1. Due to non-disjunction of chromosomes during spermatogenesis, sperms carry both sex chromosomes (22A + XY) and some sperms do not carry any sex chromosome (22A + 0). If these sperms fertilise normal eggs (22A + X), what types of genetic disorders appear among the offspring ?
Ans: (1) Turner’s syndrome and Klinefelter’s syndrome
2. When a fresh water protozoan is placed in marine water, …………….
Ans: (3) The contractile vacuoles disappear
3. Which one of the following pairs is an example for lateral meristem ?
Ans: (4) Phellogen and fascicular cambium
4. In peritoneal dialysis, ……………
Ans: (2) The blood is not removed from the body and a natural filter is used
5. The following is a scheme showing the electron transport system. Identify the electron carrier molecules indicated as A and B. Choose the correct option.
Ans: (1) A = Coenzyme Q, B = Cytochrome C
Note: The quality of the diagram (scanned) is very bad
6. The sugar present in milk is …………
Ans: (3) Glucose
(4) Lactose
Note: Low concentrations of free glucose (about 0.1 mM) and free galactose (about 0.2 mM) are found in cow milk and milk of other species. Other carbohydrates found free in milk include amino sugars, sugar phosphates, neutral and acid oligosaccharides, and nucleotide sugars. Some of the complex oligosaccharides are thought to be important in helping establish the microflora of the neonate intestine (such as the bifidis factor identified in human milk; see Human Lactation Lesson). Some milk proteins are glycosylated and some milk lipids contain carbohydrate moieties. (Milk Composition and Resource Library – Internet)
The answer would have been lactose if the question were to be “ The chief / principal sugar present in milk …….”.
7. According to Steward’s starch hydrolysis theory, which one of the following is the principal reason for the opening of stomata during daytime ?
Ans: (4) Photosynthetic utilization of CO2 in guard cells
8. Which one of the following processes results in the formation of a clone of bacteria ?
Ans: (1) Binary fission
9. Match the types of immunity listed in Column - I with the examples listed in Column - II. Choose the answer that gives the correct combination of alphabets of the two columns.
Column I Column II
Types of immunity Example
A. Natural active p. Immunity developed by heredity
B. Artificial passive q. From mother to foetus through placenta
C. Artificial active r. Injection of antiserum to travelers
D. Natural passive s. Fighting infections naturally
t. Induced by vaccination
Ans: (4) A = s, B = r, C = t, D = q
10. How do you differentiate a butterfly from a moth ?
Ans: (1) Moth has feathery antennae but butterfly has club-shaped antennae
11. Which one of the following statements is NOT correct ?
Ans: (3) UAA codon codes for Lysine
12. In the absence of acrosome, the sperm ………….
Ans: (1) Can not penetrate the egg
13. The diagram of the ultrasturcture of a plant cell is given below. Identify the functions of the organelles labeled A, B, C, D and E in the diagram.
Ans: (2) A = Principal director of macromolecular traffic, B = Site of oxidative phosphorylation, C = Intracellular transport, D = Site of photophosphorylation, E = Storage of cell sap
14. Which one of the following species of earthworm is NOT recommended for vermicomposting ?
Ans: (4) Pheretima posthuma
15. The main aim of the human genome project is ……………
Ans: (2) To identify and sequence all the genes present in human DNA
16. The species, though insignificant in number, determine the existence of many other species in a given ecosystem. Such species is known as ………….
Ans: (4) Keystone species
17. Compare the statements A and B.
Statement A: Sclerenchyma cells do not have plasmodesmata.
Statement B: The cell walls of some permanent tissues are heavily lignified.
Select the correct description;
Ans: (3) Both the statements A and B are correct
18. Which one of the following is NOT a reason for very high load of bilirubin in a newborn ?
Ans: (3) Mother’s milk contains a high amount of bilirubin
19. Which one of the following diseases is caused by Nosema bombycis in mulberry silkworm ?
Ans: (2) Pebrine
20. The following is the scheme showing the path of reflex arc. Identify the different labellings A, B, C, D, E and F in the reflex arc.
Ans: (2) A = Stimulus, B = Receptor, C = Sensory nerve, D = Motor nerve, E = Effector, F = Response
21. Identify the pair that exhibits “circinate vernation”.
Ans: (3) Nephrolepis and Cycas
22. Pyruvate dehydrogenase complex needed for the conversion of Pyruvic acid to Acetyl Co-A is located in …………….
Ans: (1) Matrix of mitochondria
23. The presence of corollary corona, sagittate anthers and dumb-bell shaped stigma are the characteristic features of …………….
Ans: (3) Catheranthus roseus
24. Every time, when the dosage of a drug has to be increased to achieve the same ‘kick’ that initially occurred in response to a smaller dose, this condition is known as …………..
Ans: (2) Tolerance
25. DNA gyrase, the enzyme that participates in the process of DNA replication, is a type …………….
Ans: (1) DNA topoisomerase
26. Mendel found that the reciprocal crosses yielded identical results. From that, he concluded that ………….
Ans: (4) Sex has no influence on the dominance of traits
27. Compare the statements A and B.
Statement A: Synthesis of DNA takes place in the S - phase of interphase.
Statement B: Every chromosome, during metaphase, has two chromatids.
Select the correct description;
Ans: (3) Both the statements A and B are correct and A is the reason for B
28. Match the animals listed in Column - I with their nature of blood listed in Column – II. Choose the answer which gives the correct combination of alphabets of the two columns.
Column I Column II
A. Man p. Plasma and cells are colourless
B. Earthworm q. Plasma is colourless and nucleated RBC
C. Cockroach r. Plasma is colourless and enucleated RBC
D. Frog s. Plasma is red and nucleated, colourless RBC
t. Plasma RBC have haemoglobin
Ans: None of the option is correct
Note: Plasma is a straw yellow coloured component of the blood. It is not colourless. Moreover, blood cells in invertebrates like earthworm and cockroach are not called RBCs as they do not contain the respiratory pigment. They are called corpuscles or haemocytes.
29. During lactic acid fermentation, ……………
Ans: (2) Neither O2 is used nor CO2 is liberated
30. Which one of the following is NOT the function of insulin ?
Ans: (3) Initiates the conversion glycogen to glucose
31. Which one of the following is a unicellular, non-motile desmid ?
Ans: (4) Cosmarium
32. Some of the steps involved in the production of humulin are given below. Choose the correct sequence.
(i) Synthesis of gene (DNA) for human insulin artificially
(ii) Culturing recombinant E.coli in bioreactors
(iii) Purification of humulin
(iv) Insertion of human insulin gene into plasmid
(v) Introduction of recombinant plasmid into E.coli
(vi) Extraction of recombinant gene product from E.coli
Ans: (3) i, iv, v, ii, vi, iii
33. Cockroaches can climb smooth and steep surfaces due to the presence of adhesive pads found on the tarsus of their legs. They are called …………..
Ans: (1) Plantulae
(4) Arolium
Note: Arolium is also a pat of the pretarsus. But it is present between the two claws and is attached to the pretarsus.
34. Gastrula has a pore which is known as ……………
Ans: (1) Blastopore
35. The diagram of Labeo rohita is given below. Identify the parts labelled A, B, C, D, E and G.
Ans: (3) A = Nostril, B = Eye, C = Dorsal fin, D = Caudal fin, E = Anal fin, F = Pelvic fin, G = Pectoral fin
36. Compare the statements A and B.
Statement A: To counteract the increase in turgor pressure in plant cells, the cell wall produces an equal and opposite pressure, i.e., wall pressure.
Statement B: When plant cells undergo endosmosis, they swell but do not burst.
Ans: (1) Both the statements A and B are correct and A is the reason for B
37. When red blood corpuscles containing both A and B antigens are mixed with your blood serum, they agglutinate. Hence, your blood group is ……….. type.
Ans: (2) O
38. Bovine spongiform encephalopathy is a disease caused by prions in a …………..
Ans: (2) Cow
39. Compare the statements A and B.
Statement A: When the urine moves through the descending limb, it becomes hypertonic and as it passes through the ascending limb of Henle’s loop, it becomes hypotonic.
Statement B: The descending limb is permeable to sodium ions, while ascending limb is impermeable to sodium ions.
Ans: (1) Both the statements A and B are wrong
Note: It is not urine which moves in the pct, Henle’s loop and dct, but primary urine or glomerular filtrate.
40. To meet the demands of the society, in vitro production of a large number of plantlets in a short duration is practiced in floriculture and horticulture industry today. This is called …………..
Ans: (4) Micropropagation
41. With reference to enzymes, turnover number means ………………
Ans: (3) The number of substrate molecules that a molecule of an enzyme converts into products per minute
42. The following diagrams A, B, C, D and E show the different types of arrangement of stamens based on their cohesion of their parts in different plants. Assign the stamens to their respective plants. Choose the correct answer.
Ans: (4) A = Hibiscus rosa-sinensis, B = Crotalaria juncea, C = Bombax ceiba, D = Helianthus annus, E = Cucurbita pepo
43. The single horned Rhinoceros is protected at ……………
Ans: (2) Kaziranga National Park
44. According to Boyle’s law, the product of pressure and volume is a constant. Hence,
Ans: (1) if volume of lungs is increased, the pressure decreases proportionately
45. According to Darwin, evolution is …………..
Ans: (2) A slow, gradual and continuous process
46. Succus entericus is secreted by ………….
Ans: None of the option is correct
Note: Secretion of intestinal glands (both crypts of Lieberkuhn and Brunner’s gland) is called intestinal juice (succus entericus), i.e., succus entericus is the combined secretion of both crypts of Lieberkuhn and Brunner’s gland. The Brunner’s glands open into the crypts of Lirberkuhn. Their secretion contains alkaline mucus which prevents the action of acidic chyme on the mucosal lining of the small intestine.
47. Cell A and cell B are adjacent plant cells. In cell A, Ψs = - 20 bars and Ψp = 8 bars. In cell B, Ψs = -12 bars and Ψp = 2 bars. Then,
Ans: (3) Water moves from cell B to cell A
48. Populations are said to be sympatric when …………….
Ans: (4) Two populations share the same environment but can not interbreed
49. In which of the following situations, is there a risk factor for children of incurring erythroblastosis foetalis ?
Ans: (2) Mother is Rh –ve and father is Rh +ve
50. Compare the statements A and B.
Statement A: RNA produced during transcription in eukaryotic cells can not be straight away used in photosynthesis.
Statement B: RNA splicing phenomenon helps in the removal of exons.
Choose the correct description;
Ans: (1) Both the statements A and B are wrong
51. The diagram of large intestine of man is given below. Identify the parts labeled A, B, C, D, E and F.
Ans: (3) A = Caecum, B = Vermiform appendix, C = Ascending colon, D = Transverse colon, E = Descending colon, F = Sigmoid
52. In genetic fingerprinting, the ‘probe’ refers to ……….
Ans: (1) A radioactively labelled single stranded DNA molecule
53. Which one of the following is NOT a method of soil conservation ?
Ans: (2) Overgrazing
54. In C4 pathway, the CO2 fixation in mesophyll cells is carried out by the enzyme …
Ans: (1) PEP carboxylase
55. Which one of the following is a driving force for the process of passive absorption of water in roots ?
Ans: (4) Transpiration in leaves
56. If the systolic pressure is 120 mm Hg and diastolic pressure is 80 mm Hg, the pulse pressure is
Ans: (3) 120 – 80 = 40 mm Hg
57. Tyloses are found in …………….
Ans: (2) Secondary xylem
58. Which one of the following growth regulators is used to promote synchronized flowering in pineapple ?
Ans: (3) Indolebutyric acid
(4) 2-chloroethylphosphonic acid
Note: Auxins like NAA, 2,4-D and IBA are also used for synchronised flowering in pineapple and litchi. However, NAA and 2,4-D are preferred over IBA
59. Sporopollenin, a chemical substance is found in ……………..
Ans: (2) Exine of pollen grain
60. Which one of the following statements about the events of non-cyclic photophosphorylation is NOT correct ?
Ans: (1) Only one photosystem participates
(2) ATP and NADPH are not produced
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Thursday, April 29, 2010
Friday, March 26, 2010
Few sample questions for CET in MOLECULAR BIOLOGY
DNA:
1.In Griffith’s experiment with Pneumococcus and mice,
(1) living non – virulent bacteria transformed the living virulent strains
(2) living non – virulent bacteria transformed the heat killed virulent strains
(3) living virulent bacteria transformed heat killed non- virulent strains
(4) heat killed virulent bacteria transformed living non – virulent strains
2.Purines and pyrimidines donot contain
(1) carbon
(2) nitrogen
(3) phosphorous
(4) oxygen
3.Purines are
(1) large sized, dicyclic, heterocyclic nitrobases
(2) large sized, monocyclic, heterocyclic nitrobases
(3) small sized, monocyclic, heterocyclic nitrobases
(4) small sized, dicyclic, heterocyclic nitrobases
4.In a nucleotide, the nitrogenous base and phosphate group are attached to the
(1) 1’ and 3’ carbon atoms of sugar respectively
(2) 2’ and 3’ carbon atoms of sugar respectively
(3) 1’ and 5’ carbon atoms of sugar respectively
(4) 2’ and 5’ carbon atoms of sugar respectively
5.Guanosine differs from Guanylic acid in not having
(1) phosphate
(2) sugar
(3) nitro-base
(4) phosphate and sugar
6.The lengths of major groove and minor groove are
(1) 22 Å and 12 Å respectively
(2) 20 Å and 10 Å respectively
(3) 34 Å and 3.4 Å respectively
(4) 20 Å and 12 Å respectively
7.Which of the following is not relevant to the structure of double helical DNA ?
(1) the helix makes one complete spiral turn every 34 Ǻ
(2) the distance between adjacent nucleotides is 3.4 Ǻ
(3) each strand of helix has a backbone of alternating ribose sugar and nitrobase
(4) the two adjacent sugar molecules are joined with phosphate by phosphodiester bond
8.The two strands of DNA are held together by H - bonds between
(1) nitrogen bases
(2) phosphoric acid
(3) sugar and phosphate
(4) nitrobases and sugar
9.Identify the incorrect statement;
(1) the two strands of DNA are antiparallel because of phosphodiester bonds
(2) the two strands of DNA coil helically around a common axis
(3) the distance between two base pairs in B-DNA is 34 Ǻ
(4) the diameter of DNA molecule is 20 Ǻ
10.Read the two statements A and B. Choose the correct answer from those given.
Statement A: B-DNA is a right handed helical molecule which is like a twisted ladder.
Statement B: Sugar and phosphate molecules make up the uprights of the ladder and hydrogen bonded bases make up the rungs.
(1) statement A is correct, B is wrong
(2) statement B is correct, A is wrong
(3) both the statements A and B are correct
(4) both the statements A and B are wrong
11.Which of the following statements does not apply to the Watson – Crick model for DNA ?
(1) the two strands of the helix are antiparallel
(2) the double ringed purines of one strand pair with the single ringed pyrimidines of the other strand
(3) the two strands of the helix are held together by covalent bonds
(4) the verticals of the helix are composed of alternately arranged sugar and phosphate units
12.Statement A: A DNA duplex has a constant diameter of 20 À.
Statement B: Dicyclic purines of one strand always pair with monocyclic pyrimidines of the other strand in a DNA molecule.
(1) Both the statements A and B are correct and A is the reason for B
(2) Both the statements A and B are correct and B is the reason for A
(3) Statement A is correct and statement B is wrong
(4) Statement A is wrong and statement B is correct
13.When the base composition of DNA from a bacterium was analysed, 22% of the bases were found to be Adenine. What is the G + C content ?
(1) 44%
(2) 33%
(3) 56%
(4) 22%
14.A sample of DNA is found to have the base composition (mole ratio) of A = 35, T = 25, G = 24 and C = 16. It suggests that
(1) DNA is circular duplex
(2) DNA is linear duplex
(3) DNA is single stranded
(4) it is Z - DNA
15.Identify the correct statement;
(1) DNA is always a right handed helix
(2) mitochondrial DNA is left handed double helix
(3) according to Chargaff’s rule, A + G = T + C
(4) DNA is double stranded in all organisms
16.Z - DNA is
(1) right handed helix with 9 base pairs per turn
(2) right handed helix with 10 base pairs per turn
(3) left handed helix with 11 base pairs per turn
(4) left handed helix with 12 base pairs per turn
17.A DNA strand is directly involved in the synthesis of all the following except a
(1) polypeptide chain
(2) mRNA strand
(3) rRNA strand
(4) complementary DNA strand
18.DNA replication occurs
(1) whenever a cell makes protein
(2) before a cell divides
(3) whenever a cell needs RNA
(4) in the cytoplasm of an eukaryotic cell
19.Which of the following processes occurs in an eukaryotic cell ?
(1) DNA replication only
(2) translation only
(3) DNA replication and translation
(4) DNA replication, translation and transcription
20.The enzymes responsible for unwinding of DNA helix during replication is / are
(1) helicases
(2) topoisomerases
(3) DNA polymerase
(4) primase
21.The enzyme DNA polymerase can add nucleotide to the ________ of sugar of another nucleotide.
(1) 3’ carbon position
(2) 5’ carbon position
(3) 4’ carbon position
(4) both 3’ and 5’ carbon positions
22.Okazaki fragments are small pieces of DNA that are produced at the time of the formation of
(1) leading strand
(2) lagging strand
(3) sense strand
(4) antisense strand
23.Statement A: RNA primers are synthesized using both DNA strands as templates during DNA replication.
Statement B: DNA polymerases do not initiate replication but only elongate DNA from an available –OH group at its 3’ end.
(1) Both the statements A and B are correct and A is the reason for B
(2) Both the statements A and B are correct and B is the reason for A
(3) Statement A is correct and statement B is wrong
(4) Statement A is wrong and statement B is correct
24.DNA having labelled thymidine is allowed to replicate in a medium having non-radioactive thymidine. After three duplications, the number of DNA molecules having labelled thymidine shall be
(1) one
(2) two
(3) four
(4) eight
25.In a bacterial cell, in the absence of ligase enzyme
(1) the accumulation of Okazaki fragments occurs
(2) the RNA primer is not synthesised
(3) replicatioin is not at all initiated
(4) replication becomes continuous
26.In vitro synthesis of RNA and DNA was carried out respectively first by
(1) Kornberg and Nirenberg
(2) Ochoa and Kornberg
(3) Ochoa and Nirenberg
(4) Nirenberg and Khorana
27.DNA ligase was discovered by
(1) H G Khorana
(2) Okazaki
(3) Kornberg
(4) Nirenberg
28.DNA replication requires
(1) helicase, gyrase, DNA polymerase, primase
(2) helicase, RNA primer, ligase, DNA polymerase
(3) RNA primer, DNA polymerase, helicase, ligase, primase
(4) RNA primer, DNA polymerase, primase, helicase, ligase, gyrase
RNA:
29.Ribose sugar is covalently linked to uracil in RNA by
(1) hydrogen bond
(2) amide linkage
(3) phosphodiester bond
(4) disulphide bond
30.RNA and DNA are similar in that they
(1) are double stranded
(2) are polynucleotide chains
(3) have similar pentose sugars
(4) have similar pyrimidines
31.Which nitrogenous base is normally used in the synthesis of ribonucleic acid but not in the synthesis of deoxyribonucleic acid ?
(1) guanine
(2) cytosine
(3) uracil
(4) thymine
32.Which of the following has the highest molecular weight ?
(1) Mrna
(2) rRNA
(3) tRNA
(4) snRNA
33.The prokaryotic mRNA has
(1) poly A tail at 5’ end
(2) poly A tail at 3’ end
(3) poly A tail at both the ends
(4) no poly A tail
34.After the mRNA molecule is transcribed from an eukaryotic gene, segments called _______ are removed and the remaining ________ are spliced together to produce a mRNA molecule with a continuous coding sequence.
(1) exons ………….. introns
(2) exons ……………. promoters
(3) introns ………… exons
(4) introns …………... regulators
35.What is not true about RNA ?
(1) It has an important role in protein synthesis
(2) mRNA is the most abundant of all RNA in the cell
(3) tRNA selects amino acids as per codon for peptide assembly
(4) There are atleast 20 types of tRNA
36.Which is true about RNA ?
(1) tRNA acts as adapter for transferring amino acids to mRNA template during protein synthsis
(2) rRNA is the smallest RNA
(3) rRNA directs the sequence of amino acids in a polypeptide
(4) RNA has higher molecular weight than DNA
37.In the synthesis of proteins, what is the function of mRNA molecules ?
(1) to act as a template for the synthesis of deoxyribonucleic acid
(2) to carry information that determines the sequence of amino acids
(3) to carry the anticodons and amino acids to the ribosomes
(4) to carry specific enzymes for the formation of peptide bonds between amino acids
38.This virus has double stranded genomic RNA
(1) TMV
(2) wound tumor virus
(3) HIV
(4) HBV
39.DHU loop in tRNA is
(1) amino acid binding site
(2) ribosomal binding site
(3) amino-acyl binding site
(4) mRNA binding site
40.The greek letter in in the TC loop of tRNA molecule denotes
(1) pseudouridine
(2) dihydrouridine
(3) inosine
(4) ribothymidine
41.The 5’ end of a tRNA always starts with and 3’ end terminates with
(1) A and GGA respectively
(2) G and CCA respectively
(3) GGA and G respectively
(4) A and CAC respectively
42.NODOC is
(1) a unit of three nucleotides on tRNA
(2) a unit of three nucleotides on mRNA
(3) amino acid binding site in the ribosome
(4) the initiation factor required for protein synthesis
43.tRNA molecule does not have one of the following;
(1) anticodon
(2) amino acid binding site
(3) aminoacyl synthetase site
(4) ATP site
44.A RNA polymerase among the following is
(1) reverse transcriptase
(2) primase
(3) RNAase
(4) topoisomerase
45.RNA directed DNA synthesis was first reported in
(1) polio virus
(2) rhabdovirus
(3) retrovirus
(4) TMV
GENE CONCEPT
46.`One gene-one enzyme’ concept was first proposed by
(1) Beadle & Tatum
(2) Morgan & Sutton
(3) Jacob & Monad
(4) Mendel
47.A gene composed of coding part of exons and non – coding part of introns is called
(1) split gene
(2) clustered gene
(3) multiple gene
(4) cistron
48.The functional unit of a gene which directs the synthesis of a polypeptide chain during protein synthesis is called
(1) replicon
(2) recon
(3) muton
(4) cistron
49.This part of the gene has the maximum number of nucleotides
(1) cistron
(2) recon
(3) replicon
(4) muton
50.The portion of gene which is transcribed but not translated is
(1) cistron
(2) exon
(3) intron
(4) meson
51.Prokaryotic genes are different from eukaryotic genes in that
(1) they produce mono-cistronic mRNA
(2) they produce poly-cistronic mRNA
(3) they are split genes
(4) they are recognised & activated by transcription factors
GENETIC CODE:
52.The stop signals contain the following bases
(1) uracil, guanine and adenine
(2) adenine, guanine and cytosine
(3) adenine, cytosine and uracil
(4) uracil, guanine and thymine
53.A triplet codon which is functionally related to UAA and UAG is
(1) AUG
(2) UGA
(3) AUG
(4) UUU
54.Genetic code is degenerate because
(1) a single triplet codon recognises different amino acids
(2) a single amino acid is recognized by different triplet codons
(3) triplet codons undergo degeneration during translation
(4) triplet codons undergo reduction during translation
55.The first triplet codon to be deciphered by Nirenberg and Matthaei was
(1) UUU
(2) AAA
(3) CCC
(4) AUG
56.Wobble hypothesis establishes
(1) peptide chain formation
(2) initiation of peptide chain
(3) termination of peptide chain
(4) economy of tRNA molecules
57.A messenger RNA molecule for making a protein is made in the nucleus and sent out to a ribosome. The ribosome reads the mRNA message and makes a protein containing 120 amino acids. The mRNA consisted of atleast how many codons ?
(1) 40
(2) 120
(3) 240
(4) 360
PROTEIN SYNTHESIS:
58.All of the following are accurate regarding the processes of transcription and translation except
(1) transcription and translation take place outside the nucleus
(2) transcription and translation are the processes for making enzymes
(3) transcription and translation are the processes for making proteins
(4) transcription and translation occur when a cell is doing its regular cell work
59.The translation machinery consists of
(1) ribosomes and mRNA
(2) ribosomes, mRNA and tRNA
(3) ribosomes, mRNA, tRNA and amino acids
(4) robosomes, mRNA, tRNA, amino acids and amino-acyl tRNA synthetase
60.Central dogma of molecular biology has been modified after the discovery of
(1) reverse transcriptase
(2) DNA polymerase
(3) RNA polymerase
(4) Restriction endonuclease
61.Pick the wrong statement with respect to protein synthesis in eukaryotes
(1) during initiation, methionine is formylated
(2) synthesis occurs in 80 S ribosomes
(3) translation begins after the completion of transcription
(4) usually monoribosomes are employed for synthesis
62.The sequence of nitrogen bases in a portion of the sense strand of DNA was AAT GCT TAG GCA. What will be the sequence of nitrogen bases in the corresponding region of the transcripted mRNA ?
(1) UUT CGT TUC CGU
(2) AAU GCU UAG GCA
(3) UUA CGA AUC CGU
(4) TTA CGA ATC CGT
63.Choose the answer that has the following events of protein synthesis in the proper sequence;
(a) large sub-unit of ribosome gets attached to small sub-unit to form a complex
(b) a charged tRNA with the anticodon UAC binds to AUG of mRNA at P site of ribosome
(c) a second tRNA with an amino acid occupies the A site
(d) small ribosomal sub-unit associates with mRNA
(e) peptidyl transferase catalyses the formation of peptide bond
(1) a, c, b, d, e
(2) b, d, e, a, c
(3) d, b, a, c, e
(4) d, a, c, b, e
OPERON CONCEPT:
64.In E.coli, gene action is regulated
(1) at the time of transcription
(2) at the time of translation
(3) after transcription
(4) after translation
65.Wild type of E.coli cells are growing in normal medium with glucose. They are transferred to a medium containing only lactose as the sugar. Which one of the following changes takes place ?
(1) the lac – operon is repressed
(2) all operons are induced
(3) E.coli cells stop dividing
(4) the lac – operon is induced
66.In Lac – operon, the structural gene ‘z’ is responsible for the synthesis of the enzyme
(1) β – galactosidase
(2) galactoside permease
(3) galactoside transacetylase
(4) lactase
67.During transcription by a cistron, the DNA site at which RNA polymerase binds/interacts is
(1) promoter
(2) regulator
(3) operator
(4) repressor
68.According to operon concept, the regulatory gene regulates biochemical reactions in a cell by
(1) inhibiting transcription
(2) inactivating the enzymes
(3) inactivating the substrates
(4) inhibiting the migration of mRNA
69.The recognition site in the promoter region of a Lac - operon is sometimes referred to as
(1) TATA (TATAA box)
(2) Pribnow box
(3) Hognesse box
(4) poly A tail
70.Match the components of Lac operon of E. coli and choose the correct combination
Column I Column II
A. Structural gene p. Binding site for
repressor protein
B. Operator gene q. Codes for repressor
protein
C. Promoter gene r. Induces lactose transport
from the medium
D. Regulator gene s. Codes for enzyme proteins
t. Binding site of RNA
polymerase
(1) A = q, B = t, C = p, D = r
(2) A = r, B = s, C = t, D = p
(3) A = s, B = p, C = t, D = q
(4) A = t, B = s, C = q, D = p
---------
1.In Griffith’s experiment with Pneumococcus and mice,
(1) living non – virulent bacteria transformed the living virulent strains
(2) living non – virulent bacteria transformed the heat killed virulent strains
(3) living virulent bacteria transformed heat killed non- virulent strains
(4) heat killed virulent bacteria transformed living non – virulent strains
2.Purines and pyrimidines donot contain
(1) carbon
(2) nitrogen
(3) phosphorous
(4) oxygen
3.Purines are
(1) large sized, dicyclic, heterocyclic nitrobases
(2) large sized, monocyclic, heterocyclic nitrobases
(3) small sized, monocyclic, heterocyclic nitrobases
(4) small sized, dicyclic, heterocyclic nitrobases
4.In a nucleotide, the nitrogenous base and phosphate group are attached to the
(1) 1’ and 3’ carbon atoms of sugar respectively
(2) 2’ and 3’ carbon atoms of sugar respectively
(3) 1’ and 5’ carbon atoms of sugar respectively
(4) 2’ and 5’ carbon atoms of sugar respectively
5.Guanosine differs from Guanylic acid in not having
(1) phosphate
(2) sugar
(3) nitro-base
(4) phosphate and sugar
6.The lengths of major groove and minor groove are
(1) 22 Å and 12 Å respectively
(2) 20 Å and 10 Å respectively
(3) 34 Å and 3.4 Å respectively
(4) 20 Å and 12 Å respectively
7.Which of the following is not relevant to the structure of double helical DNA ?
(1) the helix makes one complete spiral turn every 34 Ǻ
(2) the distance between adjacent nucleotides is 3.4 Ǻ
(3) each strand of helix has a backbone of alternating ribose sugar and nitrobase
(4) the two adjacent sugar molecules are joined with phosphate by phosphodiester bond
8.The two strands of DNA are held together by H - bonds between
(1) nitrogen bases
(2) phosphoric acid
(3) sugar and phosphate
(4) nitrobases and sugar
9.Identify the incorrect statement;
(1) the two strands of DNA are antiparallel because of phosphodiester bonds
(2) the two strands of DNA coil helically around a common axis
(3) the distance between two base pairs in B-DNA is 34 Ǻ
(4) the diameter of DNA molecule is 20 Ǻ
10.Read the two statements A and B. Choose the correct answer from those given.
Statement A: B-DNA is a right handed helical molecule which is like a twisted ladder.
Statement B: Sugar and phosphate molecules make up the uprights of the ladder and hydrogen bonded bases make up the rungs.
(1) statement A is correct, B is wrong
(2) statement B is correct, A is wrong
(3) both the statements A and B are correct
(4) both the statements A and B are wrong
11.Which of the following statements does not apply to the Watson – Crick model for DNA ?
(1) the two strands of the helix are antiparallel
(2) the double ringed purines of one strand pair with the single ringed pyrimidines of the other strand
(3) the two strands of the helix are held together by covalent bonds
(4) the verticals of the helix are composed of alternately arranged sugar and phosphate units
12.Statement A: A DNA duplex has a constant diameter of 20 À.
Statement B: Dicyclic purines of one strand always pair with monocyclic pyrimidines of the other strand in a DNA molecule.
(1) Both the statements A and B are correct and A is the reason for B
(2) Both the statements A and B are correct and B is the reason for A
(3) Statement A is correct and statement B is wrong
(4) Statement A is wrong and statement B is correct
13.When the base composition of DNA from a bacterium was analysed, 22% of the bases were found to be Adenine. What is the G + C content ?
(1) 44%
(2) 33%
(3) 56%
(4) 22%
14.A sample of DNA is found to have the base composition (mole ratio) of A = 35, T = 25, G = 24 and C = 16. It suggests that
(1) DNA is circular duplex
(2) DNA is linear duplex
(3) DNA is single stranded
(4) it is Z - DNA
15.Identify the correct statement;
(1) DNA is always a right handed helix
(2) mitochondrial DNA is left handed double helix
(3) according to Chargaff’s rule, A + G = T + C
(4) DNA is double stranded in all organisms
16.Z - DNA is
(1) right handed helix with 9 base pairs per turn
(2) right handed helix with 10 base pairs per turn
(3) left handed helix with 11 base pairs per turn
(4) left handed helix with 12 base pairs per turn
17.A DNA strand is directly involved in the synthesis of all the following except a
(1) polypeptide chain
(2) mRNA strand
(3) rRNA strand
(4) complementary DNA strand
18.DNA replication occurs
(1) whenever a cell makes protein
(2) before a cell divides
(3) whenever a cell needs RNA
(4) in the cytoplasm of an eukaryotic cell
19.Which of the following processes occurs in an eukaryotic cell ?
(1) DNA replication only
(2) translation only
(3) DNA replication and translation
(4) DNA replication, translation and transcription
20.The enzymes responsible for unwinding of DNA helix during replication is / are
(1) helicases
(2) topoisomerases
(3) DNA polymerase
(4) primase
21.The enzyme DNA polymerase can add nucleotide to the ________ of sugar of another nucleotide.
(1) 3’ carbon position
(2) 5’ carbon position
(3) 4’ carbon position
(4) both 3’ and 5’ carbon positions
22.Okazaki fragments are small pieces of DNA that are produced at the time of the formation of
(1) leading strand
(2) lagging strand
(3) sense strand
(4) antisense strand
23.Statement A: RNA primers are synthesized using both DNA strands as templates during DNA replication.
Statement B: DNA polymerases do not initiate replication but only elongate DNA from an available –OH group at its 3’ end.
(1) Both the statements A and B are correct and A is the reason for B
(2) Both the statements A and B are correct and B is the reason for A
(3) Statement A is correct and statement B is wrong
(4) Statement A is wrong and statement B is correct
24.DNA having labelled thymidine is allowed to replicate in a medium having non-radioactive thymidine. After three duplications, the number of DNA molecules having labelled thymidine shall be
(1) one
(2) two
(3) four
(4) eight
25.In a bacterial cell, in the absence of ligase enzyme
(1) the accumulation of Okazaki fragments occurs
(2) the RNA primer is not synthesised
(3) replicatioin is not at all initiated
(4) replication becomes continuous
26.In vitro synthesis of RNA and DNA was carried out respectively first by
(1) Kornberg and Nirenberg
(2) Ochoa and Kornberg
(3) Ochoa and Nirenberg
(4) Nirenberg and Khorana
27.DNA ligase was discovered by
(1) H G Khorana
(2) Okazaki
(3) Kornberg
(4) Nirenberg
28.DNA replication requires
(1) helicase, gyrase, DNA polymerase, primase
(2) helicase, RNA primer, ligase, DNA polymerase
(3) RNA primer, DNA polymerase, helicase, ligase, primase
(4) RNA primer, DNA polymerase, primase, helicase, ligase, gyrase
RNA:
29.Ribose sugar is covalently linked to uracil in RNA by
(1) hydrogen bond
(2) amide linkage
(3) phosphodiester bond
(4) disulphide bond
30.RNA and DNA are similar in that they
(1) are double stranded
(2) are polynucleotide chains
(3) have similar pentose sugars
(4) have similar pyrimidines
31.Which nitrogenous base is normally used in the synthesis of ribonucleic acid but not in the synthesis of deoxyribonucleic acid ?
(1) guanine
(2) cytosine
(3) uracil
(4) thymine
32.Which of the following has the highest molecular weight ?
(1) Mrna
(2) rRNA
(3) tRNA
(4) snRNA
33.The prokaryotic mRNA has
(1) poly A tail at 5’ end
(2) poly A tail at 3’ end
(3) poly A tail at both the ends
(4) no poly A tail
34.After the mRNA molecule is transcribed from an eukaryotic gene, segments called _______ are removed and the remaining ________ are spliced together to produce a mRNA molecule with a continuous coding sequence.
(1) exons ………….. introns
(2) exons ……………. promoters
(3) introns ………… exons
(4) introns …………... regulators
35.What is not true about RNA ?
(1) It has an important role in protein synthesis
(2) mRNA is the most abundant of all RNA in the cell
(3) tRNA selects amino acids as per codon for peptide assembly
(4) There are atleast 20 types of tRNA
36.Which is true about RNA ?
(1) tRNA acts as adapter for transferring amino acids to mRNA template during protein synthsis
(2) rRNA is the smallest RNA
(3) rRNA directs the sequence of amino acids in a polypeptide
(4) RNA has higher molecular weight than DNA
37.In the synthesis of proteins, what is the function of mRNA molecules ?
(1) to act as a template for the synthesis of deoxyribonucleic acid
(2) to carry information that determines the sequence of amino acids
(3) to carry the anticodons and amino acids to the ribosomes
(4) to carry specific enzymes for the formation of peptide bonds between amino acids
38.This virus has double stranded genomic RNA
(1) TMV
(2) wound tumor virus
(3) HIV
(4) HBV
39.DHU loop in tRNA is
(1) amino acid binding site
(2) ribosomal binding site
(3) amino-acyl binding site
(4) mRNA binding site
40.The greek letter in in the TC loop of tRNA molecule denotes
(1) pseudouridine
(2) dihydrouridine
(3) inosine
(4) ribothymidine
41.The 5’ end of a tRNA always starts with and 3’ end terminates with
(1) A and GGA respectively
(2) G and CCA respectively
(3) GGA and G respectively
(4) A and CAC respectively
42.NODOC is
(1) a unit of three nucleotides on tRNA
(2) a unit of three nucleotides on mRNA
(3) amino acid binding site in the ribosome
(4) the initiation factor required for protein synthesis
43.tRNA molecule does not have one of the following;
(1) anticodon
(2) amino acid binding site
(3) aminoacyl synthetase site
(4) ATP site
44.A RNA polymerase among the following is
(1) reverse transcriptase
(2) primase
(3) RNAase
(4) topoisomerase
45.RNA directed DNA synthesis was first reported in
(1) polio virus
(2) rhabdovirus
(3) retrovirus
(4) TMV
GENE CONCEPT
46.`One gene-one enzyme’ concept was first proposed by
(1) Beadle & Tatum
(2) Morgan & Sutton
(3) Jacob & Monad
(4) Mendel
47.A gene composed of coding part of exons and non – coding part of introns is called
(1) split gene
(2) clustered gene
(3) multiple gene
(4) cistron
48.The functional unit of a gene which directs the synthesis of a polypeptide chain during protein synthesis is called
(1) replicon
(2) recon
(3) muton
(4) cistron
49.This part of the gene has the maximum number of nucleotides
(1) cistron
(2) recon
(3) replicon
(4) muton
50.The portion of gene which is transcribed but not translated is
(1) cistron
(2) exon
(3) intron
(4) meson
51.Prokaryotic genes are different from eukaryotic genes in that
(1) they produce mono-cistronic mRNA
(2) they produce poly-cistronic mRNA
(3) they are split genes
(4) they are recognised & activated by transcription factors
GENETIC CODE:
52.The stop signals contain the following bases
(1) uracil, guanine and adenine
(2) adenine, guanine and cytosine
(3) adenine, cytosine and uracil
(4) uracil, guanine and thymine
53.A triplet codon which is functionally related to UAA and UAG is
(1) AUG
(2) UGA
(3) AUG
(4) UUU
54.Genetic code is degenerate because
(1) a single triplet codon recognises different amino acids
(2) a single amino acid is recognized by different triplet codons
(3) triplet codons undergo degeneration during translation
(4) triplet codons undergo reduction during translation
55.The first triplet codon to be deciphered by Nirenberg and Matthaei was
(1) UUU
(2) AAA
(3) CCC
(4) AUG
56.Wobble hypothesis establishes
(1) peptide chain formation
(2) initiation of peptide chain
(3) termination of peptide chain
(4) economy of tRNA molecules
57.A messenger RNA molecule for making a protein is made in the nucleus and sent out to a ribosome. The ribosome reads the mRNA message and makes a protein containing 120 amino acids. The mRNA consisted of atleast how many codons ?
(1) 40
(2) 120
(3) 240
(4) 360
PROTEIN SYNTHESIS:
58.All of the following are accurate regarding the processes of transcription and translation except
(1) transcription and translation take place outside the nucleus
(2) transcription and translation are the processes for making enzymes
(3) transcription and translation are the processes for making proteins
(4) transcription and translation occur when a cell is doing its regular cell work
59.The translation machinery consists of
(1) ribosomes and mRNA
(2) ribosomes, mRNA and tRNA
(3) ribosomes, mRNA, tRNA and amino acids
(4) robosomes, mRNA, tRNA, amino acids and amino-acyl tRNA synthetase
60.Central dogma of molecular biology has been modified after the discovery of
(1) reverse transcriptase
(2) DNA polymerase
(3) RNA polymerase
(4) Restriction endonuclease
61.Pick the wrong statement with respect to protein synthesis in eukaryotes
(1) during initiation, methionine is formylated
(2) synthesis occurs in 80 S ribosomes
(3) translation begins after the completion of transcription
(4) usually monoribosomes are employed for synthesis
62.The sequence of nitrogen bases in a portion of the sense strand of DNA was AAT GCT TAG GCA. What will be the sequence of nitrogen bases in the corresponding region of the transcripted mRNA ?
(1) UUT CGT TUC CGU
(2) AAU GCU UAG GCA
(3) UUA CGA AUC CGU
(4) TTA CGA ATC CGT
63.Choose the answer that has the following events of protein synthesis in the proper sequence;
(a) large sub-unit of ribosome gets attached to small sub-unit to form a complex
(b) a charged tRNA with the anticodon UAC binds to AUG of mRNA at P site of ribosome
(c) a second tRNA with an amino acid occupies the A site
(d) small ribosomal sub-unit associates with mRNA
(e) peptidyl transferase catalyses the formation of peptide bond
(1) a, c, b, d, e
(2) b, d, e, a, c
(3) d, b, a, c, e
(4) d, a, c, b, e
OPERON CONCEPT:
64.In E.coli, gene action is regulated
(1) at the time of transcription
(2) at the time of translation
(3) after transcription
(4) after translation
65.Wild type of E.coli cells are growing in normal medium with glucose. They are transferred to a medium containing only lactose as the sugar. Which one of the following changes takes place ?
(1) the lac – operon is repressed
(2) all operons are induced
(3) E.coli cells stop dividing
(4) the lac – operon is induced
66.In Lac – operon, the structural gene ‘z’ is responsible for the synthesis of the enzyme
(1) β – galactosidase
(2) galactoside permease
(3) galactoside transacetylase
(4) lactase
67.During transcription by a cistron, the DNA site at which RNA polymerase binds/interacts is
(1) promoter
(2) regulator
(3) operator
(4) repressor
68.According to operon concept, the regulatory gene regulates biochemical reactions in a cell by
(1) inhibiting transcription
(2) inactivating the enzymes
(3) inactivating the substrates
(4) inhibiting the migration of mRNA
69.The recognition site in the promoter region of a Lac - operon is sometimes referred to as
(1) TATA (TATAA box)
(2) Pribnow box
(3) Hognesse box
(4) poly A tail
70.Match the components of Lac operon of E. coli and choose the correct combination
Column I Column II
A. Structural gene p. Binding site for
repressor protein
B. Operator gene q. Codes for repressor
protein
C. Promoter gene r. Induces lactose transport
from the medium
D. Regulator gene s. Codes for enzyme proteins
t. Binding site of RNA
polymerase
(1) A = q, B = t, C = p, D = r
(2) A = r, B = s, C = t, D = p
(3) A = s, B = p, C = t, D = q
(4) A = t, B = s, C = q, D = p
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Thursday, March 25, 2010
Sample questions for CET in GENETICS
GENETICS
1. Which one of the following phenomena of hereditary characters was not known to Mendel ?
(1) Assortment according to probability
(2) Independent assortment
(3) Dominant – recessive traits
(4) Linkage
2. In a monohybrid cross, the phenotypic ratio in the F2 generation will be
(1) 1 : 2 : 1
(2) 3 : 1
(3) 1 : 1
(4) 9 : 3 : 3 : 1
3. A garden pea plant heterozygous for tallness is selfed and 1200 seeds are subsequently germinated. How many seedlings would have the parental phenotype ?
(1) 300
(2) 600
(3) 900
(4) 1200
4. Five out of 20 plants obtained by selfing a red flowered plant were having white flowers. This is an indication that the plant is
(1) Homozygous
(2) Heterozygous
(3) Hemizygous
(4) Homogeneous
5. A pea plant has two genes for height. But, each of its sperm has only one. This illustrates
(1) independent assortment
(2) linked genes
(3) codominance
(4) segregation
6. When a plant heterozygous for tallness is selfed, the F2 generation will have both tall and dwarf plants. This proves the principle of
(1) Dominance
(2) Independent assortment
(3) Segregation
(4) Incomplete dominance
7. To determine heterzyogosity or homozygosity, a plant must be crossed with
(1) Recessive plant
(2) Dominant plant
(3) Homozygous plant
(4) Heterozygous plant
8. A hybrid tall plant of pea can be distinguished from a pure tall pea plant by
(1) Selfing and noting that all plants are short
(2) Selfing and noting that all plants are tall
(3) Crossing with dwarf plant & noting that all plants are tall
(4) Crossing with dwarf plant and noting that 50% are tall and 50% are short
9. In a plant, red flower colour is dominant over white flower colour. A true breeding plant producing red flower is crossed with a pure plant producing white flowers. After selfing the F1 plants, the proportion of plants producing white flowers in the total progeny would be
(1) One third
(2) One fourth
(3) Three fourth
(4) Half
10. If a male who is heterozygous for an autosomal trait mates with a female who is also heterozygous for that trait, what percent of their offsprings are likely to be heterozygous for this trait as well ?
(1) 0
(2) 25
(3) 50
(4) 75
11. When a particular characteristic of an individual like flower colour shows variation among the offspring produced after the individual is selfed, it is said to be
(1) Pure breeding
(2) True breeding
(3) Homozygous
(4) Heterozygous
12. If a heterozygous tall plant is crossed with homozygous dwarf plant, the proportion of the dwarf progeny will be
(1) 25%
(2) 50%
(3) 75%
(4) 100%
13. Two pea plants were subjected to cross pollination. Of the 183 plants produced in the next generation, 94 plants were found to be tall and 89 plants were found to be dwarf. The genotypes of the two parental plants are likely to be
(1) TT and tt
(2) Tt and Tt
(3) Tt and tt
(4) TT and TT
14. The allele for coloured seed coat (S) is dominant over that of white seed coat (s). When flowers of a plant with white seed coat (ss) are pollinated with those of coloured seed coat (Ss), the colour of the coat in the developing seeds will be
(1) White
(2) Coloured
(3) Mosaic
(4) White and coloured in 1 : 1 ratio
15. A monohybrid produces 2 kinds of gametes. A dihybrid produces 4 kinds of gametes. If ‘n’ is the number of characters, how many kinds of gametes are produced by a hybrid ?
(1) 2 x n
(2) 2n
(3) 2/n
(4) n/2
16. Tallness (T) and round shape of seed (R) are dominant over dwarfness (t) and wrinkled shape (r). What will be the ratio of tall plants producing wrinkled seeds in the cross TtRr x TtRr ?
(1) 2/32
(2) 4/32
(3) 6/32
(4) 8/32
17. If a cell of an individual with the genotype AaBb undergoes meiosis, the possible genotypes of the gametes will be
(1) Aa, Bb, AB, ab
(2) AA, BB, aa, bb
(3) AB, Aa, BB, AA
(4) AB, Ab, aB, ab
18. If ‘B’ and ‘b’ are the two alternate forms (alleles) of a gene for a particular trait, which of the following crosses would produce 50% homozygotes and 50% heterozygotes ?
(1) BB x Bb only
(2) Bb x Bb only
(2) Bb x Bb and Bb x bb only
(4) Bb x Bb, BB x Bb and Bb x bb
19. The genetic ratio (phenotypic) of 9 : 3 : 3 : 1 is due to
(1) Incomplete dominance
(2) Independent assortment of genes
(3) Crossing over of chromosomes
(4) Synapsis between homologous chromosomes
20. The F2 generation of a cross produced identical phenotypic and genotypic ratios. It is not an expected Mendelian ratio, and can be attributed to
(1) Independent assortment
(2) Linkage
(3) Incomplete dominance
(4) Principle of dominance
21. In monohybrid crosses, absence of complete dominance is indicated by the F1 plants that exhibit intermediate characters. This can be further confirmed if the phenotypic ratio in F2 is
(1) 3 : 1
(2) 1 : 1
(3) 1 : 2 : 1
(4) 2 : 1 : 1
22. In Mirabilis jalapa, true bred red and white flowered plants are crossed. In F2 generation, they form
(1) 75% red and 25% white flowered plants
(2) 25% red and 75% white flowered plants
(3) 50% red and 50% white flowered plants
(4) 25% red, 50% pink and 25% white flowered plants
23. When two pink flowered four ‘O’ clock (Mirabilis jalapa) plants are selfed, the proportion of the progeny having pink flowers is
(1) 0
(2) ¼
(3) ½
(4) ¾
24. There is a plant in which genes for black flower colour and white flower colour donot show complete dominance or recessiveness. If a plant carrying black flower colour genes is crossed with plant having only white flower colour genes, all the off-springs have grey coloured flowers. If two of these grey flowered plants are crossed, the theoretical progeny ratio would be
(1) All grey
(2) Either all black or all white
(3) 50 % black and 50 % white
(4) 50% grey, 25% black and 25% white
25. In Mirabilis jalapa, RR, Rr and rr determine red, pink and white colours (of flowers) respectively. When F1 hybrid of RR and rr was crossed with dominant parent, the ratio produced is
(1) All red
(2) 2 red : 2 pink
(3) All white
(4) 2 pink : 2 white
26. The molecules on the membranes of RBCs that make up a person’s blood type are
(1) phosphates
(2) lipids
(3) nucleotides
(4) proteins
27. Agglutinogens are present
(1) In the plasma of blood
(2) On the plasma membrane of RBC
(3) In the W B C
(4) In the cytoplasm
28. Agglutinogens are not found in the following blood group;
(1) A
(2) B
(3) AB
(4) O
29. A person is heterozygous for blood group A. His blood group is determined by
(1) Different alleles on the same chromosome
(2) Different alleles on the chromosomes of a homologous pair
(3) Similar alleles on the same chromosome
(4) Similar alleles on the chromosomes of a homologous pair
30. The blood comes under group ‘O’ if:
(1) The blood shows coagulation with both antisera A & B
(2) The blood does not coagulation with antisera A or B
(3) The blood shows coagulation with antiserum A
(4) The blood shows coagulation with antiserum B
31. If both father and mother have blood group `O’, the possible blood group of the children will be
(1) O
(2) A, B, AB, O
(3) A, AB
(4) B, AB
32. If the blood groups of parents are AB and B, what would be the possible blood group of their children ?
(1) A, B and O
(2) A, B and AB
(3) A and B
(4) O
33. If the genotypes for blood groups of parents are IA IB and IB IO, what would be the possible blood groups of their children ?
(1) A, B and AB
(2) A, B and O
(3) A and B
(4) All possible
34. Three children in a family have blood types O, AB and B. What could be the genotypes of their parents ?
(1) IA IB and i i
(2) IA i and IB i
(3) IA IA and IB I i
(4) IB IB and IA IA
35. A child with blood group genotype IAIB is born of a woman with genotype IBIB. The father could not be a man of genotype
(1) IAIB
(2) IBIO
(3) IAIO
(4) IAIA
36. A woman sues a man for the support of her child. She has blood group ‘A’, her child ‘O’ and the man ‘B’. It implies that
(1) The man could be the father of this child
(2) Both woman and man could together produce such a child
(3) They cannot have a child with ‘O’ group
(4) Both (a) and (b) are possible
37. A poor woman and a rich woman delivered babies at the same time in a local nursing home. There was a mix-up. Both the women claimed the male baby as theirs and nobody claimed the female one. The blood group of the disputed female baby is ‘O’, that of poor woman is AB and that of rich woman is A. Who is the actual mother of the female baby ?
(1) Poor woman
(2) Rich woman
(3) Neither of the two
(4) Data insufficient
38. Biologically marriage should be avoided between
(1) Rh+ male and Rh+ female
(2) Rh+ male and Rh- female
(3) Rh- male and Rh+ female
(4) Rh- male and Rh- female
39. When a Rh– mother and Rh+ father produce a Rh– baby,
(1) mother produces antibodies which enter foetal blood causing erythroblastosis foetalis
(2) there will be still birth if the mother is not administered Rhogam antibodies
(3) mother does not produce any antibody and the child will be norml
(4) the baby will always have congenital defects
40. Usually, the recessive character is expressed only when the allele is present in double recessive condition. However, single recessive gene can express itself when the gene is present on
(1) Either an autosome or X – chromosome
(2) The X - chromosome of the male
(3) The X - chromosome of the female
(4) Any autosome
41. A sex linked character is inherited
(1) Only in males
(2) Only in females
(3) In both males and females
(4) Through the autosomes in one sex only
42. Statement A: Colourblindness is more likely to occur in females than in males.
Statement B: The Y chromosome of the male has the genes for distinguishing colour.
(1) Statement A is correct and B is wrong
(2) Statement B is correct and A is wrong
(3) Both the statements A and B wrong
(4) Both the statements A and B correct
43. A person whose father is colour blind marries a lady whose mother is the daughter of a colour blind man. Then,
(1) All the sons are colour blind
(2) Some of the sons are colour blind some are normal
(3) All children are colour blind
(4) All daughters are normal
44. Expression of recessive genes on X - chromosome is definite in males because of
(1) Homozygous nature
(2) Hemizygous nature
(3) Polyzygous nature
(4) Inverted condition
45. When a colour blind man marries the daughter of a colour blind person, in the progeny
(1) 50% of the sons are colour blind
(2) All the daughters are colour blind
(3) All the sons are colour blind
(4) 50% of the daughters & 50% of the sons are colour blind
46. The probability of the daughters of husband and wife having normal vision (their fathers being colour blind) becoming colour blind is
(1) 0%
(2) 25%
(3) 50%
(4) 75%
47. When a haemophilic / colour blind female marries a normal male, in the progeny
(1) Females are normal but carriers and males are haemophilic / colour blind
(2) Females are haemophilic / colour blind and males are normal
(3) Both males and females are haemophilic / colour blind
(4) Both males and females are normal
48. A colour blind boy has two sisters. One of them is normal and the other is colour blind. What can be the possible nature of their parents ?
(1) Colour blind father and normal mother
(2) Colour blind father and carrier mother
(3) Both father and mother are colour blind
(4) Normal father and colour blind mother
49. Holandric genes are the genes present on
(1) The X – chromosome
(2) The Y - chromosome
(3) The autosome
(4) Both X and Y chromosomes
50. Holandric genes are always transmitted to the son
(1) From X chromosome of male
(2) From Y chromosome of male
(3) From X chromosome of female
(4) From X chromosome of either female or male
51. A boy has hairy ears which is Y-linked inheritance. What is the chance that his grandson will inherit the trait from him ?
(1) 0%
(2) 25%
(3) 50%
(4) 100%
52. Haemophilia is
(1) Due to the lack of either clotting factors VIII or IX
(2) Due to recessive gene on X chromosome of male
(3) Failure in coagulation of blood
(4) All these
53. Haemoglobin of patients suffering from sickle cell anemia differs from normal haemoglobin by one
(1) Monosaccharide
(2) Disaccharide
(3) Amino acid
(4) Lipid
54. Sickle cell anaemia is
(1) A sex linked disorder
(2) Iron deficiency disease
(3) An organic disorder
(4) Inherited by a recessive gene on the autosome
55. Statement A: Heterozygous individuals for sickle cell trait of RBC have 50% of their haemoglobin defective.
Statement B: Sickle cell anaemia is due to a defective gene in the sex chromosome.
(1) Both the statements are correct and B is the reason for A
(2) Both the statements are correct and B is not the reason A
(3) Statement A is correct and statement B is wrong
(4) Statement B is correct and statement A is wrong
56. Sickle celled anaemia is due to
(1) Mutation of gene directing the synthesis of a - chain of haemoglobin
(2) Mutation of gene directing the synthesis of b - chain of haemoglobin
(3) Mutation of a gene located on the X – chromosome
(4) Lack of haemoglobin
57. Sickle celled anaemia is caused due to the substitution of
(1) Valine at 6th position of beta - chain of haemoglobin by glutamic acid
(2) Glutamic acid at 6th position of beta - chain of haemoglobin by valine
(3) Glutamic acid at 6th position of alpha - chain of haemoglobin by valine
(4) Valine at 6th position of alpha - chain of haemoglobin by glutamic acid
58. Sickle cell anemia is a disorder caused due to the change in the chemical nature of
(1) Alpha chain of haemoglobin
(2) Beta chain of haemoglobin
(3) Both alpha and beta chains of haemoglobin
(4) Plasma proteins
59. Gynaecomastia is a condition seen in
(1) Down’s syndrome
(2) Turner’s syndrome
(3) Klinefelter’s syndrome
(4) Cri-du-chat syndrome
60. Find out the correct statement
(1) Monosomy and trisomy are the two types of euploidy
(2) Polyploidy is more common in animals than in plants
(3) Polyploids occur due to the failure in complete separation of sets of chromosomes
(4) 2n – 1 condition results in trisomy
61. Statement A: An additional X chromosome in humans results in Klinefelter’s syndrome.
Statement B: Non-disjunction of chromosomes is the chief cause for aneuploidy.
(1) Both the statements are correct and B is the reason for A
(2) Both the statements are correct and B is not the reason A
(3) Statement A is correct and statement B is wrong
(4) Statement B is correct and statement A is wrong
62. The chromosome complement in an individual suffering from Klinefelter’s syndrome is
(1) 44AA + XX
(2) 44AA + XXY
(3) 44AA+XO
(4) 22A+Y
63. A person with Klinefelter’s syndrome is considered a
(1) monosomic
(2) nullisomic
(3) trisomic
(4) triploid
64. One of the following is not synonymous to the rest:
(1) Mongoloid idiocy
(2) Down’s syndrome
(3) 21-trisomy
(4) Turner’s syndrome
65. Down’s syndrome is due to
(1) Crossing over
(2) Sex-linkage
(3) Linkage
(4) Non-disjunction of chromosomes
66. Turner’s syndrome is due to
(1) Monosomic chromosome
(2) Bisomic chromosomes
(3) Trisomic chromosomes
(4) Ploysomic chromosomes
67. Deletion of a part of the 5th chromosome is responsible for
(1) Underdeveloped breasts, sparse pubic hair, small uterus, reduced ovaries, sterility
(2) Small prostate glands, small testes, sparse body hair, long stature due to long limb bones, enlarged breasts
(3) Clogging of blood capillaries, anaemia, reduced blood supply to vital organs, jaundice, muscle cramps, headache, etc.
(4) Very small head, increased inter-orbital distance, severe mental deficiency, moon face, cat cry of new born
68. Cri du chat syndrome is due to
(1) Addition of an autosome
(2) Deletion of a part of an autosome
(3) Addition of a X – chromosome
(4) Deletion of a part of X – chromosome
69. These are the characteristics of Down’s syndrome
(1) Sparse body hair, gynaecomastia, azoospermia, male phenotype
(2) Female phenotype, short stature, absence of menstrual cycle
(3) Microcephaly, hypertelorism, underdeveloped larynx, moon – face
(4) Short stature, broad forehead, epicanthal folds, short phalanges, short flat nose
70. Webbing in the neck, short stature, pigeon chest, underdeveloped breasts, rudimentary ovaries, sterility, etc., are characters of
(1) Down’s syndrome
(2) Klinefelter’s syndrome
(3) Ullrich’s syndrome
(4) Turner’s syndrome
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1. Which one of the following phenomena of hereditary characters was not known to Mendel ?
(1) Assortment according to probability
(2) Independent assortment
(3) Dominant – recessive traits
(4) Linkage
2. In a monohybrid cross, the phenotypic ratio in the F2 generation will be
(1) 1 : 2 : 1
(2) 3 : 1
(3) 1 : 1
(4) 9 : 3 : 3 : 1
3. A garden pea plant heterozygous for tallness is selfed and 1200 seeds are subsequently germinated. How many seedlings would have the parental phenotype ?
(1) 300
(2) 600
(3) 900
(4) 1200
4. Five out of 20 plants obtained by selfing a red flowered plant were having white flowers. This is an indication that the plant is
(1) Homozygous
(2) Heterozygous
(3) Hemizygous
(4) Homogeneous
5. A pea plant has two genes for height. But, each of its sperm has only one. This illustrates
(1) independent assortment
(2) linked genes
(3) codominance
(4) segregation
6. When a plant heterozygous for tallness is selfed, the F2 generation will have both tall and dwarf plants. This proves the principle of
(1) Dominance
(2) Independent assortment
(3) Segregation
(4) Incomplete dominance
7. To determine heterzyogosity or homozygosity, a plant must be crossed with
(1) Recessive plant
(2) Dominant plant
(3) Homozygous plant
(4) Heterozygous plant
8. A hybrid tall plant of pea can be distinguished from a pure tall pea plant by
(1) Selfing and noting that all plants are short
(2) Selfing and noting that all plants are tall
(3) Crossing with dwarf plant & noting that all plants are tall
(4) Crossing with dwarf plant and noting that 50% are tall and 50% are short
9. In a plant, red flower colour is dominant over white flower colour. A true breeding plant producing red flower is crossed with a pure plant producing white flowers. After selfing the F1 plants, the proportion of plants producing white flowers in the total progeny would be
(1) One third
(2) One fourth
(3) Three fourth
(4) Half
10. If a male who is heterozygous for an autosomal trait mates with a female who is also heterozygous for that trait, what percent of their offsprings are likely to be heterozygous for this trait as well ?
(1) 0
(2) 25
(3) 50
(4) 75
11. When a particular characteristic of an individual like flower colour shows variation among the offspring produced after the individual is selfed, it is said to be
(1) Pure breeding
(2) True breeding
(3) Homozygous
(4) Heterozygous
12. If a heterozygous tall plant is crossed with homozygous dwarf plant, the proportion of the dwarf progeny will be
(1) 25%
(2) 50%
(3) 75%
(4) 100%
13. Two pea plants were subjected to cross pollination. Of the 183 plants produced in the next generation, 94 plants were found to be tall and 89 plants were found to be dwarf. The genotypes of the two parental plants are likely to be
(1) TT and tt
(2) Tt and Tt
(3) Tt and tt
(4) TT and TT
14. The allele for coloured seed coat (S) is dominant over that of white seed coat (s). When flowers of a plant with white seed coat (ss) are pollinated with those of coloured seed coat (Ss), the colour of the coat in the developing seeds will be
(1) White
(2) Coloured
(3) Mosaic
(4) White and coloured in 1 : 1 ratio
15. A monohybrid produces 2 kinds of gametes. A dihybrid produces 4 kinds of gametes. If ‘n’ is the number of characters, how many kinds of gametes are produced by a hybrid ?
(1) 2 x n
(2) 2n
(3) 2/n
(4) n/2
16. Tallness (T) and round shape of seed (R) are dominant over dwarfness (t) and wrinkled shape (r). What will be the ratio of tall plants producing wrinkled seeds in the cross TtRr x TtRr ?
(1) 2/32
(2) 4/32
(3) 6/32
(4) 8/32
17. If a cell of an individual with the genotype AaBb undergoes meiosis, the possible genotypes of the gametes will be
(1) Aa, Bb, AB, ab
(2) AA, BB, aa, bb
(3) AB, Aa, BB, AA
(4) AB, Ab, aB, ab
18. If ‘B’ and ‘b’ are the two alternate forms (alleles) of a gene for a particular trait, which of the following crosses would produce 50% homozygotes and 50% heterozygotes ?
(1) BB x Bb only
(2) Bb x Bb only
(2) Bb x Bb and Bb x bb only
(4) Bb x Bb, BB x Bb and Bb x bb
19. The genetic ratio (phenotypic) of 9 : 3 : 3 : 1 is due to
(1) Incomplete dominance
(2) Independent assortment of genes
(3) Crossing over of chromosomes
(4) Synapsis between homologous chromosomes
20. The F2 generation of a cross produced identical phenotypic and genotypic ratios. It is not an expected Mendelian ratio, and can be attributed to
(1) Independent assortment
(2) Linkage
(3) Incomplete dominance
(4) Principle of dominance
21. In monohybrid crosses, absence of complete dominance is indicated by the F1 plants that exhibit intermediate characters. This can be further confirmed if the phenotypic ratio in F2 is
(1) 3 : 1
(2) 1 : 1
(3) 1 : 2 : 1
(4) 2 : 1 : 1
22. In Mirabilis jalapa, true bred red and white flowered plants are crossed. In F2 generation, they form
(1) 75% red and 25% white flowered plants
(2) 25% red and 75% white flowered plants
(3) 50% red and 50% white flowered plants
(4) 25% red, 50% pink and 25% white flowered plants
23. When two pink flowered four ‘O’ clock (Mirabilis jalapa) plants are selfed, the proportion of the progeny having pink flowers is
(1) 0
(2) ¼
(3) ½
(4) ¾
24. There is a plant in which genes for black flower colour and white flower colour donot show complete dominance or recessiveness. If a plant carrying black flower colour genes is crossed with plant having only white flower colour genes, all the off-springs have grey coloured flowers. If two of these grey flowered plants are crossed, the theoretical progeny ratio would be
(1) All grey
(2) Either all black or all white
(3) 50 % black and 50 % white
(4) 50% grey, 25% black and 25% white
25. In Mirabilis jalapa, RR, Rr and rr determine red, pink and white colours (of flowers) respectively. When F1 hybrid of RR and rr was crossed with dominant parent, the ratio produced is
(1) All red
(2) 2 red : 2 pink
(3) All white
(4) 2 pink : 2 white
26. The molecules on the membranes of RBCs that make up a person’s blood type are
(1) phosphates
(2) lipids
(3) nucleotides
(4) proteins
27. Agglutinogens are present
(1) In the plasma of blood
(2) On the plasma membrane of RBC
(3) In the W B C
(4) In the cytoplasm
28. Agglutinogens are not found in the following blood group;
(1) A
(2) B
(3) AB
(4) O
29. A person is heterozygous for blood group A. His blood group is determined by
(1) Different alleles on the same chromosome
(2) Different alleles on the chromosomes of a homologous pair
(3) Similar alleles on the same chromosome
(4) Similar alleles on the chromosomes of a homologous pair
30. The blood comes under group ‘O’ if:
(1) The blood shows coagulation with both antisera A & B
(2) The blood does not coagulation with antisera A or B
(3) The blood shows coagulation with antiserum A
(4) The blood shows coagulation with antiserum B
31. If both father and mother have blood group `O’, the possible blood group of the children will be
(1) O
(2) A, B, AB, O
(3) A, AB
(4) B, AB
32. If the blood groups of parents are AB and B, what would be the possible blood group of their children ?
(1) A, B and O
(2) A, B and AB
(3) A and B
(4) O
33. If the genotypes for blood groups of parents are IA IB and IB IO, what would be the possible blood groups of their children ?
(1) A, B and AB
(2) A, B and O
(3) A and B
(4) All possible
34. Three children in a family have blood types O, AB and B. What could be the genotypes of their parents ?
(1) IA IB and i i
(2) IA i and IB i
(3) IA IA and IB I i
(4) IB IB and IA IA
35. A child with blood group genotype IAIB is born of a woman with genotype IBIB. The father could not be a man of genotype
(1) IAIB
(2) IBIO
(3) IAIO
(4) IAIA
36. A woman sues a man for the support of her child. She has blood group ‘A’, her child ‘O’ and the man ‘B’. It implies that
(1) The man could be the father of this child
(2) Both woman and man could together produce such a child
(3) They cannot have a child with ‘O’ group
(4) Both (a) and (b) are possible
37. A poor woman and a rich woman delivered babies at the same time in a local nursing home. There was a mix-up. Both the women claimed the male baby as theirs and nobody claimed the female one. The blood group of the disputed female baby is ‘O’, that of poor woman is AB and that of rich woman is A. Who is the actual mother of the female baby ?
(1) Poor woman
(2) Rich woman
(3) Neither of the two
(4) Data insufficient
38. Biologically marriage should be avoided between
(1) Rh+ male and Rh+ female
(2) Rh+ male and Rh- female
(3) Rh- male and Rh+ female
(4) Rh- male and Rh- female
39. When a Rh– mother and Rh+ father produce a Rh– baby,
(1) mother produces antibodies which enter foetal blood causing erythroblastosis foetalis
(2) there will be still birth if the mother is not administered Rhogam antibodies
(3) mother does not produce any antibody and the child will be norml
(4) the baby will always have congenital defects
40. Usually, the recessive character is expressed only when the allele is present in double recessive condition. However, single recessive gene can express itself when the gene is present on
(1) Either an autosome or X – chromosome
(2) The X - chromosome of the male
(3) The X - chromosome of the female
(4) Any autosome
41. A sex linked character is inherited
(1) Only in males
(2) Only in females
(3) In both males and females
(4) Through the autosomes in one sex only
42. Statement A: Colourblindness is more likely to occur in females than in males.
Statement B: The Y chromosome of the male has the genes for distinguishing colour.
(1) Statement A is correct and B is wrong
(2) Statement B is correct and A is wrong
(3) Both the statements A and B wrong
(4) Both the statements A and B correct
43. A person whose father is colour blind marries a lady whose mother is the daughter of a colour blind man. Then,
(1) All the sons are colour blind
(2) Some of the sons are colour blind some are normal
(3) All children are colour blind
(4) All daughters are normal
44. Expression of recessive genes on X - chromosome is definite in males because of
(1) Homozygous nature
(2) Hemizygous nature
(3) Polyzygous nature
(4) Inverted condition
45. When a colour blind man marries the daughter of a colour blind person, in the progeny
(1) 50% of the sons are colour blind
(2) All the daughters are colour blind
(3) All the sons are colour blind
(4) 50% of the daughters & 50% of the sons are colour blind
46. The probability of the daughters of husband and wife having normal vision (their fathers being colour blind) becoming colour blind is
(1) 0%
(2) 25%
(3) 50%
(4) 75%
47. When a haemophilic / colour blind female marries a normal male, in the progeny
(1) Females are normal but carriers and males are haemophilic / colour blind
(2) Females are haemophilic / colour blind and males are normal
(3) Both males and females are haemophilic / colour blind
(4) Both males and females are normal
48. A colour blind boy has two sisters. One of them is normal and the other is colour blind. What can be the possible nature of their parents ?
(1) Colour blind father and normal mother
(2) Colour blind father and carrier mother
(3) Both father and mother are colour blind
(4) Normal father and colour blind mother
49. Holandric genes are the genes present on
(1) The X – chromosome
(2) The Y - chromosome
(3) The autosome
(4) Both X and Y chromosomes
50. Holandric genes are always transmitted to the son
(1) From X chromosome of male
(2) From Y chromosome of male
(3) From X chromosome of female
(4) From X chromosome of either female or male
51. A boy has hairy ears which is Y-linked inheritance. What is the chance that his grandson will inherit the trait from him ?
(1) 0%
(2) 25%
(3) 50%
(4) 100%
52. Haemophilia is
(1) Due to the lack of either clotting factors VIII or IX
(2) Due to recessive gene on X chromosome of male
(3) Failure in coagulation of blood
(4) All these
53. Haemoglobin of patients suffering from sickle cell anemia differs from normal haemoglobin by one
(1) Monosaccharide
(2) Disaccharide
(3) Amino acid
(4) Lipid
54. Sickle cell anaemia is
(1) A sex linked disorder
(2) Iron deficiency disease
(3) An organic disorder
(4) Inherited by a recessive gene on the autosome
55. Statement A: Heterozygous individuals for sickle cell trait of RBC have 50% of their haemoglobin defective.
Statement B: Sickle cell anaemia is due to a defective gene in the sex chromosome.
(1) Both the statements are correct and B is the reason for A
(2) Both the statements are correct and B is not the reason A
(3) Statement A is correct and statement B is wrong
(4) Statement B is correct and statement A is wrong
56. Sickle celled anaemia is due to
(1) Mutation of gene directing the synthesis of a - chain of haemoglobin
(2) Mutation of gene directing the synthesis of b - chain of haemoglobin
(3) Mutation of a gene located on the X – chromosome
(4) Lack of haemoglobin
57. Sickle celled anaemia is caused due to the substitution of
(1) Valine at 6th position of beta - chain of haemoglobin by glutamic acid
(2) Glutamic acid at 6th position of beta - chain of haemoglobin by valine
(3) Glutamic acid at 6th position of alpha - chain of haemoglobin by valine
(4) Valine at 6th position of alpha - chain of haemoglobin by glutamic acid
58. Sickle cell anemia is a disorder caused due to the change in the chemical nature of
(1) Alpha chain of haemoglobin
(2) Beta chain of haemoglobin
(3) Both alpha and beta chains of haemoglobin
(4) Plasma proteins
59. Gynaecomastia is a condition seen in
(1) Down’s syndrome
(2) Turner’s syndrome
(3) Klinefelter’s syndrome
(4) Cri-du-chat syndrome
60. Find out the correct statement
(1) Monosomy and trisomy are the two types of euploidy
(2) Polyploidy is more common in animals than in plants
(3) Polyploids occur due to the failure in complete separation of sets of chromosomes
(4) 2n – 1 condition results in trisomy
61. Statement A: An additional X chromosome in humans results in Klinefelter’s syndrome.
Statement B: Non-disjunction of chromosomes is the chief cause for aneuploidy.
(1) Both the statements are correct and B is the reason for A
(2) Both the statements are correct and B is not the reason A
(3) Statement A is correct and statement B is wrong
(4) Statement B is correct and statement A is wrong
62. The chromosome complement in an individual suffering from Klinefelter’s syndrome is
(1) 44AA + XX
(2) 44AA + XXY
(3) 44AA+XO
(4) 22A+Y
63. A person with Klinefelter’s syndrome is considered a
(1) monosomic
(2) nullisomic
(3) trisomic
(4) triploid
64. One of the following is not synonymous to the rest:
(1) Mongoloid idiocy
(2) Down’s syndrome
(3) 21-trisomy
(4) Turner’s syndrome
65. Down’s syndrome is due to
(1) Crossing over
(2) Sex-linkage
(3) Linkage
(4) Non-disjunction of chromosomes
66. Turner’s syndrome is due to
(1) Monosomic chromosome
(2) Bisomic chromosomes
(3) Trisomic chromosomes
(4) Ploysomic chromosomes
67. Deletion of a part of the 5th chromosome is responsible for
(1) Underdeveloped breasts, sparse pubic hair, small uterus, reduced ovaries, sterility
(2) Small prostate glands, small testes, sparse body hair, long stature due to long limb bones, enlarged breasts
(3) Clogging of blood capillaries, anaemia, reduced blood supply to vital organs, jaundice, muscle cramps, headache, etc.
(4) Very small head, increased inter-orbital distance, severe mental deficiency, moon face, cat cry of new born
68. Cri du chat syndrome is due to
(1) Addition of an autosome
(2) Deletion of a part of an autosome
(3) Addition of a X – chromosome
(4) Deletion of a part of X – chromosome
69. These are the characteristics of Down’s syndrome
(1) Sparse body hair, gynaecomastia, azoospermia, male phenotype
(2) Female phenotype, short stature, absence of menstrual cycle
(3) Microcephaly, hypertelorism, underdeveloped larynx, moon – face
(4) Short stature, broad forehead, epicanthal folds, short phalanges, short flat nose
70. Webbing in the neck, short stature, pigeon chest, underdeveloped breasts, rudimentary ovaries, sterility, etc., are characters of
(1) Down’s syndrome
(2) Klinefelter’s syndrome
(3) Ullrich’s syndrome
(4) Turner’s syndrome
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