Biology

General

Easy

### Question

#### Evolutionary convergence is the development of:

- common set of characters in groups of different ancestry
- common set of characters in closely related groups
- dissimilar characters in closely related groups
- random mating

#### The correct answer is: common set of characters in groups of different ancestry

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### Related Questions to study

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#### Darwin's finches represents:

#### Darwin's finches represents:

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#### Organisms which obtain energy by the oxidation of reduced inorganic compounds are called:

#### Organisms which obtain energy by the oxidation of reduced inorganic compounds are called:

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#### Finding of Miller's experiment on origin of life has provided evidence for:

#### Finding of Miller's experiment on origin of life has provided evidence for:

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#### The number of codons that code different amino acids is -

#### The number of codons that code different amino acids is -

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#### Assertion (A):If $a>0,b>0$ and $c>0$, then $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\ge 9$Reason (R): For positive numbers $a,$ $b,$ $c,$ $AM\ge GM$

$AM\ge GM$$=\succ \frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3}\ge \left(\begin{array}{cc}111& \\ -& -\\ \overline{b}a.& c\end{array}\right)\to \left(1\right)$$D:{x}^{2}-3x-4<0\supset (x+1)(x-4)<0\Rightarrow 1<x<4$${x}^{2}-3x+2>0Rejectx-2\left)\right(x-1)>0=x<1{a}^{r}x>2$$\frac{(a+b+c)}{3}\ge (abc{)}^{\frac{1}{3}}\to (2)$multiply (1) and (2)

#### Assertion (A):If $a>0,b>0$ and $c>0$, then $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\ge 9$Reason (R): For positive numbers $a,$ $b,$ $c,$ $AM\ge GM$

maths-General

$AM\ge GM$$=\succ \frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3}\ge \left(\begin{array}{cc}111& \\ -& -\\ \overline{b}a.& c\end{array}\right)\to \left(1\right)$$D:{x}^{2}-3x-4<0\supset (x+1)(x-4)<0\Rightarrow 1<x<4$${x}^{2}-3x+2>0Rejectx-2\left)\right(x-1)>0=x<1{a}^{r}x>2$$\frac{(a+b+c)}{3}\ge (abc{)}^{\frac{1}{3}}\to (2)$multiply (1) and (2)

maths-

#### The length of the chord joining the points $\left(4\mathrm{cos}\theta ,4\mathrm{sin}\theta \right)$ and $(4\mathrm{cos}\left(\theta +60\xb0\right),4\mathrm{sin}(\theta +60\xb0)$of the circle ${x}^{2}+{y}^{2}=16$ is

#### The length of the chord joining the points $\left(4\mathrm{cos}\theta ,4\mathrm{sin}\theta \right)$ and $(4\mathrm{cos}\left(\theta +60\xb0\right),4\mathrm{sin}(\theta +60\xb0)$of the circle ${x}^{2}+{y}^{2}=16$ is

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maths-

#### Observe the following statements and choose correct answer Statement-I:A particle p moves along a straight line, starting from a fixed point ‘O’, obeying s=16+48t-${t}^{3}.$the direction of p when t=5 is along $\overline{OP}$ Statement-II: For a particle moving on a line, if velocity v<0 then the body moves towards the initial point.

Statement -I:

#### Observe the following statements and choose correct answer Statement-I:A particle p moves along a straight line, starting from a fixed point ‘O’, obeying s=16+48t-${t}^{3}.$the direction of p when t=5 is along $\overline{OP}$ Statement-II: For a particle moving on a line, if velocity v<0 then the body moves towards the initial point.

maths-General

Statement -I:

maths-

#### Statement ‐ I : The equation $z\overline{z}+\overline{a}z+a\overline{z}+\lambda =0$ where a is a complex number, represents a circle in Argand plane if $\lambda $ is real. Statement ‐II : The radius of the circle $z\overline{z}+\overline{a}z+a\overline{z}+\lambda =0$ is $\sqrt{\lambda -a\overline{a}}$

Result

#### Statement ‐ I : The equation $z\overline{z}+\overline{a}z+a\overline{z}+\lambda =0$ where a is a complex number, represents a circle in Argand plane if $\lambda $ is real. Statement ‐II : The radius of the circle $z\overline{z}+\overline{a}z+a\overline{z}+\lambda =0$ is $\sqrt{\lambda -a\overline{a}}$

maths-General

Result

GeneralVideo

#### Divide the following fractions:

Here, we have to divide the given fractions.

Now, 2/11 ÷ 9/22

= 2/11 × 22/9

= 2/1 × 2/9

= 2 × 2 / 1 × 9

= 4/9

Hence, the correct option is (a).

Now, 2/11 ÷ 9/22

= 2/11 × 22/9

= 2/1 × 2/9

= 2 × 2 / 1 × 9

= 4/9

Hence, the correct option is (a).

#### Divide the following fractions:

GeneralGeneral

Here, we have to divide the given fractions.

Now, 2/11 ÷ 9/22

= 2/11 × 22/9

= 2/1 × 2/9

= 2 × 2 / 1 × 9

= 4/9

Hence, the correct option is (a).

Now, 2/11 ÷ 9/22

= 2/11 × 22/9

= 2/1 × 2/9

= 2 × 2 / 1 × 9

= 4/9

Hence, the correct option is (a).

biology

#### The third cleavage in frog's development is

#### The third cleavage in frog's development is

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#### The reason why the right kidney is slightly lower than the left is:

Excretion involves the processes in which substances of no further use or those present in excess qualities are thrown out of the body.

#### The reason why the right kidney is slightly lower than the left is:

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Excretion involves the processes in which substances of no further use or those present in excess qualities are thrown out of the body.

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#### This happens if the proximal convoluted tubule is removed from nephron:

#### This happens if the proximal convoluted tubule is removed from nephron:

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Biology

#### Which is the highest structural organization found in all enzymes?

#### Which is the highest structural organization found in all enzymes?

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#### The 20 different common amino acids have different-

#### The 20 different common amino acids have different-

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Maths-

#### A box contains 100 tickets numbered 1, 2 ...... 100. Two tickets are chosen at random. It is given that the maximum number on the two chosen tickets is not more than 10. The minimum number on them is 5 with probability

Let be the event that the maximum number on the two chosen tickets is not more than 10 i.e., the number on them and be the event that the maximum number on them is 5, i.e., the number on them is we have to find .

Now

Now the number of ways of getting a number on the two tickets is the coefficient of in the expansion of

Thus coefficient of of of of is 9.

Hence

and

Hence required probability

Now

Now the number of ways of getting a number on the two tickets is the coefficient of in the expansion of

Thus coefficient of of of of is 9.

Hence

and

Hence required probability

#### A box contains 100 tickets numbered 1, 2 ...... 100. Two tickets are chosen at random. It is given that the maximum number on the two chosen tickets is not more than 10. The minimum number on them is 5 with probability

Maths-General

Let be the event that the maximum number on the two chosen tickets is not more than 10 i.e., the number on them and be the event that the maximum number on them is 5, i.e., the number on them is we have to find .

Now

Now the number of ways of getting a number on the two tickets is the coefficient of in the expansion of

Thus coefficient of of of of is 9.

Hence

and

Hence required probability

Now

Now the number of ways of getting a number on the two tickets is the coefficient of in the expansion of

Thus coefficient of of of of is 9.

Hence

and

Hence required probability