Monthly Mock Test (December 31) - NEET MCQ

• The process of segregation that occurs during meiosis is a random process.
• Thus the gametes will have 50% of each allele.
• Hence Vv heterozygous plant would produce 50% V and 50% v containing gametes.

The double helix describes the appearance of double-stranded DNA, which is composed of two linear strands that run opposite to each other, or anti-parallel, and twist together. Each DNA strand within the double helix is a long, linear molecule made of smaller units called nucleotides that form a chain.

In stretching, specific resistance remains unchanged.
After stretching, specific resistance (rho) will remain same.
Original resistance of the wire,

Clearly, 2Ω, 4Ω and ( 1 + 5) Ωresistors are
in parallel. Hence, potential difference is
same across each of them.
∴ I₁ × 2 = I₂ × 4 = I₃ × 6

Given I₁ = 3A ∴ I₁ × 2 = I₃ × 6
Given I₁ = 3A.
∴ I₁ × 2 = I₃ × 6 provides

Now, the potential across the 5Ω resistor is
V = I3 × 5 = 1 × 5 = 5V.
∴ the power dissipated in the 5Ωresistor

Fossils are dead remains of plants and animals buried inside the earth crust millions of year ago. Fossils are commonly preserved in sedimentary rock as this kinds of rocks are formed layer by layer in course of time.

Rewriting the given data for the reaction

Actually this reaction is autocatalyzed and involves complex calculation for concentration terms.
We can look at the above results in a simple way to find the dependence of reaction rate (i.e. rate of disappearance of Br₂).
From data (1) and (2) in which concentration of CH₃COCH₃ and H⁺ remain unchanged and only the concentration of Br₂ is doubled, there is no change in rate of reaction. It means the rate of reaction is independent of concentration of Br₂.
Again from (2) and (3) in which (CH₃CO CH₃) and (Br₂) remain constant but H⁺ increases from 0.05 M to 0.10 i.e. doubled, the rate of reaction changes from 5.7×10^–5 to 1.2 × 10^–4 (or 12 × 10^–5), thus it also becomes almost doubled. It shows that rate of reaction is directly proportional to [H⁺].
From (3) and (4), the rate should have doubled due to increase in conc of [H⁺] from 0.10 M to 0.20 M but the rate has changed from 1.2× 10^–4 to 3.1×10^–4. This is due to change in concentration of CH₃ CO CH₃ from 0.30 M to 0.40 M. Thus the rate is directly proportional to [CH₃ COCH₃].
We now get rate = k [CH₃COCH₃]¹[Br₂]⁰[H⁺]1        
= k [CH₃COCH₃][H⁺].

Fossils are generally found in sedimentary rocks.

In human genome project, commonly used host were bacteria and yeast and their vectors are called as Bacterial artificial chromosome (BAC) and yeast artificial chromosome (YAC)

(i) When key between the terminals 1
and 2 is plugged in,
P.D. across R = IR = k l₁
⇒ R = k l₁ as I = 1A
(ii) When key between terminals 1 and 3 is
plugged in,
P.D. across (X + R) = I(X + R) = k l₂
⇒ X + R = k l₂
∴ X = k (l₂ – l₁)
∴ R = kl₁ and X = k (l₂ – l₁)

According to question, tallness (T) is dominant over dwarfism (t) and red colour (R) is dominant over white (r) fruit colour.

Parent Generation; P₁ : RRTt  x  rrtt

F₁ generation :

Phenotypic ratio = 1  (tall plant, red fruit)         
: 1 (dwarf plant, red fruit)
; thus, percent of tall plant with red fruit is 50%.

• Natural selection mainly leads to three selections.
• They are: Stabilizing selection, directional selection, and disruptive selection.
• Technical selection does not belong to natural selection.
• This classification is based on different organism-environmental relationship.

The power dissipated in the circuit

v = 10 volt

Substituting the values in equation (i)

Given initial concentration (a) = 2.00 m; Time taken (t) = 200 min and final concentration (a – x)    = 0.15 m. For a first order reation rate constant,

Further

Codon AAA code for amino acid lysine but UUA, AUG and CCC do not code for Valine, cysteine and alanine respectively. So, AAA and lysine pair is correct option out of given options.

Current from D to C = 1A
∴ V_D – V_C = 2 × 1 = 2V
V_A = 0     ∴ VC = 1V, ∴ V_D – V_C = 2
⇒V_D – 1 = 2     ∴ V_D = 3V
∴ V_D – V_B = 2 ∴ 3 – V_B = 2 ∴ V_B = 1V

Rate of disappearance of H₂ = rate of formation of NH₃.

= 3 ×10 ^–4 mol L^–1s ^–1

UAA of mRNA do not code for any amino acids so it is a termination codon. If termination codon is present on mRNA, the protein synthesis stops abruptly at that point.

UGA, UAG and UAAare called termination or stop codon because they do not code for any amino acid. As soon as any one of them codon is located on m-RNA translation process is stopped.

The current through the resistance R

The potential difference across R

Thus V increases as R increases upto certain
limit, but it does not increase further.

For a first order reaction, A → products

= 3 × 10^–2

Further, 

The terminal potential difference of a cell is
given by V + Ir = E

V= V_A – V_B
or V = E – I_r

 Also for, i = 0 then V = E

∴ slope = – r, intercept = E

VNTR of two persons may be similar at certain sites but could be different at other sites because a child inherits 50% of chromosomes from mother and remaining 50% from father. VNTR genes also undergoes different kinds of mutations.

AB blood group of mother have two alleles IA and IB. The heterozygous B blood group of father has alleles IB and i. The possible outcome are IA IB,IAi,IBIB and IBi. That form AB, A, B and B blood group respectively.

If we write rate of reaction in terms of concentration of NH₃ and H₂,then Rate of reaction

So, 

• Stabilizing selection operates in a constant environment.
• It favors the average or complex phenotype and eliminates the extreme values.
• The mean value never changes. Also, the peak gets higher and narrower.

• Chimpanzee is closely related to man. DNA content and DNA matching are the same in both.
• This similarity is more than 99% with chimpanzee whereas 94% with a gibbon.

Rate of appearance of B is equal to rate of disappearance of A.

• The regulation of a lac operon by the repressor is known as negative regulation.
• At rare occasions, lac operons are also observed to be under the control of positive regulation.
• In negative regulation, the operon cannot transcribe the RNA polymerase enzyme.

RateI  = k [A]^x[B]^y ... (1)

or Rate1 = 4k[A]^x[2B]^y

From (1) and (2) we get

or    y = –2.

Galvanometer is converted into ammeter,
by connected a shunt, in parallel with it.

Here S << G so
S = 0.001 Ω