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(D) The bacterium divides every $35$ minutes. Given total time = $175$ minutes. Number of divisions $(n)$ = $\frac{175}{35} = 5$. Since one bacterium

Vedclass · Accepted Answer

$32 	imes 10^5$ cells/ml

Vedclass · Answer

(D) The bacterium divides every $35$ minutes. Given total time = $175$ minutes. Number of divisions $(n)$ = $\frac{175}{35} = 5$. Since one bacterium produces two cells per division (binary fission),the population grows exponentially according to the formula $N_t = N_0 	imes 2^n$,where $N_0$ is the initial concentration and $n$ is the number of divisions. Initial concentration $(N_0)$ = $10^5$ cells/ml. Final concentration $(N_t)$ = $10^5 	imes 2^5 = 10^5 	imes 32 = 32 	imes 10^5$ cells/ml.

Column-$I$	Column-$II$
$(A)$ Butterfly	$(1)$ $60 \text{ years}$
$(B)$ Crow	$(2)$ $140 \text{ years}$
$(C)$ Parrot	$(3)$ $15 \text{ years}$
$(D)$ Crocodile	$(4)$ $1-2 \text{ weeks}$

Column-$I$ (Reproductive Structures)	Column-$II$ (Examples)
$P$. Conidia	$I$. Hydra
$Q$. Buds	$II$. Penicillium
$R$. Gemmules	$III$. Sponges

Column – $I$	Column – $II$
$(1)$ Conidia	$(p)$ Hydra
$(2)$ Buds	$(q)$ Penicillium
$(3)$ Gemmules	$(r)$ Amoeba
$(4)$ Binary fission	$(s)$ Sponge

$A$ bacterium divides every $35$ minutes. If a culture containing $10^5$ cells/ml is grown for $175$ minutes,what will be the cell concentration/ml after $175$ minutes?

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Match the following and choose the correct combination from the options given:

Column-$I$ Column-$II$

$(A)$ Butterfly $(1)$ $60 \text{ years}$

$(B)$ Crow $(2)$ $140 \text{ years}$

$(C)$ Parrot $(3)$ $15 \text{ years}$

$(D)$ Crocodile $(4)$ $1-2 \text{ weeks}$

Match the following columns:

Column-$I$ (Reproductive Structures) Column-$II$ (Examples)

$P$. Conidia $I$. Hydra

$Q$. Buds $II$. Penicillium

$R$. Gemmules $III$. Sponges

Match the following:

Column – $I$ Column – $II$

$(1)$ Conidia $(p)$ Hydra

$(2)$ Buds $(q)$ Penicillium

$(3)$ Gemmules $(r)$ Amoeba

$(4)$ Binary fission $(s)$ Sponge

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