Botany Test Paper by TEACHING CARE online tuition and coaching classes

  1. The most commonly used method of choice for gene transfer using immature

embryos as explants is

(a) Microprojectile

(b) Electroporation

(c) Liposome mediated

(d) Chemically stimulated

Ans a. Microprojectile bombardment is a method by which foreign substances such as DNA

are introduced into living cells and tissues via high-velocity microprojectiles using a

shotgun.

 

  1. A pericentric inversion in chromosome involves

(a) One arm of a chromosome

(b) Both the arms of a chromosome

(c) Two different chromosomes

(d) More than two chromosomes

Ans b. Pericentric means around centromere. Pericentric inversions include the centromere and

there is a break point in each arm.

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  1. Who of the following made the first artificial recombinant DNA?

(a) Boyer and Cohen

(b) H. Smith

(c) Nirenberg and Khorana

(d) S.B. Weiss

Ans a. Smith is related to discovered restriction endonuclease.

 

  1. Microtubules are cylindrical structures having alpha and beta tubulin proteins.

They are the constituents of which one of the following groups?

(a) Nucleus, Nucleolus, Nucleoid, Nuclear membrane

(b) Centrioles, Spindle fibres, Flagella, Cilia

(c) Mitochondria, Lysosome, Chloroplast, Golgi apparatus

(d) Chromosome, Chromocentre, Chromatid, Chromatin

Ans b. Centrioles have 9+0 pattern of microtubules,flagella and cilia have 9+2 arrangement.

  1. Restriction enzymes cut DNA at specific sites known as

(a) Telomeric sequences

(b) Palindromic sequences

(c) Terminator sequences

(d) Attenuator sequences

Ans b. A palindromic sequence is a nucleic acid sequence (DNA or RNA) that is the same whether read 5′ to 3′ on one strand or 5′ to 3′ on the complementary strand with which it forms adouble helix.

 

  1. The proteins that reproduce within the living cells are termed as

(a) Plasmids

(b) Phages

(c) Prions

(d) Prophages

Ans c. Prions are unprecedented infectious pathogens that cause a group of invariably fatal neurodegenerative diseases. Bovine spongiform encephalopathy (BSE), scrapie of sheep, and Creutzfeldt–Jakob disease (CJD) of humans are among the most notable prion diseases. Prions are transmissible particles that are devoid of nucleic acid and seem to be composed exclusively of a modified protein (PrPSc)

 

  1. Which one of the following is involved in self-splicing of introns?

(a) Primase

(b) Ribozyme

(c) Reverse transcriptase

(d) RNA polymerase

Ans b. Many natural ribozymes cleave one of their own phosphodiester bonds (self-cleaving ribozymes), or bonds in other RNAs.

 

  1. Which one of the following plasmid vectors was consrtucted by combining

centomere, telomere and autonomously replicating sequence?

(a) pBR 322

(b) Cosmid

(c) Yeast Artificial Chromosome (pYAC)

(d) Bacterial Artificial Chromosome (BAC)

Ans. c. YAC is an artificially constructed chromosome and contains the telomericcentromeric, and replication origin sequences named autonomous replicating sequence needed for replication and preservation in yeast cells.

 

  1. Consider the following statements:
  2. Transfer RNA contains a number of rare bases that are not found in other

nucleic acids.

  1. During translation, the frame shift errors many times result in the synthesis of

very useful proteins.

Which of the statements given above is/are correct?

(a) 1 only

(b) 2 only

(c) Both 1 and 2

(d) Neither 1 nor 2

Ans a. The polypeptide being created could be abnormally short or abnormally long, and will most likely not be functional.

  1. Consider the following statements:
  2. Anabaena lives symbiotically with Azolla
  3. Azotobacter is a free living organism in the soil.

Which of the statements given above is/are correct?

(a) 1 only.

(b) 2 only

(c) Both 1 and 2

(d) Neither 1 nor 2

Ans. c.

 

  1. Which one of the following is a rod-shaped bacterium?

(a) Bacillus subtilis

(b) Pneumococcus pneumoniae

(c) Streptococcus nigricans

(d) Vibrio cholera

Ans a. Pneumococcus pneumonia-spherical

Streptococcus nigricans-spherical

Vibrio cholera-comma shaped

 

  1. Which one of the following phytohormones is known to contribute to hud

dormancy?

(a) Ehtylene

(b) Coumarin

(c) Cytokinin

(d) ABA

Ans. d. ABA (abscissic acid)

 

  1. Perisperm is a post-fertilization modification of

(a) Nucellus

(b) Outer integument

(c). Central cell

(d) Inner integument

Ans a. Perisperm is persistant nucellus in the seed

 

  1. Which one of the following pairs is correctly matched?

(a) Pistia— Sciophyte

(b) Lemna —Xerophyte

(c) Rhizophora— Halophyte

(d) Casuarina— Hydophyte

Ans c. Sciophyte-shade loving plant

 

  1. Consider the following:
  2. Hydathodes
  3. Salt glands
  4. Nectaries
  5. Lenticels

Which of these are secretory structures?

(a) I, 2 and 4  (b) 3 and 4

(c) 1, 2 and 3 (d) 1, 2, 3 and 4

Ans c. Lenticels are for gaseous exchange

  1. The taxonomic category ‘Class’ is between

(a) Order and Family

(b) Order and Genus

(c) Kingdom and Phylum

(d) Division and Order

Ans d.

kingdom
phylum (in zoology)division (in botany)
class
order
family
genus
species

 

 

  1. The fixation of CO2 to malate and its decarboxylation are common to both

C4 and CAM plants, but in CAM plants these events

(a) Are separated spatially

(b) Are separated temporally

(c) Require high light intensity

(d) Require high CO2 concentration

Ans b.

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  1. What is the primary acceptor of CO2 in Hatch-Slack cycle?

(a) Phosphoenol pyruvic acid

(b) Ribulose biphosphate

(c) Phosphoglyceric acid

(d) Diphosphoglyceric acid

Ans a. Hatch-Slack cycle is C4 cycle. For details of this cycle refer to figure of previous

question.

 

  1. A universal hydrogen acceptor in an electron transport system is

(a) ATP

(b) UDP

(c) NAD

(d) FMN

Ans c

 

  1. Consider the following statements:
  2. Bacterial DNA is not bound by histones.
  3. Histones are basic proteins.

Which of the statements given above is/are correct

(a) 1 only

(b) 2 only

(c) Both 1 and 2

(d) Neither 1 nor 2

 

Ans. c. Histones are basic proteins associated with eukaryotic DNA

  1. In photorespiration, RuBP carboxylase combines with oxygen to yield

(a) Two molecules of phosphoglycerate

(b) Two molecules of phosphoglycolate

(c) One molecule of phosphoglycerate and one molecule of phosphoglycolate

(d) Two molecules of glucose

Ans c. The first reaction of photorespiration produces phosphoglycerate (PGA) and phosphoglycolate

 

Directions: The following  items consist of two statement, one labeled as

‘Assertion (A)’ and the other as ‘Reason (R)’. You are to examine these two

statements carefully and select the answers to these items using the code given

below:

(a) Both A and R are individually true and R is the correct explanation of A

(b) Both A and R are individually true and R is not the correct explanation of A

(c) A is true but R is false

(d) A is false but R is true Assertion

 

  1. Assertion (A): Some plant species grow poorly in nature if the soil fungi are

killed with a fungicide.

Reason (R): The mycorrhizal fungi transport nitrate into the plant.

Ans a. Mycorrhizal fungi absorb water and minerals for the plant.

 

  1. Periderm includes

(a) Phelloderm, collenchyma and cortex

(b) Phellem, cambium and cortex

(c) All the tissues between epidermis and pith

(d) Phellogen, phellem and phelloderm

Ans d

  1. Consider the following:
  2. Succinic dehydrogenase
  3. Aconitase
  4. alpha-ketoglutarate dehydrogenase
  5. Isocitric dehydrogenase

What is the correct order in which the above enzymes catalyse the reactions in

Kreb’s cycle?

(a) 1 – 2 – 3 – 4 (b) 2 – 4 – 1 – 3

(c) 3 – 2 – 4 – 1 (d) 2 – 4 – 3 – 1

Ans. d.

 

  1. Which one of the following pairs is not matched?

(a) Copper: Plastocyanin

(b) Sulphur: Serine

(c) Molybdenum: Nitrate reductase

(d) Zinc: Alcohol dehydrogenase

Ans b. S is found in cystein and methionine

 

  1. Addition of KCN reduces the rate of water absorption in the root. This

indicates that water absorption is a/an

(a) Passive process

(b) Energy dependent process

(c) Osmotic difference, dependent process

(d) Exchange diffusion process

Ans b. CN is a respiratory inhibitor. When respiration stops, no energy is generated which is

required for active absorption of water.

 

  1. Consider the following events involved in stomatal opening:
  2. Turgor pressure of guard cells increases.
  3. K+ ions move into guard cells.
  4. pH of guard cells decreases.
  5. Water moves into guard cells.

What is the correct sequence of these events leading to stomatal opening:

(a) 2 – 4 – 3 – 1 (b) 3 – 2 – 4 – 1

(c) 2 – 3 – 4 – 1 (d) 3 – 1 – 2 – 4

Ans  b.

 

  1. Which one of the following is popularly associated with the name of J.

Liebig?

(a) Law of the Minimum

(b) Law of Tolerance

(c) Ecological Niche

(d) Speciation

Ans a.  Liebig’s Law of minimum states that growth is controlled not by the total amount

of resources available, but by the scarcest resource

 

  1. Which one of the following ecological pyramids is always upright?

(a) Pyramid of numbers

(b) Pyramid of biomass

(c) Pyramid of energy

(d) Age pyramid of plant populations

Ans. c. Energy pyramids is always upright because energy flow is always unidirectional.

 

  1. Somaclonal variation can be advantageous because

(a) There are chromosomal abnormalities

(b) Monosomics are produced

(c) Enrichment of genetic diversity occurs

(d) It gives high genetic uniformity

Ans. c

 

  1. Consider the following statments :
  2. The first living organisms on planet Earth originated in water.
  3. When life originated on planet Earth, the atmosphere contained nitrogen,

ammonia, oxygen, hydrogen, carbon dioxide, methane and water vapour.

Which of the statements given above is/are correct?

(a) 1 only (b) 2 only

(c) Both 1 and 2 (d) Neither 1 nor 2

Ans a. Molecular oxygen was absent in primitive atmosphere

 

  1. In crop improvement programmes, haploids are of great importance

because they

(a) Grow better even under adverse climatic conditions

(b)Are useful in studies of meiosis.

(c) Have less requirement for energy inputs

(d) Give homozygous lines following diploidization

Ans d. Colchicine is used for diploidization

 

  1. Which one of the following pairs is not correctly matched?

(a) Production of sulphur dioxide: Burning of coal

(b) Depletion of ozone layer: Release of chlorofluorocarbons in the

atmosphere

(c) Eutrophication: Increase in nitrogen and phosphorus

contents in aquatic bodies

(d) Decrease in B.O.D. of pond water: Increase in global temperature

Ans d. Increase in global temperature is due to green house gases

 

  1. Which one of the following was the objective of signing the ‘Montreal

Protocol?

(a) Protection of Wild life

(b) Protection of ozone layer

(c) Control over the use of insecticides

(d) Control of noise pollution

Ans b.

 

  1. Which one of the following is correct to measure beta-diversity ?

(a) Species richness within an ecosystem

(b) Species evenness equitability

(c) Degree of change in species composition along an environmental gradient

(d) Species diversity of several habitats in a large geographical region

Ans c.

Alpha diversity:

the diversity within a particular area or ecosystem; usually expressed by the number of species (i.e., species richness) in that ecosystem

Beta diversity:

a comparison of of diversity between ecosystems, usually measured as the amount of species change between the ecosystems

Gamma diversity:

a measure of the overall diversity within a large region. Geographic-scale species diversity according to Hunter 

 

  1. The number of individuals in each population that can live in a particular

ecosystem is limited; and that number is known as .

(a) Biotic potential

(b) Carrying capacity

(c) Intrinsic natural increase

(d) Reproductive capacity

Ans b. Biotic potential is reproductive potential

 

  1. Consider the following codons:
  2. UAA 2. UAC
  3. UAG

Which of these are considered to be the termination codons in protein synthesis?

(a) 1, 2 and 3

(b) 1 and 2

(c) 2 and 3

(d) 1 and 3

Ans d

 

  1. The function of ‘reverse transcriptase’ is to

(a) Transcribe a complementary DNA from an RNA strand

(b) Transcribe a complementary RNA from an RNA strand

(c) Translate messages for protein synthesis

(d) Replicate DNA from a DNA strand

Ans a

 

  1. In which type of tapetum do the protoplasts of tapetal cells mix or fuse and

surround the developing microscopes in the anther?

(a) Glandular tapetum

(b) Secretory tapetum

(c) Amoeboid tapetum

(d) Dual tapetum

Ans c

 

  1. The fertilization in which the entry of pollen tube into the ovule through the

funiculus is known as

(a) Syngamy

(b) Chalazogamy

(c) Mesogamy

(d) Porogamy

Ans c. Syngamy refers to fertilization

 

  1. 41. A wild-type fruit fly (heterozygous for gray body color and normal wings was mated with a black fly with vestigial wings. The offspring had the following phenotypic distribution: wild type, 778; black-vestigial, 785; black-normal, 158; gray-vestigial, 162. What is the recombination frequency between these genes for body color and wing type.

First count the total number of offspring 778+785+158+162 = 1883

In all dihybrid test crosses (a cross between a known heterozygote for two wild type traits and a homozygous recessive individual for both traits) the expected ratio of phenotypes if the genes are on separate chromosomes must be:

wild type, 25%; black-vestigial, 25% black-normal, 25%; gray-vestigial, 25%. These results do not fit the experimental data above (778+785+158+162).

In fact the black-normal (158) and gray-vestigial (162) offspring represent recombinant individuals.

Calculation of recombination frequency:

778 – wild type

785 – black-vestigial

778/1883 = 41.3%

785/1883 = 41.7%

83% are non-recombinant
158 – black-normal

162 – gray-vestigial

158/1883 = 8.4%

162/1883 = 8.6%

17% are due to recombination

Recombination frequency = 17%

The generally accepted method of symbolizing the genotypes for a dihybrid cross of linked genes is as follows:

  • b+= wild-type (gray body)
  • b = black body
  • vg+= normal wing shape
  • vg = vestigial wings

The testcross is symbolized as follows:

If no cross over occurs then only two phenotypes should be seen. That is 50% of the offspring should be dominant for both traits —- and the other 50% should be homozygous recessive —- just as in the parents above.

Cross over in the heterozygous parent  results in 50% recombinants:

 

 

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