1902, the chromosome movement during meiosis had been worked out.
Walter Sutton and Theodore Boveri (1902) noted that the behaviour of chromosomes was parallel to the behaviour of genes and used chromosome movement to explain Mendel’s laws.
They studied the behaviour of chromosomes during mitosis (equational division) and during meiosis (reduction division). The chromosomes as well as genes occur in pairs and the two alleles of gene pair are located on homologous sites of homologous chromosomes.

Chromosomes segregate when germ cells are formed.

Question 8.
Write short notes on –
(i) Phenylketonuria
(ii) Aneuploidy
Answer:
(i) Phenylketonuria : It is an inborn error of metabolism. The affected individual lack an enzyme called phenylalanine hydroxylase that converts the amino acid phenylalanine into tyrosine. As a result, phenylalanine gets accumulated and converted into phenylpyruvic acid and other derivatives in brain, causing mental retardation. These are also excreted through urine due to their absorption by kidney.

(ii) Aneuploidy: It is a phenomenon which occurs due to non¬disjunction resulting into gain or loss of one or more chromosomes, during meiosis.

Long answer type questions

Question 1.
(a) A garden pea plant bearing terminal, violet flowers, when crossed with another pea plant bearing axial, violet flowers, produced axial, violet flowers and axial, white flowers in the ratio of 3 : 1. Work out the cross showing the genotypes of the parent pea plants and their progeny.
(b) Name and state the law that can be derived from this cross and not from a monohybrid cross.
Answer:

(b) Law of Independent Assortment : When two pairs of traits are combined in a hybrid, segregation of one pair of characters is independent of the other pair of character.

Question 12.
(a) Write the blood group of people with genotype IAIB. Give reasons in support of your answer.
(b) In one family, the four children each have a different blood group. Their mother has blood group A and their father has blood group B. Work out a cross to explain how it is possible.
Answer:
(a) Blood group AB. Both the alleles I^A and I^B are co-dominant and express themselves completely.
(b) A cross is carried out between heterozygous father (of blood group B) and heterozygous mother (of blood group A) to get four children with different blood groups.

All the four blood groups are controlled by three allelic genes I^A, I^B, i
and thus it shows phenomena of multiple allelism. Both I^A and I^B are dominant over i. However, when together, both are dominant and show the phenomena of co-dominance forming the blood group AB. Six genotypes are possible with combination of these three alleles.

Question 3.
(A) Why are colourblindness and thalassemia categorised as Mendelian disorders? Write the symptoms of these diseases seen in people suffering from them.
(B) About 8% of human male population suffers from colourblindness whereas only about 0.4% of human female population suffers from this disease. Write an explanation to show how it is possible.
Answer:
(A) Colourblindness and thalassemia are categorised as Mendelian disorders because they are (i) due to alteration or mutation in a single gene (ii) transmission to the offspring follow principle of inheritance (iii) can be studied by pedigree analysis.
Symptoms of Colourblindness
(a) Difficulty in distinguishing between colours

• Protanopia-red colourblindness
• Deuternopia-green colourblindness
• Tritanopia-blue colourblindness

(b) Rapid eye movement (in rare cases)
(c) Inability to see shades or tones of the same colour.
(d) In rare cases, some people see only black, white and grey.

Symptoms of Thalassemia
(a) Formation of abnormal haemoglobin molecules resulting into haemolytic anaemia.
(b) Slow growth, delayed puberty.
(c) Bones become wider than normal, brittle and break easily
(d) Poor appetite
(e) A pale and listless appearance
(f) Dark urine
(g) Enlarged spleen, liver or heart.

(B) Colourblindness is a recessive sex-linked disorder in which the patient cannot distinguish red-green colour. The gene for colourblindness is present on X-chromosome. Presence of colourblindness in 8% human male population, is due to the presence of a single X-chromosome in male. There is no chance of dominant and recessive condition, as there is a single gene. Hemizygous condition is enough for the occurrence of defect in males. In females, due to presence of two X-chromosomes, the single gene of colourblindness cannot express and such females are carriers (XX^c) for a female to be colourblind, both of her X-chromosome, carry gene for colourblindness(X^cX^c).

Punjab State Board PSEB 12th Class Biology Important Questions Chapter 4 Reproductive Health Important Questions and Answers.

PSEB 12th Class Biology Important Questions Chapter 4 Reproductive Health

Very short answer type questions

Question 1.
Reproductive health refers only to healthy reproductive functions. Comment. [NCERT Exemplar]
Answer:
Reproductive health refers to the total well-being in all aspects of reproduction, i.e., physical, behavioural, psychological and social.

Question 2.
The present population growth rate in India is alarming. Suggest ways to check it. [NCERT Exemplar]
Answer:

• By increasing marriageable age.
• By promoting use of birth control measure.
• By educating people about consequences of un- controlled population growth.

Question 3.
Why do intensely lactating mothers not generally conceive? [NCERT Exemplar]
Answer:
Due to suppression of gonadotropins, ovulation and menstrual cycle do not take place.

Question 4.
Name an IUD that you would recommend to promote the cervix hostility to sperms.
Answer:
The hormone releasing IUD’s, e.g. progestasert, LNG-20 are , recommended to promote the cervix hostility to sperms.

Question 5.
Mention any tvgo events that are exhibited by the intake of oral contraceptive pills to prevent pregnancy in humans.
Answer:
Two events that are exhibited by the intake of oral contraceptive pills to prevent pregnancy in humans are ovulation and implantation.

Question 6.
Why is tubectomy considered a contraceptive method?
Answer:
Tubectomy involves cutting a piece of the fallopian tube and tying its ends. This way, the sperms are not able to reach the egg and it acts as a contraceptive method.

Question 7.
Mention the primary aim of the ‘Assisted Reproductive Technology’ (ART) programme. [NCERT Exemplar]
Answer:
‘Assisted Reproductive Technology’ (ART) is the collection of certain I1 special techniques. The primary aim of the ART programmes is to assist infertile couples to have children through certain special techniques (like ZIFT, IUT, GIFT, ICSI, AI, etc.), when corrective treatment for infertility problems is not possible.

Question 8.
Expand GIFT and ICSI.
Answer:
GIFT: Gamete Intra Fallopian Transfer.
ICSI: Intra Cytoplasmic Sperm Injection.

Short answer type questions

Question 1.
Comment on the RCH programme of the government to improve the reproductive health of the people. [NCERT Exemplar]
Answer:
The basic aims of the RCH programmes are creating public awareness f regarding reproduction-related aspects and providing facilities to build up a healthy society with added emphasis on the health of mother and child.

Question 2.
(a) List any four characteristics of an ideal contraceptive.
(b) Name two intrauterine contraceptive devices that effect the motility of sperms.
Answer:
(a) An ideal contraceptive should be:

• user friendly,
• easily available,
• effective,
• reversible with no or least side-effects,
• non-interfering with the sexual drive/desire and/or the sexual act of the user, (any four)

(b) Intrauterine Devices (IUDs): Lippes loop, copper releasing IUDs (Cu-T, Cu-7, multiload 375).

Question 3.
Name two hormones that are constituents of contraceptive pills. Why do they have high and effective contraceptive value? Name a commonly prescribed non-steroidal oral pill.
Answer:
Hormonal preparations (progestogens or progesterons and estrogens) are highly effective contraceptive because they inhibit ovulation and implantation, e.g., Mala-D, Mala-L. Morning after pills are used as emergency contraceptives, to avoid pregnancy due to rape or casual unprotected intercourse.

“Saheli”, a new oral pill is used “once-a-week” a non-steroidal preparation With very less side effects and high contraceptive value developed by CDRI in Lucknow, India.

Question 4.
Name and explain the surgical method advised to human males and females as a means of birth control. Mention its one advantage and one disadvantage.
Answer:
In the males-Vasectomy. In this method, a small part of the vas deferens is removed or tied up through a small incision on the scrotum. In the females-Tubectomy. In this method, a small part of the fallopian tube is removed or tied up through a small incision in the abdomen or through vagina.
Advantage: Highly effective
Disadvantage: Reversibility is very poor

Question 5.
A childless couple has agreed for a test tube baby programme. List only the basic steps of the procedure would involve to conceive the baby.
Answer:
IVF is the technique used in the case of childless couple. In IVF or In Vitro fertilisation, fertilisation is carried out in a glass container outside the body of the mother. Purified semen is poured over the mature retrieved oocytes. The fertilised eggs are separated and allowed to remain in culture medium, maintained in incubator for 48-72 hours. During the period, the fertilised egg undergoes cleavage and reach 4-8 celled stage, 2-3 fertilised 4-8 celled embryos are transferred or implanted .into the uterus of the recipient surrogate mother, for further development up to delivery.
Note: Excess fertilised oocytes are cryopreserved for use in case of implantation failure.

Question 6.
Why is medical termination of pregnancy (MTP) carried out?
Answer:
MTP is carried out to get rid of unwanted pregnancies. It is also essential when the foetus is suffering from an incurable disease or when continuation of the pregnancy could be harmful or even fatal to the mother and/or foetus.

Long answer type questions

Question 1.
Your school has been selected by the Department of Education to organise and host an interschool seminar on “Reproductive Health-Problems and Practices.” However, many parents are reluctant to permit their wards to attend it. Their argument is that the topic is “too embarrassing.”
Put forth four arguments with appropriate reasons and explanation to justify the topic to be very essential and timely.
Answer:
The selection of your school, to host a seminar on “Reproductive Health-Problems and Practices” is a matter great pride for the prestigious institute. The students will have an opportunity to listen to the diverse ideas, suggested by the learned speakers.

It is sad that many parents are reluctant to permit their wards to attend the seminar assuming that the topic is too embarrassing. The following arguments will justify the relevance of the topic in the present time:
(i) Introduction of sex education and the proper information about reproductive organs, adolescence and related changes will protect the youth from social evils like sex-abuse and sex-related crimes.

(ii) Right information about safe, healthy and hygienic sexual practices, sexually transmitted diseases (STDs) would help the people to lead a reproductive healthy life.

(iii) Decline in sex-ratio is a matter of great concern. The Govt, has put a I statutory ban on female foeticide. Both girls and boys have equal rights and equal opportunities in all spheres of life.

(iv) India is facing another problem of population explosion. It is eating, almost all the benefits of overall development. The benefits of development are not trickling down to the poor at lower strata. There is need for family planning, socially conscious healthy families of desired size i.e., Hum Do Humare Do.

Punjab State Board PSEB 12th Class Biology Important Questions Chapter 3 Human Reproduction Important Questions and Answers.

PSEB 12th Class Biology Important Questions Chapter 3 Human Reproduction

Very short answer type questions

Question 1.
Why are the human testes located outside the abdominal cavity? Name the pouch in which they are present.
Answer:
The human testes need lower temperature, 2-2.5°C less than the body temperature, for the formation of sperms which is provided outside the body. Testes are present in scrotal sac or scrotum.

Question 2.
Write the location and function of the sertoli cells in humans.
Answer:
Sertoli cells are present in seminiferous tubules. They provide nutrition to the germ cells or sperms.

Question 3.
Mention the location and the function of leydig cells in humans.
Answer:
Leydig cells are present in seminiferous tubules. They synthesise and secrete androgens.

Question 4.
The path of sperm transport is given below. Provide the missing steps in blank boxes. [NCERT Exemplar]

Answer:
Vasa efferentia, Vas deferens.

Question 5.
Female reproductive organs and associated functions are given below in column A and B. Fill in the blank boxes. [NCERT Exemplar]

Column A	Column B
Ovaries	Ovulation
Oviduct	A
B	Pregnancy
Vagina	Birth

Answer:
A – Fertilisation
B – Uterus

Question 6.
What is the role of cervix of the human female system in reproduction? [NCERT Exemplar]
Answer:
Cervix helps in regulating the passage of sperms into the uterus and forms the birth canal to facilitate parturition.

Question 7.
When do the oogenesis and the spermatogenesis initiate in human females and males respectively?
Answer:
In females; oogenesis initiates during foetal life. In males, spermatogenesis begins at the time of puberty.

Question 8.
Mention the importance of LH surge during menstrual cycle. [NCERT Exemplar]
Answer:
LH surge is essential for the events leading to ovulation.

Question 9.
Menstrual cycles are absent during pregnancy. Why? [NCERT Exemplar]
Answer:
The high levels of progesterone and estrogens during pregnancy suppress the release of gonadotropins required for the development of new follicles. Therefore, new cycle cannot be initiated.

Question 10.
How does the sperm penetrate through the zona pellucida in [ human ovum?
Answer:
The sperm penetrates through zona pellucida with the help of secretions from acrosome.

Question 11.
Mention the function of trophoblast in human embryo.
Answer:
Trophoblast is the outer layer of blastocyst which helps in the attachment of blastocyst to the endometrium of the uterus.

Question 12.
Explain the function of umbilical cord.
Answer:
It transports nutrients and respiratory gases and metabolic wastes to and from mother and foetus.

Question 13.
Given below are the stages in human reproduction. Write them ‘ in correct sequential order.
Insemination, Gametogenesis, Fertilisation, Parturition, Gestation, Implantation. [NCERT Exemplar]
Answer:
Gametogenesis, Insemination, Fertilisation, Implantation, Gestation, Parturition.

Question 14.
What stimulates pituitary to release the hormone responsible for parturition? Name the hormone.
Answer:
The signal from the fully developed foetus and placenta or the foetal , ejection reflex induces mild uterine contraction. The hormone released is oxytocin.

Question 15.
Name the important mammary gland secretions that help in resistance of the new bom baby. [NCERT Exemplar]
Answer:
Colostrum

Question 16.
Mention the function of mitochondria in sperm.
Answer:
It provide energy for the movement of sperm tail.

Short answer type questions

Question 1.
Write the function of each of the following:
(a) Middle piece in human sperm.
(b) Luteinising hormone in human males.
Answer:
(a) Provides energy for movement.
(b) Stimulates synthesis and secretion of androgens or male hormones for spermatogenesis.

Question 2.
Differentiate between mayor structural changes in the human ovary during the follicular and luteal phase of the menstrual cycle.
Answer:

Follicular phase	Luteal phase
1. During this, primary follicles grow to become fully mature Graafian follicle.	During this, remaining part of Graafian follicle transforms into corpus luteum.
2. Endometrium regenerates through proliferation.	Endometrium further thickens secreting progesterone for implantation after fertilisation. If fertilisation does not occur, corpus luteum degenerates.

Question 3.
Explain the events in a normal woman during her menstrual cycle on the following days :
(a) Pituitary hormone levels from 8 to 12 days.
(b) Uterine events from 13 to 15 days.
(c) Ovarian events from 16 to 23 days.
Answer:
(a) The level of LH and FSH secreted by anterior lobe of pituitary, stimulated by GnRH, increases.

(b) The endometrium of the uterus regenerates through proliferation. It grows and become thickened. There is repair of ruptured blood vessels and new blood capillaries develop. Uterine glands elongate.

(c) The remnant of Graafian follicle forms corpus luteum which secretes large amount of progesterone essential for maintenance of endometrium for implantation and for pregnancy.

Question 4.
What happens to corpus luteum in human female if the ovum is (a) fertilised, (b) not fertilised?
Answer:
(a) In case the ovum is fertilised, the corpus luteum persists and secretes a large amount of progesterone. The progesterone is essential for maintenance of endometrium, a necessity for implantation and for pregnancy.

(b) In the absence of fertilisation, corpus luteum degenerates. The level of LH and progesterone decreases very low. The absence of hormonal support, leads to the disintegration of endometrium and results in menstrual flow.

Question 5.
A large number of married couples the world over are childless. It is shocking to know that in India the female partner is often blamed for the couple being childless.
(a) Why in your opinion the female partner is often blamed for such situations in India? Mention any two values that you as a biology student can promote to check this Social evil.
(b) State any two reasons responsible for the cause of infertility.
(c) Suggest a technique that can help the couple to have a child where the problem is with the male partner.
Answer:
(a) The female partner is wrongly blamed for not bearing the child. It is due to the lack of proper information and knowledge about the reproduction and reproductive organs. Being a biology student I will advise and explain the couple as well as their other family members. Both male and female are equally responsible for bearing child. The two value, promoted are (i) concern for others (ii) scientific
temperament.

(b) Infertility is the inability to produce children inspite of unprotected sex and sexual co-habitation. It may be due to (i) physical/congenital disease (ii) immuno- logical or even physiological reason.

(c) In case, the problem is with male partner, artificial insemination (AI) is adopted. Semen collected either from the husband or a healthy donor is artificially introduced in the vagina or into the uterus of the female.

Question 6.
Name and explain the role of inner and middle walls of the human uterus.
Answer:
The inner wall of the uterus is called endometrium. It supports foetal growth and helps in placenta formation after implantation.
The middle wall of the uterus is called myometrium, exhibits strong contraction during delivery of baby.

Question 7.
Write a brief account of the structure and functions of placenta. [NCERT Exemplar]
Answer:
Placenta connects the foetus to the uterus through an umbilical chord. Both the foetal and the maternal tissues contribute to its formation. The foetal part is the chorionic villi and the maternal part is the uterine mucosa.

Long answer type questions

Question 1.
(a) Briefly explain the events of fertilisation and implantation in an adult human female.
(b) Comment on the role of placenta as an endocrine gland.
Or
Name the stage of human embryo of which it gets implanted.
Explain the process of implantation.
Or Draw a labelled diagram of a human blastocyst. How does it get implanted in the uterus?
Answer:
(a) (i) Fertilisation : Fertilisation occurs, if the ovum and sperms are transported simultaneously to the ampullary-isthmic junction and involve fusion of sperm with an ovum.
Secretions of acrosome of sperm help it to enter into the cytoplasm of ovum through zona pellucida and the plasma membrane. It induces meiotic division-II to form haploid ovum (ootid) and secondary polar body. The fusion of sperm with ovum to form diploid zygote is called fertilisation.

(ii) Implantation: Zygote undergoes cleavage to form a solid mass of 16 cells-morula, with daughter cells called blastomeres. Morula develops into a embryo with about 64 cells and with a cavity called blastocoel and the embryo is termed as blastocyst. It consists of outer layer of cells-trophoblast and inner cell mass.

The trophoblast gets attached to the endometrium- uterine wall of mother, after 7 days of fertilisation by a process called implantation leading to pregnancy. The uterine cells divide rapidly and cover blastocyst. The blastocyst gets embedded in the endometrium. Inner cell mass forms embryo.

(b) Placenta beside providing nutrients to the foetus also act as an endocrine gland. Placenta secretes human chorionic gonadotrophin (hCG), human placental lactogen (hPL), estrogen, progesterone etc., and later relaxin is secreted by ovary which facilitates parturition. Increased level of hormones like cortisol, prolactin and thyroxine etc., help in foetal growth, metabolic changes in the mother and maintenance of pregnancy.

Punjab State Board PSEB 12th Class Biology Important Questions Chapter 15 Biodiversity and Conservation Important Questions and Answers.

PSEB 12th Class Biology Important Questions Chapter 15 Biodiversity and Conservation

Very short answer type questions

Question 1.
Why is genetic variation important in the plant Rauwolfia vomitoria? [NCERT Exemplar]
Answer:
Genetic variation affects the variation in potency and concentration of me drug reserpine in the medicinal plant Rauwolfia vomitoria.

Question 2.
Name the type of biodiversity represented by the following:
(i) 50,000 different strains of rice in India
(ii) Estuaries and alpine meadows in India.
Answer:
(i) Genetic diversity
(ii) Ecological diversity

Question 3.
Identify ‘a’ and ‘b’ in the figure given below representing proportionate number of major vertebrate taxa.

Answer:
(a) Mammals
(b) Amphibians

Question 4.
Suggest two practices giving one example by each, that help protect rare or threatened species.
Answer:

1. By using cryopreservation (preservation at -196°C) technique, sperms, eggs, tissues, and embryos can be stored for long period in gene banks, seed banks etc.
2. Plants are propagated in vitro using tissue culture methods.

Question 5.
According to David Tilman, greater the diversity greater is the primary productivity. Can you think of a very low diversity man-made ecosystem that has high productivity? [NCERT Exemplar]
Answer:
Agricultural fields like wheat field or paddy field which are also examples of monoculture practices.

Question 6.
What is Red Data Book? [NCERT Exemplar]
Answer:
The Red Data Book is a compilation of data on species threatened with extinction and is maintained by IUCN. ‘

Question 7.
What is the expanded form of IUCN? [NCERT Exemplar]
Answer:
International Union for Conservation of Nature.

Question 8.
Why Western Ghats in India have been declared as biological hotspots?
Answer:
Western Ghats are biological hotspots because they have species richness and species evenness.

Question 9.
What is a national park?
Answer:
It is a protected area reserved for wildlife where human activities are not permitted.

Question 10.
What is the difference between endemic and exotic species? [NCERT Exemplar]
Answer:
Endemic species are native species restricted to a particular geographical region. Exotic species are species which are introduced from other geographical regions into an area.

Question 11.
How is the presently occurring species extinction different from the earlier mass extinctions? [NCERT Exemplar]
Answer:
Species extinction occurring at present is due to anthropogenic or man-made causes whereas the earlier extinction was due to natural causes.

Question 12.
Differentiate between in situ and ex situ approaches of conservation of biodiversity.
Answer:

In situ approach	Ex-situ approach
1. It involves protection of endangered species of plants and animals.	It involves protection of endangered species by removing them from the natural habitat.
2. This is done by protecting the natural habitat or ecosystem.	This is done by placing the species under special care.

Short answer type questions

Question 1.
What is biodiversity? Why is it a matter of concern now?
Answer:
Biodiversity is the occurrence of different types of genes, gene pools, species, habitats and ecosystems at a particular place and various parts of earth. It is a matter of concern because species are continuously lost, limiting the diversity and this will affect our survival and well-being on earth due to the changes in environment.

Question 2.
Where would you expect more species biodiversity-in tropics or in polar regions? Give reasons in support of your answer.
Answer:
More biodiversity is found in the tropics. This is because tropical regions remain undisturbed from frequent glaciations as in polar regions. Also, the tropics are less seasonal/more constant.

Question 3.
Is it true that there is more solar energy available in the tropics? Explain briefly. [NCERT Exemplar]
Answer:
As one moves from the equator to the polar regions, the length of the day decreases and the length of the night increases. The length of day and night are same at the equator. Therefore, it is true that there is more solar energy available in the tropics.

Question 4.
The given graph alongside shows species-area relationship. Write the equation of the curve ‘a’ and explain.
Answer:

The equation of the curve ‘a’ is S = CA^Z.
(i) Within a region, species richness increases with increasing explored area but only up to a limit.
(ii) Relationship between species richness and area for a wide variety of taxa turns out to be a rectangular hyperbola.

Question 5.
Explain ‘rivet popper hypothesis. Name the ecologist who proposed it.
Answer:
Paul Ehrlich proposed the rivet popper hypothesis. This hypothesis states that in an airplane (ecosystem) all parts are joined together using thousands of rivets (species). If every passenger traveling in it starts popping a rivet to take home (causing a species to become extinct), it may not affect flight safety (proper functioning of the ecosystem) initially but as more and more rivets are removed, the plane becomes dangerously weak over a period of time. Also, which rivet is removed may also be critical like loss of rivets on the wings (key species) is more serious threat to flight safety than loss of few rivets on the seats or windows inside the plane.

Question 6.
How do human activities cause desertification?
Answer:
Human activities like over-cultivation, unrestricted grazing, deforestation, and poor irrigation practices result in arid patches of land. The fertile topsoil that may take centuries to develop is eroded due to these activities. When large barren patches extend and meet over time, a desert is created. Increased urbanization is also one of the causes of desertification.

Question 7.
Why are conventional methods not suitable for the assessment of biodiversity of bacteria? [NCERT Exemplar]
Answer:
Many bacteria are not culturable under normal conditions in the laboratory. This becomes a problem in studying their morphological, biochemical, and other characterizations which are useful for their assessment.

Question 8.
List any four techniques where the principle of ex-situ conservation of biodiversity has been employed.
Answer:
Ex-situ Conservation (off-site conservation)

1. Zoological parks and botanical gardens.
2. Wildlife safari parks, aquaria.
3. Preservation of germplasm-seed gene banks, tissue culture, cryopreservation.
4. Sacred plants grown in homes, villages, and religious places.

Long answer type questions

Question 1.
(a) Why should we conserve biodiversity? How can we do it?
(b) Explain the importance of biodiversity hotspots and sacred groves.
Or
Why should biodiversity be conserved? List any two ethical arguments in its support. ‘
Answer:
(a) Need for Conservation of Biodiversity: Reasons for conservation of biodiversity can be grouped into three categories :

1. narrowly utilitarian
2. broadly utilitarian and
3. ethical.

(i) Narrowly Utilitarian: The reasons for conserving biodiversity are obvious because of their :
(a) direct economic benefits such as

• food (cereals, pulses, fruits)
• firewood
• fiber
• construction material
• products of medicinal importance
• industrial products (tannins, gums, lubricants, dyes, resins, perfumes).

(b) More than 25% of the drugs are derived from plants.

(c) 25,000 species of plants are used as traditional medicines by native people.

(ii) Broadly Utilitarian: Biodiversity plays a major role in providing ecosystem services that nature provides and which cannot be given a price tag are :

• Production of Oxygen
• Pollination of flowers by bees, bumblebees, birds, and bats, etc.
• Resulting in the formation of fruits and seeds.
• Aesthetic pleasures like bird watching, walking through the thick forests, waking up to bulbul’s song, etc.

(iii) Ethical :
(a) We share this planet with millions of plants, animals, and microbe species. Every species has an intrinsic value even if it is not of current or any economic value to us.

(b) We have an essential duty to care for their well-being and pass on the biological legacy in a proper form to our future generations. We can conserve biodiversity by two major approaches

• In situ conservation (on site/ conservation)
• Ex situ conservation (off-site conservation).

(c) Hotspots are the areas identified by conservationists for the very high level of species richness and high degree of endemism (species confined to a particular area and not found anywhere else). Hotspots help in protection of certain biodiversity-rich regions.

Sacred groves are the tracts of forest set aside where all the trees and wildlife within are given religious sanctity and total protection. Such sacred groves are found in Khasi and Jaintia hills in Meghalaya, Aravalli Hills of Rajasthan etc.

Punjab State Board PSEB 12th Class Biology Important Questions Chapter 2 Sexual Reproduction in Flowering Plants Important Questions and Answers.

PSEB 12th Class Biology Important Questions Chapter 2 Sexual Reproduction in Flowering Plants

Very short answer type questions

Question 1.
An anther with malfunctioning tapetum often fails to produce viable male gametophytes. Give any one reason.
Answer:
A malfunctioning tapetum does not provide enough nourishment to the developing male gametophytes and thus fail to produce viable male gametophytes.

Question 2.
Complete the following flow chart
Pollen mother cell → Pollen tetrad

[NCERT Exampler]
Answer:
Generative cell.

Question 3.
Gynoecium of a flower may be apocarpous or syncarpous. Explain with the help of an example each.
Answer:
Gynoecium of a flower may be apocarpous means the carpels are free from one another and there is no fusion of any part e.g., Ranunculus, Rose. Gynoecium of a flower is syncarpous, means the carpels are fused by their ovaries. The number of fusing carpels may vary from 2 (Petunia) to ∞ (Hibiscus).

Question 4.
Name the parts of the gynoecium which develop into fruit and seeds. [NCERT Exemplar]
Answer:
Ovary develops into fruit and ovules develop into seeds.

Question 5.
How many haploid cells are present in a mature female gametophyte of a flowering plant? Name them.
Answer:
One dikaryotic polar cell with two haploid nuclei and six haploid cells, viz, 3 antipodal, 2 synergids and 1 egg.

Question 6.
Name the type of pollination in self-incompatible plants. [NCERT Exemplar]
Answer:
Xenogamy.

Question 7.
How do flowers of Vallisneria get pollinated?
Answer:
In Vallisneria, the female flower stalk is coiled to reach the water surface to receive the pollen grains carried by water currents.

Question 8.
What is pollen-pistil interaction and how is it mediated?
Answer:
The pistil accepts the right type (compatible) of pollen and promotes fertilisation and rejects the pollen of other species and incompatible pollen of the same species. It is the result of interaction between the chemicals of the pollen and those of stigma.

Question 9.
State the function of filiform apparatus found in mature embryo sac of an angiosperm.
Answer:
Filiform apparatus plays an important role in guiding the path of pollen tubes into the synergids.

Question 10.
Normally one embryo develops in one seed but when an orange seed is squeezed many embryos of different shapes and sizes are seen. Mention how it has happened?
Answer:
An orange seed has many embryos because of polyembryony.

Question 11.
Name the component cells of the ‘egg apparatus’ in an embryo sac. [NCERT Exemplar]
Answer:
Two synergids and an egg.

Question 12.
Name the common function that cotyledons and nucellus perform. [NCERT Exemplar]
Answer:
Cotyledons and nucellus provide nourishment.

Question 13.
In the embryos of a typical dicot and a grass, which are the true homologous structures? [NCERT Exemplar]
Answer:
Cotyledons and scutellum.

Question 14.
In a case of polyembryony, if an embryo develops from the synergid and *another from the nucellus, which is haploid and which is diploid? [NCERT Exemplar]
Answer:
Synergid embryo is haploid and nucellar embryo is diploid.

Short answer type questions

Question 1.
Name the organic materials the exine and intine of an angiosperm pollen grains are made up of. Explain the role of exine.
Answer:
Exine is made up of sporopollenin and intine is made up of cellulose and pectin.
Due to the sporopollenin, exine can withstand high temperature and strong acids. It is also not affected by enzymes. It is because of this reason that pollen grains are well preserved as fossils.

Question 2.
Differentiate between geitonogamy and xenogamy in plants. Which one between the two will lead to inbreeding depression and why?
Answer:

Geitonogamy	Xenogamy
1. It is transfer of pollen grains from the anther to the stigma of another flower of same plant.	It is transfer of pollen grains from the anther to the stigma of a different plant.
2. The pollen grains are genetically similar to the plant.	The pollen grains are genetically different from the plant.

Geitonogamy will lead to inbreeding depression because the pollen grains are genetically similar, which results in inbreeding. Continued inbreeding will thus reduce fertility and productivity.

Question 3.
Double fertilisation is reported in plants of both, castor and groundnut. However, the mature seeds of groundnut are non-albuminous and castor are albuminous. Explain the post fertilisation events that are responsible for it.
Answer:
The development of endosperm (preceding the embryo) takes place from primary endosperm nucleus (PEN) in both, castor and groundnut. The developing embryo derives nutrition from endosperm.

PEN undergoes repeated division to give free nuclei. Subsequently cell wall is formed and endosperm becomes cellular. At this stage endosperm is retained in castor or is not fully consumed but in groundnut endosperm is consumed by growing embryo.

Question 4.
Differentiate between albuminous and non-albuminous seeds, giving one example of each.
Answer:

• Albuminous seeds have residual endosperm ip them. For example, maize.
• Non-albuminous seeds do not have any residual endosperm. For example, pea.

Question 5.
A non biology person is quite shocked to know that apple is a false fruit, mango is a true fruit and banana is a seedless fruit. As a biology student how would you satisfy this person?
Answer:
In apple only the thalamus (along with ovary) portion contributes to fruit. Therefore, it is a false fruit. Mango develops only from the ovary, therefore it is a true fruit.
Banana develops from ovary but without fertilisation. The method is known as parthenocarpy. Since there is no fertilisation, no seeds are formed.

Question 6.
Why are some seeds referred to as apomictic seeds? Mention one advantage and one disadvantage to a farmer who uses them.
Answer:
Seeds produced without fertilisation are referred to as apomictic.
Advantage: Desired characters are retained in offspring (progeny) as there is no segregation of characters in offspring (progeny). Seed production is assured in absence of pollinators.

Disadvantage: Cannot control accumulation of deleterious genetic mutation. These are usually restricted to narrow ecological niches and lack ability to adapt to changing environment.

Long answer type questions

Question 1.
A flower of brinjal plant following the process of sexual reproduction produces 360 viable seeds.
Answer the following questions giving reasons:
(a) How many ovules are minimally involved?
(b) How many megaspore mother cells are involved?
(c) What is the minimum number of pollen grains that must land on stigma for pollination?
(d) How many male gametes are involved in the above case?
(e) How many microspore mother cells must have undergone reduction division prior to dehiscence of another in the above case?
Answer:
(a) 360 ovules are involved. One ovule after fertilisation forms one seed.
(b) 360 megaspore mother cells are involved. Each megaspore mother cell forms four megaspores out of which only one remains functional.
(c) 360 pollen grains. One pollen grain participates in fertilisation of one ovule.
(d) 720 male gametes are involved. Each pollen grain carries two male gametes (which participate in double fertilisation) (360 × 2 = 720).
(e) 90 microspore mother cells undergo reduction division. Each microspore mother cell meiotically divides to form four pollen grains (360/4 = 90).

Punjab State Board PSEB 12th Class Biology Important Questions Chapter 16 Environmental Issues Important Questions and Answers.

PSEB 12th Class Biology Important Questions Chapter 16 Environmental Issues

Very short answer type questions

Question 1.
Why are lichens regarded as pollution indicators?
Answer:
Lichens are regarded as pollution indicators because they do not grow in areas that are polluted. So their presence indicates no pollution in that area and their absence indicates that the area is polluted.

Question 2.
Why is the use of unleaded petrol recommended for motor vehicles equipped with catalytic converters?
Or
Why are owners of motor vehicles equipped with catalytic converters advised to use unleaded petrol?
Or
Why is it desirable to use unleaded petrol in vehicles fitted with catalytic converters?
Answer:
Vehicles fitted with catalytic converters need to use unleaded petrol because lead inactivates the catalyst in the catalytic converter and increases hydrocarbon emission, thereby harming the environment.

Question 3.
In which year was the Air (Prevention and Control of Air Pollution) Act amended to include noise as air pollutant? [NCERT Exemplar]
Answer:
The Air (Prevention and Control of Air Pollution) Act 1981 was amended in 1987 to include noise as an air pollutant.

Question 4.
Name an industry which can cause air pollution, thermal pollution, and eutrophication. [NCERT Exemplar]
Answer:
Fertilizer factory.

Question 5.
Mention two advantages for preferring CNG over diesel as an automobile fuel.
Answer:
Advantages of CNG :

1. It burns more efficiently than petrol or diesel.
2. It is cheaper than petrol or diesel and cannot be siphoned off by thieves or adulterated.

Question 6.
What is an algal bloom? [NCERT Exemplar]
Answer:
The excessive growth of algae (free-floating) that causes coloration of water bodies is called algal bloom.

Question 7.
Write the name of the organism that is referred to as the ‘Terror of Bengal’.
Or
Why is Eichhomia crassipes nicknamed as* ‘Terror of Bengal’?
Answer:
Eichhornia crassipes is nicknamed as ‘Terror of Bengal’ because it grows very fast in the water body and depletes the dissolved oxygen. Hence, disturbing the ecosystem dynamics of the water body.

Question 8.
What is the raw material for polyblend? [NCERT Exemplar]
Answer:
Polyblends are natural man-made fibres, made by the mixture of two or more polymers, especially plastic waste products.

Question 9.
Mention the effect of UV rays on DNA and proteins in living organisms.
Answer:
The high energy of UV rays breaks the chemical bonds within DNA and protein molecules.

Question 10.
Write the unit used for measuring ozone thickness.
Answer:
Dobson unit.

Question 11.
State the purpose of signing the Montreal Protocol.
Answer:
Montreal Protocol was signed at Montreal, in 1987 to curb the emission of ozone-depleting substances.

Question 12.
What is reforestation? [NCERT Exemplar]
Answer:
Reforestation is the process of restoring a forest that once existed but was removed at some point of time in the past.

Short answer type questions

Question 1.
It is a common practice to plant trees and shrubs near the boundary walls of buildings. What purpose do they serve?
[NCERT Exemplar]
Answer:
The practice of growing and maintaining trees and shrubs near the boundary wall of residential or official buildings is a common practice. This is because it acts as a barrier for sound and check noise pollution. This green belt of trees and shrubs also acts as an effective measure to check primary air pollutants like dust, fly ash, etc.

Question 2.
‘Determination of Biological Oxygen Demand (BOD) can help in suggesting the quality of a water body. Explain.
Answer:
Biochemical Oxygen Demand (BOD) is the amount of dissolved oxygen required for microbial breakdown of biodegradable organic matter. Aerobic organisms use a lot of oxygen and as a result, there is a sharp decline in Dissolved Oxygen (DO) in the water body. This Can cause death of fishes and other aquatic species. The relationship between BOD and sewage can be understood from the graph given below :

Determination of BOD is thus an important parameter in suggesting the quality of a water body. The presence of more organic waste increases the microbial activity thus decreasing the DO. BOD is higher in polluted water and lesser in clean water.

Question 3.
Explain the process of secondary treatment given to the primary effluent up to the point it shows a significant change in the level of biological oxygen demand (BOD) in it.
Answer:
The primary effluent is passed into large aeration tanks where it is constantly agitated. Air is pumped into it mechanically. This allows vigorous growth of useful aerobic microbes into floes. These microbes consume the major part of organic matter in the effluent (this significantly reduces the BOD of the effluent).

Question 4.
Excessive nutrients in a freshwater body cause fish mortality. Give two reasons.
Answer:
Excessive nutrients in the water body result in the excessive and rapid growth of plants and animals, resulting in formation of increased organic matter, death of plants and animals increased the organic matter at the bottom which decomposes, increased BOD deplete the oxygen content, resulting in fish mortality.

Question 5.
Is it true that, if the dissolved oxygen level drops to zero, the water will become septic? Give an example which could lower the dissolved oxygen content of an aquatic body. [NCERT Exemplar]
Answer:
Yes, it is true. In case of zero level of Dissolved Oxygen (DO), the water becomes septic. Organic pollutants like fertilizer in aquatic bodies are responsible for lowering (up to zero) the level of dissolved oxygen.

Question 6.
Explain giving reasons why thermal power plants are not considered eco-friendly?
Answer:
Thermal power plants release particulate and gaseous pollutants in the environment. Inhalation of these pollutants can cause breathing or respiratory symptoms^irritation, inflammation, damage to lungs, and even premature death.

Question 7.
Blend of polyblend and bitumen, when used, helps to increase road life by a factor of three. What is the reason? [NCERT Exemplar]
Answer:
Polyblend is a fine powder of recycled modified plastic. The binding property of plastic makes the road last longer besides giving added strength to withstand more loads. This is because: plastic increases the melting point of the bitumen which would prevent it from melting in India’s hot and extremely humid climate, where temperature frequently crosses 50°C. rainwater will not seep through because of the plastic in the blend.

Question 8.
What is the main idea behind ‘Joint Forest Management Concept’ introduced by the Government of India?
[NCERT Exemplar]
Answer:
The main idea behind joint forest management concept introduced by the Government of India was involving the local communities in forest conservation. This concept was adopted considering the extraordinary courage and dedication the local people showed in protecting the wildlife through the movements like Bishnoi’s movement in Jodhpur and Chipko Movement in Garhwal Himalayas.

Long answer type questions

Question 1.
Explain giving reasons the cause of appearance of peaks ‘a’ and ‘b’ in the graph shown below:
img
Answer:
‘a’-High BOD due to sewage discharge.
‘b’-Increase in dissolved oxygen due to sewage decomposition. Micro-organisms involved in biodegradation of organic matter consume a lot of oxygen, therefore, there is a sharp decline in dissolved oxygen. When the sewage is completely degraded, oxygen concentration again increases.

Question 2.
Why is the concentration of toxins found to be more in the organisms occupying the highest trophic level in the food chain in a polluted* water body? Explain with the help of a suitable example.
Answer:
The concentration of toxic materials like heavy metals and pesticides increase at each trophic level of a food chain and is more in organisms of highest trophic level due to their accumulation at each trophic level For example, when DDT was used to control mosquitoes in a lake of USA, 800 times more DDT was found in the phytoplanktons than in the water of the lake. Zooplanktons had about 13 times more DDT than phytoplanktons. It was also observed that the fishes population had 9-40 times more DDT than zooplanktons and fish-eating birds had 25 times more DDT than fish.

Question 3.
Refrigerants arc considered to be a necessity in modem living but are said to be responsible for ozone holes detected in Antarctica. Justify.
Answer:
The widely used refrigerants are CFCs or. chlorofluorocarbons. CFCs discharged in the lower part of atmosphere move upwards to the stratosphere. Here, the UV rays act on them and release chlorine atoms. These free chlorine atoms react with ozone to release molecular oxygen. Chlorine atoms are not consumed in this reaction and hence, these continuously degrade ozone and have resulted in ozone holes.

Punjab State Board PSEB 12th Class Biology Important Questions Chapter 1 Reproduction in Organisms Important Questions and Answers.

PSEB 12th Class Biology Important Questions Chapter 1 Reproduction in Organisms

Very short answer type questions

Question 1.
What is life span?
Answer:
The period between birth and the natural death of an organism represents its life span.

Question 2.
On what factors does the type of reproduction adopted by an organism depend on? [NCERT Exemplar]
Answer:
The organism’s habitat, physiology and genetic make-up determines the type of reproduction adopted by it.

Question 3.
Name an organism where cell division is itself a mode of reproduction.
Answer:
Protists/Monerans/Amoeba/Paramecium.

Question 4.
Mention two inherent characteristics of Amoeba and Yeast that enable them to reproduce asexually. [NCERT Exemplar]
Answer:

1. They are unicellular organisms.
2. They have a very simple body structure.

Question 5.
What is conidia?
Answer:
The asexual, non-motile spores produced externally/exogenously in some fungi are called conidia, e.g., Penicillium.

Question 6.
Define gemmules.
Answer:
Internal asexual reproductive units or buds in sponges are called gemmules.

Question 7.
Name the vegetative propagules in the following
(a) Agave
(b) Bryophyllum
Answer:
(a) Agave – Bulbil
(b) Bryophyllum – Leaf buds/adventitious buds.

Question 8.
Mention the unique feature with respect to flowering and fruiting in bamboo species.
Answer:
Bamboo flowers once in its life time generally after 50-100 yrs of vegetative growth. It produces large number of fruits and dies.

Question 9.
Is Marchantia monoecious or dioecious? Where are the sex 1 organs borne in this plant? [NCERT Exemplar]
Answer:
Marchantia is dioecious. The male sex organs, antheridia, are borne on j the antheridiophores and female sex organs called archegonia are borne on archegoniophores.

Question 10.
Suggest a possible explanation why the seeds in a pea are arranged in a row, whereas those in tomato are scattered in the juicy pulp. [NCERT Exemplar]
Answer:
The ovary of pea plant is monocarpellary and the ovules are arranged along one margin whereas in tomato the ovary is tricarpellary with axile placentation.

Question 11.
In a developing embryo, analyse the consequences if cell divisions are not followed by cell differentiation.
[NCERT Exemplar]
Answer:
If differentiation does not follow division, embryo will not develop and this will not develop into a new organism.

Question 12.
Name the phenomenon and one bird where the female gamete directly develops into a new organism.
Answer:
The phenomenon is called parthenogenesis. Turkey.

Question 13.
Mention the site where syngamy occurs in amphibians and reptiles, respectively.
Answer:
In amphibians, external fertilisation occurs hence, syngamy occurs in the medium of water. In reptiles, internal fertilisation occurs hence, syngamy occurs within the body of female parent.

Question 14.
Name the group of organisms that produce non-motile gametes. How do they reach the female gamete for fertilisation?
Answer:
Angiosperms produce non-moule gametes. They reach the female gamete with the help of air or water.

Question 15.
The number of taxa exhibiting asexual reproduction is drastically reduced in the higher plants (angiosperms) and higher animals (vertebrates) as compared with lower groups of plants and animals. Analyse the possible reasons for this situation. [NCERT Exemplar]
Answer:
Both angiosperms and vertebrates have a more complex structural organisation. They have evolved very efficient mechanism of sexual reproduction. Since asexual reproduction does not create new genetic pools in the offspring and consequently hampers their adaptability to external conditions, these groups have resorted to reproduction by the sexual method.

Short answer type questions

Question 1.
The cell division involved in gamete formation is not of the same type in different organisms. Justify.
Answer:
The parents may be haploid or diploid but the gametes always have to be haploid, biploid parents undergo meiosis to produce haploid gametes, whereas haploid parents undergo mitosis to produce haploid gametes.

Question 2.
Although sexual reproduction is a long drawn, energy-intensive complex form of reproduction, many groups of organisms in Kingdom Animalia and Plantae prefer this mode of reproduction. Give at least three reasons for this. [NCERT Exemplar]
Answer:

• Sexual reproduction brings about variation in the offspring.
• Since gamete formation is preceded by meiosis, genetic recombination occurring during crossing over (meiosis-I), leads to a great deal of variation in the DNA of gametes.
• The organism has better chance of survival in a changing environment.

Question 3.
Explain the importance of syngamy and meiosis in a sexual life cycle of an organism.
Answer:
Syngamy and meiosis play an important role in sexual life cycle of any diploid organisms, may be a plant or animal syngamy, i.e. fusion of haploid gametes/sex cells (n) (fertilisation) to form diploid egg cell/zygote (2n). Zygote divides repeatedly by mitotic divisons to form an embryo which develop into a new diploid organisms. After attaining sexual maturity, the organisms undergo special type of meiotic divisions – spermatogenesis/ microsporogenesis and megasporogenesis/ oogenesis to form haploid male sex cells and female sex cells. These sex cells (n) again fuse to form diploid zygote.

Question 4.
Why are mosses and liverworts unable to complete their sexual mode of reproduction in dry conditions? Give reasons.
Answer:
For sexual reproduction to take place in mosses and liverworts the motile male gametophytes, antherozoids, have to swim on the water surface to fertilise the immotile female gametophytes, egg. In dry conditions, the antherozoids do not reach the egg and hence fertilisation cannot occur.

Question 5.
Why do algae and fungi shift to sexual mode of reproduction just before the onset of adverse conditions?
Answer:
Algae and fungi shift to sexual mode of reproduction for survival during unfavourable conditions. Fusion of gametes helps to pool their resources for survival. The zygote develops a thick wall that is resistant to dessication and damage which undergoes a period of rest before germination.

Long answer type questions

Question 1.
What is gametogenesis? Describe the different types of gametes and draw labelled diagrams. ,
Answer:
The process of formation of two types of gametes – male and female inside the gametangia is called gametogenesis.

Depending upon the size and motility, the gametes are of following types :
1. Isogametes or Homogamets: The gametes are similar in shape, size, structure arid function. Their fusion is called isogamy e.g., Cladophora. However, when the isogametes are physiologically different and the gametes produced by one parent do not fuse with each other, the gametes belonging to different mating types can only fuse and their fusion is called physiological anisogamy, e.g., Ulothrvc.

2. Anisogametes or Heterogametes: The fusing gametes are different in form, size, structure and behaviour. The larger, non-motile, food-laden gamete is called ovum/egg/oosphere/ macrogamete. The smaller, motile, active gamete is called sperm/male gamete/antherozoid. Such gametes are called anisogametes or heterogametes and their fusion is termed as anisogamy or heterogamy e.g., Cladophora.

Question 2.
Write a note on sexuality in organisms.
Answer:
Sexuality in Organisms: Sexual reproduction in organisms generally involves the fusion of gametes from two different individuals. But this is not always true.

Sexuality in Plants: Plants may have both male and female reproductive structures in the same plant (bisexual) e.g., rose or on different plants (unisexual) e.g., papaya. In several fungi and plants, terms such as homothallic and monoecious are used to denote the bisexual condition and heterothallic and dioecious are the terms used to describe unisexual condition.

In flowering plants, the unisexual male flower is staminate i.e., bearing stamens, while the female is pistillate or bearing pistils. In some flowering plants, both male and female flowers may be present on the same individual (monoecious) or on separate individuals (dioecious). Some examples of monoecious plants are cucurbits and coconuts and of dioecious plants are papaya, mulberry and date palm.

Sexuality in Animals: There are species which possess both the reproductive organs (bisexual). Earthworms, sponge, tapeworm and leech are typical examples of bisexual animals that possess both male and female reproductive organs i.e., they are hermaphrodites. There are large number of animal species which are either male or female and are called as unisexual organisms, e.g., frog, lizard, crow, dog, cat, rabbit, human beings etc.

Punjab State Board PSEB 11th Class Biology Important Questions Chapter 13 Photosynthesis in Higher Plants Important Questions and Answers.

PSEB 11th Class Biology Important Questions Chapter 13 Photosynthesis in Higher Plants

Very short answer type questions

Question 1.
Photosynthetic pigments are located in which part of the chloroplast? ‘
Answer:
Photosynthetic pigments are located in the lipid part of thylakoid membrane.

Question 2.
Does moonlight support photosynthesis? Find out. [NCERT Exemplar]
Answer:
No, because the intensity of moonlight is several thousand times less than that of direct sunlight, insufficient for the light-dependent phase of photosynthesis.

Question 3.
Oxygen evolved during photosynthesis comes from H₂O or CO₂?
Answer:
Oxygen in photosynthesis evolves from H₂O (i. e., by splitting of water).

Question 4.
Mention conditions under which PS-I functions.
Answer:
Conditions necessary under which PS-I function are as follows :

• When the wavelength of light is higher than 680 nm.
• When NADPH accumulates and CO₂ fixation is retarded.

Question 5.
Succulents are known to keep their stomata closed during the day to check transpiration. How do they meet their photosynthetic CO₂ requirements? [NCERT Exemplar]
Answer:
Succulents (water-storing) plants such as cacti, fix CO₂ into organic compound using PEP carboxylase at night when the stomata are open.

Question 6.
Which compound is meant for donating hydrogen to carbohydrate in Calvin cycle?
Answer:
NADPH is responsible for donating hydrogen to carbohydrate in Calvin cycle.

Question 7.
Indicate the main steps during Calvin cycle.
Answer:
The main steps during Calvin cycle are Carboxylation, Reduction, Regenaration.

Question 8.
Which of the mechanism of photosynthesis is responsible for trapping of light energy, splitting of water, oxygen release and formation of ATP and NADPH?
Answer:
Light reaction is responsible for all the above-mentioned conditions.

Question 9.
How many molecules of carbon dioxide, ATP and NADPH are required to make one molecule of glucose?
Answer:

1. Carbon dioxide 6 molecules
2. ATP 18 molecules
3. NADPH 12 molecules

Question 10.
What would happen to the rate of photosynthesis in C₃ -plants if CO₂ concentration level almost doubles from its present level in the atmosphere.
Answer:
If the CO₂ concentration in C₃ -plants almost get doubles from its present level in the atmosphere plants will grow much faster and leads to higher productivity due to higher rate of photosynthesis.

Question 11.
Why photosynthesis is an oxidation-reduction process?
Answer:
Photosynthesis is an oxidation-reduction process because water is oxidised to oxygen and carbon is reduced to carbohydrates.

Question 12.
The type of anatomy of leaves possessed by C₄-plant is different from those C₃-plant. Explain.
Answer:
C₄ -plant possess a special anatomy of leaves called Kranz anatomy, which means that the mesophyll tissue is undifferentiated in leaves

Short answer type questions

Question 1.
Give a brief explanation of photosynthesis.
Answer:

• Carbon dioxide is converted into sugars in a process called carbon fixation.
• Carbon fixation is a redox reaction, so photosynthesis needs to supply both a source of energy to drive this process and also the electrons needed to convert carbon dioxide into carbohydrate, which is a reduction reaction.
• In general outline, photosynthesis is the opposite of cellular respiration, where glucose and other compounds are oxidised to produce carbon dioxide, water, and release chemical energy.

Question 2.
Explain some early experiments that led to a gradual development in our understanding of photosynthesis?
Answer:
Early Experiments
(i) Joseph Priestley (1733-1804):
Priestley hypothesised that plants restore to the air whatever breathing animals and burning candles remove.

(ii) Jan Ingenhousz (1730-1799)
He showed that only the green parts of the plants could release oxygen.

(iii) Julius von Sachs

• He showed that the green substance (chlorophyll) is located in special bodies (chloroplasts).
• He provided evidence (1854) for the production of glucose in the green parts of plants and stored in the form of starch.

(iv) Engelmann (1843-1909):

• He split the light using a prism into its component parts and illuminated a green alga, Cladophora placed in a suspension of aerobic bacteria.
• The bacteria were used to detect the sites of oxygen evolution.
• He observed that the bacteria accumulated mainly in the region of blue and red light of the spectrum.
• He first described the action spectrum of photosynthesis; the action spectrum resembles roughly the absorption spectrum of chlorophyll a and b.

(v) Cornelius van Niel (1897-1985):

• He conducted experiments with purple and green sulphur bacteria, he demonstrated that photosynthesis is essentially a light-dependent reaction in which hydrogen from a hydrogen-donor reduces carbon dioxide to carbohydrates.
• He gave the present-day equation of photosynthesis.

• In green plants water (H₂O) is the hydrogen donor and it is oxidised to oxygen.
• Purple and green sulphur bacteria use H₂S as the hydrogen donor and so the oxidation product is sulphur and no oxygen is produced.
• Thus, he inferred that the oxygen evolved by green plants during photosynthesis comes from water (H₂O) and not from carbon dioxide (CO₂).
• This was later proved by using radioactive isotopes of oxygen(H7180).
• The overall correct equation for photosynthesis is as follows :

Question 3.
In chloroplast what are sites for light reactions and dark reactions?
Answer:
There is a clear division of labour in chloroplasts. The membrane system is responsible for Light reactions. Light energy is trapped by the membrane system and synthesis of ATP and NADPH takes place over there. The stroma utilises CO₂ to synthesize sugar; ATP and NADPH from light reaction are also utilized by stroma.

Question 4.
Give a brief account of light reaction.
Answer:
Light reactions or the ‘Photochemical’ phase include following steps :

• light absorption,
• water splitting,
• oxygen release, and
• the formation of high-energy chemical intermediates, ATP and NADPH.

Several complexes are involved in the process. The pigments are organised into two discrete photochemical light harvesting complexes (LHC) within the Photosystem I (PS I) and Photosystem II (PS II). These are named in the sequence of their discovery, and not in the sequence in which they function during the light reaction.

The LHC are made up of hundreds of pigment molecules bound to proteins. Each photosystem has all the pigments (except one molecule of chlorophyll a) forming a light-harvesting system also called antennae. These pigments help to make photosynthesis more efficient by absorbing different wavelengths of light.

The single chlorophyll a molecule forms the reaction centre. The reaction centre is different in both the photosystems. In PS I the reaction centre chlorophyll a has an absorption peak at 700 nm, hence is called P700, while in PS II it has absorption maxima at 680 nm, and is called P680.

Question 5.
Explain the electron transport system in photosynthesis.
Answer:
Electron Transport: The whole scheme of electron transport starting from PS II, uphill to primary electron acceptor, downhill to cytochrome complex and PS I, excitation of PS I, transfer of electrons uphill to another acceptor and finally downhill to NADP+, is called Z-scheme, because of the characteristic shape, Z, formed when all the carriers are placed in a sequence according to their redox potential values.

Question 6.
What do you understand by splitting of water and its importance in photosynthesis?
Answer:
The splitting of water is associated with the PS II; water is split into H⁺, [0] and electrons. This creates oxygen, one of the net products of photosynthesis. The electrons needed to replace those removed from photosystem I are provided by photosystem II.
2H₂O → 4H⁺ +O₂ + 4e^–
2H₂O → 4H⁺ +O₂ + 4e^–.

Question 7.
Write a short note on chemiosmotic hypothesis.
Answer:
Chemiosmotic Hypothesis
ATP synthesis is linked to the development of a proton gradient across the membranes of thylakoids; it results due to the following reasons:

1. Since the splitting of the water molecules or photolysis takes place on the inner side of the thylakoid membrane, the protons -produced accumulate within the lumen of the thylakoids.
2. The primary electron acceptor is located towards the outer side of the membrane and transfers its electrons to the H carrier; so this molecule removes a proton from the stroma while transporting an electron and releases it into the lumen or inner side of the thylakoid membrane.
3. The enzyme NADP reductase is located on the stroma side of the membrane; along with the electrons coming from PS I, protons are also needed to reduce NADP and so these protons are also removed from the stroma.

The gradient is broken down due to the movement of protons across the membrane through transmembrane channel of the Fo of the ATP synthetase; the other portion of ATP synthetase, called Fi undergoes conformational changes with the energy provided by the breakdown of proton gradient and synthesises several molecules of ATP.

Question 8.
What do you understand by biosynthetic phase of photosynthesis?
Answer:
Synthesis of food (sugar) takes place during dark reaction. Since sugar is synthesised in this phase, it is also called as biosynthetic phase.

Long answer type questions

Question 1.
Is it correct to say that photosynthesis occurs only in leaves of a plant? Besides leaves, what are the other parts that may be capable of carrying out photosynthesis? Justify.
Answer:
Although all cells in the green parts of a plant have chloroplasts, most of the energy is captured in the leaves. The cells in the interior tissues of a leaf, called the mesophyll, can contain between 450000 and 800000 chloroplasts for every square millimetre of leaf.

The surface of the leaf is uniformly coated with a water-resistant waxy cuticle that protects the leaf from excessive evaporation of water and decreases the absorption of ultraviolet or blue light to reduce heating. The transparent epidermis layer allows light to pass through to the palisade mesophyll cells, where most of the photosynthesis takes place. The green stems are also capable of performing photosynthesis.

Question 2.
The entire process of photosynthesis consists of a number of reactions. Where in the cell do each of these take place?
(i) Synthesis of ATP and NADPH ……………………………………. .
(ii) Photolysis of water …………………………….. .
(iii) Fixation of CO₂ …………………………… .
(iv) Synthesis of sugar molecule ……………………………….. .
(v) Synthesis of starch ……………………………………… .
Answer:
(i) Synthesis of ATP and NADPH in thylakoids.
(ii) Photolysis of water occurs in inner side of thylakoid membrane.
(iii) Fixation of CO₂ occurs in stroma of chloroplast.
(iv) Synthesis of sugar molecule occurs in chloroplast.
(v) Synthesis of starch occurs in cytoplasm.

Question 3.
Which factors affect the process of photosynthesis? Explain in detail.
Answer:
Factors Affecting Photosynthesis
Photosynthesis is affected by both internal/plant factors and extemal/environmental factors.
The internal or plant factors that affect the rate of photosynthesis include
(i) the number, size, age and orientation of leaves,
(ii) mesophyll cells,
(iii) chloroplasts,
(iv) the amount of chlorophyll and
(v) the internal CO₂ concentration.

Blackman’s Law of Limiting Factors:
When a physiological process is controlled by a number of factors, the rate of the reaction depends on the slowest factor. This means that at a given time, only the factor which is the least (limiting) among all the factors, will determine the rate of the reaction.

(i) Light
Light quality and light intensity influence photosynthesis.
Light of wavelength between 400 nm and 700 nm is effective for photosynthesis and this light is known as photosynthetically active radiation (PAR).

As the intensity of light increases the rate of photosynthesis increases. But at higher light intensities, the rate of photosynthesis does not increase; it may be due to two reasons :
(a) Other factors needed for photosynthesis may be limiting.
(b) Destruction (photooxidation) of chlorophyll.

(ii) Temperature

• The photochemical phase is less affected by temperature.
• But the biosynthetic phase that involves enzyme-catalysed reactions, is more sensitive to temperature.
• The C₄ plants have a higher temperature optimum, while C₃ plants have a lower temperature optimum.

(iii) Carbon dioxide concentration

• In C₃ plants, the rate of photosynthesis increases with increase in CO₂ concentration, and saturation occurs beyond 450 μ/L^-1.
• In C₄ plants, the saturation is reached at a concentration of about 360 μ/L^-1.

(iv) Water
Water influences photosynthesis in two ways :

1. If available water decreases and plants show water stress, the stomata close; hence there will be a decreased supply of carbon dioxide for photosynthesis.
2. The leaves become wilted and the surface area for activities decreases.

Punjab State Board PSEB 11th Class Biology Important Questions Chapter 12 Mineral Nutrition Important Questions and Answers.

PSEB 11th Class Biology Important Questions Chapter 12 Mineral Nutrition

Very short answer type questions

Question 1.
Write the criteria for essentiality of an element.
Answer:
Criteria for Essentiality of an Element

• The element must be absolutely necessary for supporting normal growth and reproduction; in the absence of the element, the plants do not complete their life cycle.
• The requirement of the element must be specific and not replaceable by another element, i.e., deficiency of any one element cannot be met by supplying some other element.
• The element must be directly involved in the metabolism of the plant.

Question 2.
Name two imrhobile elements in plants.
Answer:
Calcium and sulphur are two immobile elements.

Question 3.
The deficiency of which element causes the death of stem and root apices?
Answer:
Boron.

Question 4.
The deficiency of this particular element causes the con dition of little leaf or mottle leaf in the plants. Name the element.
Answer:
Deficiency of element zinc.

Question 5.
Deficiency of mineral nutrition is not responsible for etiolation. Yes or No. If no then explain why?
Answer:
No, mineral deficiency is not responsible for etiolation because it is related to absence of light.

Question 6.
Where do the symptoms of deficiency of phosphorus appear first in the plant? And why?
Answer:
The deficiency of phosphorus appears first in older leaves. Because it is a mobile element.

Question 7.
A particular macroelement is obtained by the plants from both mineral and non-mineral sources. Identify the element.
Answer:
Nitrogen is the element which is obtained from both mineral and non-mineral sources.

Question 8.
Yellowish edges appear in leaves deficient in. [NCERT Exemplar]
Answer:
Magnesium.

Question 9.
Name the macronutrient which is a component of all organic compounds but is not obtained from soil. [NCERT Exemplar]
Answer:
Nitrogen.

Question 10.
Ammonification of nitrogen during nitrogen cycle is bring about by a special process. Identify the process.
Answer:
It is done by decomposition.

Question 11.
Organisms like Pseudomonas and Thiobacillus are of great significance in nitrogen cycle. How? [NCERT Exemplar]
Answer:
These organisms carry out denitrification. They help to maintain the constant level of nitrogen in the atmosphere.

Question 12.
A farmer adds Azotobacter culture to soil before sowing maize. Which mineral element is being replenished? [NCERT Exemplar]
Answer:
Nitrogen.

Short answer type questions

Question 1.
What are two macronutrients which are not obtained through soil as mineral nutrition? What is their importance?
Answer:
Oxygen and carbon are not obtained through soil as mineral nutrition. Oxygen is necessary for respiration and carbon dioxide is necessary for photosynthesis.

Question 2.
A fanner adds/supplies Na, Ca, Mg and Fe regularly to his field and yet he observes that the plants show deficiency of Ca, Mg and Fe. Give a valid reason and suggest a way to help the farmer improve the growth of plants.
Answer:
Calcium deficiency can be due to shortage of water so proper irrigation should be done. Iron deficiency can be due to high alkalinity of soil. And excessive use of potassium can result in poor absorption of magnesium. The farmer should take corrective actions.

Question 3.
Which enzyme is found in root nodules of leguminous plants? What role does it play?
Answer:
Nitrogenase is found in root nodules of leguminous plants. The enzyme binds with nitrogen and helps it in getting transported to the plant.

Question 4.
Write short notes on – Reductive animation and Transamination.
Answer:
(i) Reductive Animation: In these processes, ammonia reacts with α-Ketoglutaric acid and forms glutamic acid as indicated in the equation given below :
Glutamate

(ii) Transamination: It involves the transfer of amino group from one amino acid to the keto group of a keto acid. Glutamic acid is the main amino acid from which the transfer of NH₂, the amino group takes place and other amino acids are formed through transamination. The enzyme transaminase catalyses all such reactions. For example,

Long answer type questions

Question 1.
Hydroponics have been shown to be a successful technique for growing of plants, yet most of the crops are still grown on land. Why?
Answer:
The technique of growing plants in a nutrient solution is known as hydroponics.
This method requires purified water and mineral nutrient salts. After a series of experiments, in which the roots of the plants were immersed in nutrient solutions and an element was added/removed or given in varied concentration, a mineral solution suitable for the plant growth was obtained.

By this method, essential elements were identified and their deficiency symptoms discovered. Hydroponics has been successfully employed as a technique for the commercial production of vegetables such as tomato, seedless cucumber and lettuce.

Yet, most of the crops are still grown on land because the nutrient solutions must be adequately aerated to obtain the optimum growth in hydroponis. Moreover, the minerals must be continuously added in the solution. No, such activity is required in the soil.

Question 2.
We find that Rhizobium forms nodules on the roots of leguminous plants. Also Frankia, another microbe forms nitrogen fixing nodules on the roots of non-leguminous plant Alnus.
(i) Can we artificially induce the property of nitrogen fixation in a plant, leguminous or non-leguminous?
(ii) What kind of relationship is observed between mycorrhiza and pine trees?
(iii) Is it necessary for a microbe to be in close association with a plant to provide mineral nutrition? Explain with the help of one example. [NCERT Exemplar]
Answer:
(i) Yes, we can artificially induce the property of nitrogen fixation in a plant, leguminous or non-leguminous by genetic engineering. It involves the introduction of nif genes that cause the synthesis of nitrogenase enzyme by some vector in the plant in which we have to induce symbiosis.

(ii) Symbiotic association.

(iii) Yes, it is necessary for a-microbe to be in close association with a plant to provide mineral nutrition. For example, plants that contribute to nitrogen fixation include the legume family-Fabaceae or Leguminosae, such as clover, soyabeans, alfalfa, lupines and peanuts. They contain symbiotic bacteria called Rhizobia within nodules in their root systems, producing nitrogen compounds that help the plant to grow and compete with other plants.