Biology
General
Easy

Question

A widely accepted, improved model of cell membrane is

  1. Fluid mosaic model

         

  2. Robertson’s model    

  3. Danielli and Davson’s model

         

  4. Unit membrane model    

The correct answers are:

Fluid mosaic model

,

Danielli and Davson’s model


    The ability to distinguish different neighbouring cells is important for organism’s function
    Glycolipids are lipids with attached carbohydrate, which acts as recognition sites during cell-cell interaction, as well as sites of attachment in a tissue
    Glycoproteins are often integral membrane proteins and are also important for cell recognition

    Book A Free Demo

    +91

    Grade*

    Related Questions to study

    General
    biology

    Rough endoplasmic reticulum differs from smooth walled endoplasmic reticulum due to the presence of

    In eukaryotes, DNA is tightly bound to histones which form a DNA protein particle called nucleosome

    Rough endoplasmic reticulum differs from smooth walled endoplasmic reticulum due to the presence of

    biologyGeneral
    In eukaryotes, DNA is tightly bound to histones which form a DNA protein particle called nucleosome
    General
    biology

    Middle lamella is present

    Steps of Gram’s staining technique
    (i) Staining with weak alkaline solution of crystal violet
    (ii) Treatment with 0.5% iodine solution
    (iii) Washing with water
    (iv) Treatment with absolute alcohol/acetone

    Middle lamella is present

    biologyGeneral
    Steps of Gram’s staining technique
    (i) Staining with weak alkaline solution of crystal violet
    (ii) Treatment with 0.5% iodine solution
    (iii) Washing with water
    (iv) Treatment with absolute alcohol/acetone
    General
    biology

    A are granular structures first observed under electron microscope as dense particles by …B… (1953). Here, A and B refer to

    J D Watson and F H C Crick (1953) showed that DNA has a double helical structure with two polynucleotide chains connected by hydrogen bonds and running in opposite directions (antiparallel). The antiparallel strands of a DNA molecule means that the phosphate groups at the start of two DNA strands are in opposite position (pole).

    A are granular structures first observed under electron microscope as dense particles by …B… (1953). Here, A and B refer to

    biologyGeneral
    J D Watson and F H C Crick (1953) showed that DNA has a double helical structure with two polynucleotide chains connected by hydrogen bonds and running in opposite directions (antiparallel). The antiparallel strands of a DNA molecule means that the phosphate groups at the start of two DNA strands are in opposite position (pole).
    General
    biology

    Number of protofilament in microtubule is

    In a DNA molecule, a complete line measures 34 text Å end text times left parenthesis 3.4 times n m right parenthesis with a distance of 3.4 text Å end text times left parenthesis 0.34 n m right parenthesis between two successive base pairs.

    Number of protofilament in microtubule is

    biologyGeneral
    In a DNA molecule, a complete line measures 34 text Å end text times left parenthesis 3.4 times n m right parenthesis with a distance of 3.4 text Å end text times left parenthesis 0.34 n m right parenthesis between two successive base pairs.
    General
    biology

    Ribosomes that occur exclusively in mitochondria is

    In prokaryotes, a nucleus is absent but nucleoid is found which is equivalent to a single chromosome or prochromosome

    Ribosomes that occur exclusively in mitochondria is

    biologyGeneral
    In prokaryotes, a nucleus is absent but nucleoid is found which is equivalent to a single chromosome or prochromosome
    General
    physics-

    A T shaped object with dimension shown in the figure, is lying on a smooth floor. A force F is applied at the point P parallel to A B, such that the object has only the translational motion without rotation. Find the location of P with respect to C.

    For translator motion the force should be applied on the centre of mass of the body so we have to calculate the location of centre of mass of T shaped object.
    Let mass of rod A B is m so the mass of rod C D will be 2m.
    Let y subscript 1 end subscript is the centre of mass of rod A B and y subscript 2 end subscript is the centre of mass of rod C D. We can consider that whole mass of the rod is placed at their respective centre of mass i e, mass m is placed at y subscript 1 end subscript and mass 2 m is placed at y subscript 2 end subscript.
    Taking point c at the origin position vector of points y subscript 1 end subscript a n d blank y subscript 2 end subscript can be written as
    r subscript 1 end subscript equals 2 l j comma blank r subscript 2 end subscript equals l j
    and m subscript 1 end subscript equals m a n d m subscript 2 end subscript equals 2 m
    Position vector of centre of mass of the system
    r subscript C M end subscript equals fraction numerator m subscript 1 end subscript r subscript 1 end subscript plus m subscript 2 end subscript r subscript 2 end subscript over denominator m plus m subscript 2 end subscript end fraction equals fraction numerator m 2 l stack j with hat on top plus 2 m l stack j with hat on top over denominator m plus 2 m end fraction equals fraction numerator 4 m l stack j with hat on top over denominator 3 m end fraction equals fraction numerator 4 l stack j with hat on top over denominator 3 end fraction

    A T shaped object with dimension shown in the figure, is lying on a smooth floor. A force F is applied at the point P parallel to A B, such that the object has only the translational motion without rotation. Find the location of P with respect to C.

    physics-General
    For translator motion the force should be applied on the centre of mass of the body so we have to calculate the location of centre of mass of T shaped object.
    Let mass of rod A B is m so the mass of rod C D will be 2m.
    Let y subscript 1 end subscript is the centre of mass of rod A B and y subscript 2 end subscript is the centre of mass of rod C D. We can consider that whole mass of the rod is placed at their respective centre of mass i e, mass m is placed at y subscript 1 end subscript and mass 2 m is placed at y subscript 2 end subscript.
    Taking point c at the origin position vector of points y subscript 1 end subscript a n d blank y subscript 2 end subscript can be written as
    r subscript 1 end subscript equals 2 l j comma blank r subscript 2 end subscript equals l j
    and m subscript 1 end subscript equals m a n d m subscript 2 end subscript equals 2 m
    Position vector of centre of mass of the system
    r subscript C M end subscript equals fraction numerator m subscript 1 end subscript r subscript 1 end subscript plus m subscript 2 end subscript r subscript 2 end subscript over denominator m plus m subscript 2 end subscript end fraction equals fraction numerator m 2 l stack j with hat on top plus 2 m l stack j with hat on top over denominator m plus 2 m end fraction equals fraction numerator 4 m l stack j with hat on top over denominator 3 m end fraction equals fraction numerator 4 l stack j with hat on top over denominator 3 end fraction
    General
    physics-

    Two balls each of mass m are placed on the vertices A and B of an equilateral triangle A B C of side 1m. A ball of mass 2m is placed at vertex C. The centre of mass of this system from vertex A (located at origin) is

    The centre of mass is given by

    stack x with minus on top equals fraction numerator m subscript 1 end subscript x subscript 1 end subscript plus m subscript 2 end subscript x subscript 2 end subscript plus m subscript 3 end subscript x subscript 3 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript plus m subscript 3 end subscript end fraction
    stack x with minus on top equals fraction numerator m cross times 0 plus m cross times 1 plus 2 m cross times open parentheses fraction numerator 1 over denominator 2 end fraction close parentheses over denominator m plus m plus 2 m end fraction
    stack x with minus on top equals fraction numerator 2 m over denominator 4 m end fraction equals fraction numerator 1 over denominator 2 end fraction m
    stack y with minus on top equals fraction numerator m subscript 1 end subscript y subscript 1 end subscript plus m subscript 2 end subscript y subscript 2 end subscript plus m subscript 3 end subscript y subscript 3 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript plus m subscript 3 end subscript end fraction
    stack y with minus on top equals fraction numerator m cross times 0 plus m cross times 0 plus 2 m cross times square root of 3 divided by 2 over denominator m plus m plus 2 m end fraction
    equals fraction numerator square root of 3 over denominator 4 end fractionm
    therefore Centre of mass is open parentheses fraction numerator 1 over denominator 2 end fraction m comma fraction numerator square root of 3 over denominator 4 end fraction m close parentheses.

    Two balls each of mass m are placed on the vertices A and B of an equilateral triangle A B C of side 1m. A ball of mass 2m is placed at vertex C. The centre of mass of this system from vertex A (located at origin) is

    physics-General
    The centre of mass is given by

    stack x with minus on top equals fraction numerator m subscript 1 end subscript x subscript 1 end subscript plus m subscript 2 end subscript x subscript 2 end subscript plus m subscript 3 end subscript x subscript 3 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript plus m subscript 3 end subscript end fraction
    stack x with minus on top equals fraction numerator m cross times 0 plus m cross times 1 plus 2 m cross times open parentheses fraction numerator 1 over denominator 2 end fraction close parentheses over denominator m plus m plus 2 m end fraction
    stack x with minus on top equals fraction numerator 2 m over denominator 4 m end fraction equals fraction numerator 1 over denominator 2 end fraction m
    stack y with minus on top equals fraction numerator m subscript 1 end subscript y subscript 1 end subscript plus m subscript 2 end subscript y subscript 2 end subscript plus m subscript 3 end subscript y subscript 3 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript plus m subscript 3 end subscript end fraction
    stack y with minus on top equals fraction numerator m cross times 0 plus m cross times 0 plus 2 m cross times square root of 3 divided by 2 over denominator m plus m plus 2 m end fraction
    equals fraction numerator square root of 3 over denominator 4 end fractionm
    therefore Centre of mass is open parentheses fraction numerator 1 over denominator 2 end fraction m comma fraction numerator square root of 3 over denominator 4 end fraction m close parentheses.
    General
    physics-

    Three identical spheres, each of mass 1 kg are kept as shown in figure below, touching each other, with their centers on a straight line. If their centres are marked P,Q, R blankrespectively, the distance of centre of mass of the system from P is

    ) fraction numerator P Q plus Q R plus P R over denominator 6 end fraction
    Distance of centre of mass from P is
    r equals fraction numerator r subscript 1 end subscript plus r subscript 2 end subscript plus r subscript 3 end subscript over denominator 3 end fraction equals fraction numerator 0 plus P Q plus P R over denominator 3 end fraction equals fraction numerator P Q plus P R over denominator 3 end fraction

    Three identical spheres, each of mass 1 kg are kept as shown in figure below, touching each other, with their centers on a straight line. If their centres are marked P,Q, R blankrespectively, the distance of centre of mass of the system from P is

    physics-General
    ) fraction numerator P Q plus Q R plus P R over denominator 6 end fraction
    Distance of centre of mass from P is
    r equals fraction numerator r subscript 1 end subscript plus r subscript 2 end subscript plus r subscript 3 end subscript over denominator 3 end fraction equals fraction numerator 0 plus P Q plus P R over denominator 3 end fraction equals fraction numerator P Q plus P R over denominator 3 end fraction
    General
    chemistry-

    WhatweightofHNO3isneededtoconvert 62gmofP4 inH3PO4 ithreaction? P4 +HNO3 →H3PO4 +NO2 +H2O

    WhatweightofHNO3isneededtoconvert 62gmofP4 inH3PO4 ithreaction? P4 +HNO3 →H3PO4 +NO2 +H2O

    chemistry-General
    General
    physics-

    The distance of the centre of mass of the T-shaped plate from O is

    Co-ordinate of CM is given by
    X subscript C M end subscript equals fraction numerator m subscript 1 end subscript x subscript 1 end subscript plus m subscript 2 end subscript x subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction

    Taking parts A and B as two bodies of same system
    m subscript 1 end subscript equals l cross times b cross times sigma equals 8 cross times 2 cross times sigma equals 16 sigma
    m subscript 2 end subscript equals l cross times b cross times sigma equals 6 cross times 2 cross times sigma equals 12 sigma
    Choosing O as origin,
    x subscript 1 end subscript equals 1 blank m comma blank x subscript 2 end subscript equals 2 plus 3 equals 5 blank m
    therefore X subscript C M end subscript equals fraction numerator 16 sigma cross times 1 plus 12 sigma cross times 5 over denominator 16 sigma plus 12 sigma end fraction equals fraction numerator 19 over denominator 7 end fraction
    equals 2.7 blank m f r o m O

    The distance of the centre of mass of the T-shaped plate from O is

    physics-General
    Co-ordinate of CM is given by
    X subscript C M end subscript equals fraction numerator m subscript 1 end subscript x subscript 1 end subscript plus m subscript 2 end subscript x subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction

    Taking parts A and B as two bodies of same system
    m subscript 1 end subscript equals l cross times b cross times sigma equals 8 cross times 2 cross times sigma equals 16 sigma
    m subscript 2 end subscript equals l cross times b cross times sigma equals 6 cross times 2 cross times sigma equals 12 sigma
    Choosing O as origin,
    x subscript 1 end subscript equals 1 blank m comma blank x subscript 2 end subscript equals 2 plus 3 equals 5 blank m
    therefore X subscript C M end subscript equals fraction numerator 16 sigma cross times 1 plus 12 sigma cross times 5 over denominator 16 sigma plus 12 sigma end fraction equals fraction numerator 19 over denominator 7 end fraction
    equals 2.7 blank m f r o m O
    General
    physics-

    Look at the drawing given in the figure, which has been drawn with ink of uniform line-thickness. The mass of ink used to draw each of the two inner circles, and each of the two line segments is m. The mass of ink used to draw the outer circle is 6m. The coordinates of the centres of the different parts are : outer circle (0, 0) left inner circle (-a comma blank a), right inner circle (a comma blank a) vertical line (0, 0) and horizontal line (0, -a). The y- coordinate of the centre of mass of the ink in the drawing is

    y subscript C M end subscript equals blank fraction numerator m subscript 1 end subscript y subscript 1 end subscript plus m subscript 2 end subscript y subscript 2 end subscript plus m subscript 3 end subscript y subscript 3 end subscript plus m subscript 4 end subscript y subscript 4 end subscript plus m subscript 5 end subscript y subscript 5 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript plus m subscript 3 end subscript plus m subscript 4 end subscript plus m subscript 5 end subscript end fraction
    equals fraction numerator open parentheses 6 m close parentheses open parentheses 0 close parentheses plus m open parentheses a close parentheses plus m open parentheses a close parentheses plus m open parentheses 0 close parentheses plus m left parenthesis negative a right parenthesis over denominator 6 m plus m plus m plus m plus m end fraction equals fraction numerator a over denominator 10 end fraction

    Look at the drawing given in the figure, which has been drawn with ink of uniform line-thickness. The mass of ink used to draw each of the two inner circles, and each of the two line segments is m. The mass of ink used to draw the outer circle is 6m. The coordinates of the centres of the different parts are : outer circle (0, 0) left inner circle (-a comma blank a), right inner circle (a comma blank a) vertical line (0, 0) and horizontal line (0, -a). The y- coordinate of the centre of mass of the ink in the drawing is

    physics-General
    y subscript C M end subscript equals blank fraction numerator m subscript 1 end subscript y subscript 1 end subscript plus m subscript 2 end subscript y subscript 2 end subscript plus m subscript 3 end subscript y subscript 3 end subscript plus m subscript 4 end subscript y subscript 4 end subscript plus m subscript 5 end subscript y subscript 5 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript plus m subscript 3 end subscript plus m subscript 4 end subscript plus m subscript 5 end subscript end fraction
    equals fraction numerator open parentheses 6 m close parentheses open parentheses 0 close parentheses plus m open parentheses a close parentheses plus m open parentheses a close parentheses plus m open parentheses 0 close parentheses plus m left parenthesis negative a right parenthesis over denominator 6 m plus m plus m plus m plus m end fraction equals fraction numerator a over denominator 10 end fraction
    General
    chemistry-

    Thenthalpanentropchangfor reaction Br2(l)+Cl2(g)→2BrCl(g)are30KJmol1and 105JK1mol1respectively.Thetemperature at which threactiowilbiequilibriuis-

    Thenthalpanentropchangfor reaction Br2(l)+Cl2(g)→2BrCl(g)are30KJmol1and 105JK1mol1respectively.Thetemperature at which threactiowilbiequilibriuis-

    chemistry-General
    General
    chemistry-

    Assumeacreactioicarried out in an open container. Fowhich reactiowill DH=DE?

    Assumeacreactioicarried out in an open container. Fowhich reactiowill DH=DE?

    chemistry-General
    General
    Maths-

    Marks scored by 100 students in a 25 marks unit test of Mathematics is given below Their median is

    Marks scored by 100 students in a 25 marks unit test of Mathematics is given below Their median is

    Maths-General
    General
    Maths-

    The mode for the following frequency distribution

    The mode for the following frequency distribution

    Maths-General