Mock Test - 1 (TB) - Software Development MCQ

Given:

Here, we arrange the letters according to the given options one by one:

(1) 2, 5, 3, 7, 6, 4, 1 → NEURTAL

(2) 2, 5, 3, 6, 7, 1, 4 → NEUTRLA

(3) 2, 3, 5, 6, 7, 4, 1 → NUETRAL

(4) 2, 5, 3, 6, 7, 4, 1 → NEUTRAL

Here, NEUTRAL is a meaningful word.

Hence, the correct answer is "2, 5, 3, 6, 7, 4, 1".

Given:

Total people qualified for the driving test in 1996 and 1997 = 45,000

Number of people who took the driving test in 1996 = 25,000

Calculation:

Number of people who qualified driving test in 1996 = 60% of 25000

= 60100×25000 = 15,000

People qualified the driving test in 1997 = 45000 – 15000 = 30,000

According to the question,

⇒ People qualified the driving test in 1997 = 50% = 30,000

⇒ Total number of people who took the driving test in 1997 = 30000 + 30000 = 60,000

Therefore, the total number of people who took part in the driving test in 1997 is 60,000.

Hence, the correct answer is option 2).

Given:

The sum of the squares of the roots of the equation x2 - 14x + k = 0 = 100

Formula used:

For a quadratic equation ax² + bx + c = 0

Sum of roots = −ba

Product of roots = ca

Calculation:

x2 - 14x + k = 0

α + β = −ba = 14

αβ = ca = k

Now, the sum of the squares of the roots 

⇒ α² + β² = (α + β)² - 2αβ

⇒ α2 + β2 = (14)² - 2k = 100      (Given)

⇒ 196 - 2k = 100

⇒ 2k = 96

⇒ k = 48

∴ The required value of k is 48.

As per the given information, 

The circle represents 'Meat eaters', the triangle represents 'Vegetable eaters' and the rectangle represents 'Milk drinkers'.

The Venn diagram for the given representation is :

The shaded part represents the number of 'Milk drinkers' who are 'Vegetable eaters', but do NOT eat meat = 18

Hence, the correct answer is "18".

As per the given information:

(1). Jawed scored 23 marks in a class test.

(2).  Shakeel scored 1 mark more than Jawed, but 1 mark less than Azhar.

Therefore, Shakeel's marks = 23 + 1 = 24

Azhar marks = Shakeel's marks + 1 = 25

Thus, we get Azhar > Shakeel > Jawed

(3). Siraj scored 4 marks more than Aftab, and 2 marks less than Shakeel.

Siraj's marks = Shakeel's marks - 2 = 24 - 2 = 22

Aftab's marks = Siraj's marks - 4 = 22 - 4 = 18

Thus we get Shakeel > Siraj > Aftab

Now, The following order is:

Azhar > Shakeel > Jawed > Siraj > Aftab

Hence, the correct answer is "Azhar".

The logic followed here is, 

(First Number) - (√Second Number) = (Third Number)

(64, 100, 54):

Using the logic,

64 - √100 = 54, 54 = 54

LHS = RHS.

Similarly,

(84, 81, 75):

Using the logic,

84 - √81 = 75, 75 = 75

LHS = RHS.

Now, let's check each option,

1) (70, 49, 60):

Using the logic,

70 - √49 = 60, 63 = 60

LHS ≠ RHS.

2) (125, 64, 117):

Using the logic,

125 - √64 = 117, 117 = 117

LHS = RHS.

3) (51, 36, 43):

Using the logic,

51 - √36 = 43, 45 = 43

LHS ≠ RHS.

4) (12, 4, 9):

Using the logic,

12 - √4 = 9, 10 = 9

LHS ≠ RHS.

Here, we can see that Option 2 satisfies the logic.

Hence, the correct answer is " (125, 64, 117) ".

Given:

The average age of 25 girls in a class is 11.2 years

The remaining 15 girls is 10 years

Formula Used:

Average = sum of AgesTotal number of girls$\frac{sumofAges}{Totalnumberofgirls}$

Calculations:

The average age of 25 girls in a class is 11.2 years

⇒ Sum of ages = Average x Total number of girls

= 11.2 x 25 = 280

Now, The remaining 15 girls are 10 years

⇒ Sum of ages = Average x Total number of girls

= 10 x 15 = 150

The total sum of ages of all 40 girls 

⇒ Sum of ages = 280 + 150 = 430

⇒ Average = sum of AgesTotal number of girls=43040$\frac{sumofAges}{Totalnumberofgirls}=\frac{430}{40}$ = 10.75 years

⇒ Hence, The average age of all 40 girls is 10.75 years

Given:

A survey is done on 100 candidates,

The number of candidates like cricket = 55

The number of candidates like football = 75

Concept used:

Probability of happening an event = favourable outcomeTotal outcome$\frac{\text{favourable outcome}}{\text{Total outcome}}$

Calculation:

According to the question,

In the above figure,

a = number of candidates who likes to watch only cricket 

c = number of candidates who likes to watch only football 

b = number of candidates who likes to watch both football and cricket 

Now,

The number of candidates like to watch cricket = a + b = 55

The number of candidates like to watch football = b + c = 75

The total number of candidates = a + b + c = 100

Now,

⇒ a + b + c = 100

⇒ 55 + c = 100

⇒ c = 45

Now,

⇒ b + c = 75

⇒ b + 45 = 75

⇒ b = 30

The number of candidates who likes to watch both cricket and football = 30

Now,

The required probability = The number of candidates like both gameTotal number of candidate=30100$\frac{\text{The number of candidates like both game}}{\text{Total number of candidate}}=\frac{30}{100}$ = 0.3

Therefore, '0.3' is the required answer.

According to information given:

1) Raghav starts from Point A and drives 10 km towards the north.

2) He then takes a right turn, drives 4 km, turns right and drives 13 km.

3) He then takes a right turn and drives 7 km.

4) He takes a final right turn, drives 3 km and stops at Point P.

In the figure, we can clearly see that Point A is 3 km towards the East from point P.

Hence, the correct answer is "3 km towards the east".

Given:

The three types of milk are 403 gallons, 465 gallons, and 496 gallons

Concept used:

HCF or the highest common factor of two numbers is the highest factor that can divide the two numbers, evenly.

Calculation:

The least possible number of bottles of equal size in which different milk of different qualities can be filled without mixing will be given by the HCF of 403, 465, and 496

Prime Factors of 403, 465, and 496 will be 

403 = 13 × 31

465 = 3 × 5 × 31

496 = 2⁴ ×  31

∴ The HCF of 403, 465, and 496 = 31

Hence, the number of bottles required would be 

⇒ 403/31 + 465/31 + 496/31 = 13 + 15 + 16 = 44 bottles

∴ The number of required bottles is 44.

Alternate MethodThe HCF of two or more numbers can also be written as the difference between those numbers or a factor of that difference 

The HCF of 403, 465, and 496 can be obtained by identifying the difference between these numbers

 465 - 403 = 62 = 31 × 2

496 - 465 = 31

As 31 is a prime number and it has no further factors.

∴ The HCF of 403, 465, and 496 will be 31

Hence, the number of bottles required would be 

⇒ 403/31 + 465/31 + 496/31 = 13 + 15 + 16 = 44 bottles

∴ The number of required bottles is 44.

For this question, We have to check all the options.

Option (1):  All letters are different.

Option (2):  All letters are different.

Option (3):  All letters are NOT different There is 2 E's present. 

Option (4):  All letters are different.

Hence, Option (3) is correct.

The table shows the place value of the alphabet from A to Z.

The logic followed here is:

TRIPPLE is coded as DMOQHSS

Similarly, VICTORY coded as

Therefore, VICTORY is coded as 'XSNUBJU'.

Hence, the correct answer is "Option (4)".

Concept used:

Min value of a quadratic equation ax² + bx + c is c – (b²/4a) or –D/4a, where D = b² – 4ac

Calculation:

Method 1

Min value = c – b²/4a

⇒ Min value = 6 – [6²/(4 × 3)]

∴ Min value = 3

Method 2

3x² + 6x + 6 = 3 × (x² + 2x + 2)

⇒ 3 × (x² + 2x + 1 + 1)

⇒ 3 × ((x + 1)² + 1)

Minimum value of the expression is at x = −1

⇒ Min value = 3 × (0 + 1)

Given :

CD || EF || AB , in the figure

Concept used :

The sum of adjacent angles between two parallel lines is 180°

Calculation :

In quadrilaterals CEFD,

∠E+∠F+∠ECD+∠D=360$\angle E+\angle F+\angle ECD+\angle D=360$  

∠ECD=360−(140+2y+x)$\angle ECD=360-(140+2y+x)$

∠ECD=220−2y−x$\angle ECD=220-2y-x$

As CD || AB,

∠ABC=∠BCD=70$\angle ABC=\angle BCD=70$°

∠BCD = ∠ECD + ∠BCE = 70° 

⇒ 220 - 2y - x + y = 70°  

⇒ x + y = 150°    -----(1)

As  EF || CD,

So, 2y + x = 180°  -----(2)

On solving (1) and (2), we get

x = 120° and y = 30°

∴ The answer is 120° .

Concept used:

Linear pair is a pair of adjacent angles whose non-common arms form a straight line.

Adjacent angles are those angles that have a common vertex, and a common arm and the remaining two sides are on the adjacent side of the common arm.

Solution:

'a' and 'b' are in a straight line, therefore forming a linear pair.

'a' and 'b' form a linear pair, therefore, they are adjacent angles.

Hence, the correct option is 1.

Given:
Area of the square formed by wire = 121 cm2.

Breadth of rectangle formed = 8 cm.

Formula Used:

Area of Square = side × side

Perimeter of rectangle = 2 × (length + breadth)

Perimeter of square  = 4 × Side 

Calculations:

Area of square = Side × Side  = 121 cm2.

⇒ Side of square = 11 cm

Also, Perimeter of square  = 4 × Side 

So the length of steel wire = 4 × 11 = 44 cm

If the same wire is bent in the form of a rectangle whose breadth is 8 cm.

Then the length of wire will be equal to perimeter of the rectangle.

⇒ 2 × (length + 8) = 44

⇒ (length + 8) = 44/2 = 22

⇒ Length = 22 - 8 = 14 cm 

Hence the length of rectangle is 14 cm.

Solution:

Number of students in University 1 in the year 2017 is 20,000.

Number of students in University 2 in the year 2019 is 20,000.

Hence, the difference between both the university is 0.

Explanation:-

Let's count the squares in the given figure:
There are 9 squares as numbered in the figure.

There are 4 more squares as shown in the figures,

And
 

So there are total 14 squares in the given figure.

Hence the correct answer is option 1.

Given:

Hence, the correct answer is "Option 2".

Concept:

• Michaelis-Menten kinetics are characterized by the assumption that the enzyme and substrate participate in an equilibrium with the enzyme-substrate complex which is not disturbed by product formation during the period that the initial rate of reaction is measured.

Explanation:

• When measuring the initial rate of a reaction, Michaelis-Menten kinetics assumes that the enzyme and substrate are at an equilibrium and that the enzyme-substrate complex is not disrupted by the formation of products.

​Assumption 1:

• There is no product present at the start of the kinetic analysis.
• Therefore, as long as we monitor initial reaction rates we can ignore the reverse reaction of E+P going to ES

​Assumption 2:

• During the reaction an equilibrium condition is established for the binding and dissociation of the Enzyme and Substrate (Briggs-Haldane assumption).
• Thus, the rate of formation of the ES complex is equal to the rate of dissociation plus breakdown.

​Assumption 3:

• [E] << [S]
• The enzyme is a catalyst, it is not destroyed and can be recycled, thus, only small amounts are required
• The amount of S bound to E at any given moment is small compared to the amount of free S.
• It follows that [ES] << [S] and therefore [S] is constant during the course of the analysis (NOTE: this assumption requires that the reaction is monitored for a short period, so that not much S is consumed and [S] does not effectively change - see next assumption).

​​​Assumption 4:

• Only the initial velocity of the reaction is measured.
• [P] = 0 (reverse E + P reaction can be ignored).
• [S] » [S]initial

Assumption 5:

• The enzyme is either present as free enzyme or as the ES complex.
• [E]total = [E] + [ES]

Hence the correct answer is option 2

Concept:

• Mitochondria occupy a substantial portion of the cytoplasmic volume of eucaryotic cells, and they have been essential for the evolution of complex animals.
• Without mitochondria, present-day animal cells would be dependent on anaerobic glycolysis for all of their ATP.
• When glucose is converted to pyruvate by glycolysis, only a very small fraction of the total free energy potentially available from the glucose is released.
• In mitochondria, the metabolism of sugars is completed: the pyruvate is imported into the mitochondrion and oxidized by O2 to CO2 and H2O.
• This allows 15 times more ATP to be made than that produced by glycolysis alone.

Explanation:

• Mitochondria can use both pyruvate and fatty acids as fuel.
• Pyruvate comes from glucose and other sugars, whereas fatty acids come from fats.
• Both of these fuel molecules are transported across the inner mitochondrial membrane and then converted to the crucial metabolic intermediate acetyl CoA by enzymes located in the mitochondrial matrix.
• The acetyl groups in acetyl CoA are then oxidized in the matrix via the citric acid cycle.
• The cycle converts the carbon atoms in acetyl CoA to CO2, which is released from the cell as a waste products.
• Most importantly, the cycle generates high-energy electrons, carried by the activated carrier molecules NADH and FADH2.
• These high-energy electrons are then transferred to the inner mitochondrial membrane, where they enter the electron-transport chain
•  The loss of electrons from NADH and FADH2 also regenerates the NAD+ and FAD that is needed for continued oxidative metabolism also termed as oxidative phosphorylation.

Concept:

• The stability of the DNA double helix is mostly a result of base pairing between complementary strands and stacking between adjacent nucleotides.

Explanation:

• Stability of the DNA double helix with respect to the separation of complementary strands is known to depend on the base composition of the duplex.
• A classical study of Marmur and Doty on DNA polymer stability gives a linear relationship between the G•C content of the polymer and its melting temperature.
• Accordingly, base-stacking interactions have been thought to constitute only a small correction to the major effect of the differences in G•C and A•T pair stabilities.
• Partitioning of base pairing and stacking contributions to DNA stability not only delivers a new aspect in the fundamental understanding of DNA structure and energetics, but also it has significant implications in a number of biological processes.
• Fluctuations in local helical conformation of DNA, the phenomenon known as DNA breathing, lead to infrequent events of base pair opening thus making normally buried groups available for modification and interaction with proteins.

Hence the correct answer is option 3

Concept:

• Single-headed molecules with different length tails are from the myosin I subfamily.
• Actin-actin sliding, the development of filopodia in neuronal growth cones, the movement of membranes during endocytosis, and the attachment of actin to membranes as observed in microvilli are all functions of myosin I.

​

​Explanation:

• The family of microfilaments known as myosin is frequently grouped alongside other motor proteins.
• A heavy and a light chain make up the globular head region of myosin proteins.
• The heavy chain has a variable length -helical tail.
• The basis for myosin's ability to function as a motor protein is the head, which has an ATPase activity and can attach to and travel along actin filaments.
• Myosin II, which can be found in both muscle and many non-muscle cells, is the most well-known class.
• Two heads and two tails on each of its molecules are linked to form a long rod.
• As seen in striated and smooth muscle fibres as well as myoepithelial cells, the rods can attach to one another to create lengthy, dense filaments.

​

Concept:

• Numerous sorts of cell movements are carried out by actin filaments, frequently in conjunction with myosin.
• The first molecular motor, myosin, is a protein that transforms chemical energy in the form of ATP into mechanical energy to produce force and movement.

Explanation:

• Muscle contraction is the most notable example of such movement, and it serves as a model for studying actin-myosin interactions and the motor activity of myosin molecules.
• Actin and myosin interactions, however, play a crucial function in cell biology as they are responsible for a variety of movements of non-muscle cells, including cell division, in addition to muscle contraction.
• Furthermore, actin-myosin interactions and actin polymerization appear to be the primary drivers of the crawling movements of cells across a surface, which are regulated by the actin cytoskeleton.​

Fig 1: General structure of myosin

HMM: Heavy meromyosin 

LMM: Light meromyosin

 S1 refers to the motor domain plus the lever arm, which binds the essential and regulatory light chains (ELC and RLC respectively)

​Hence correct answer is option 1

Concept:

• DNA components called scaffold/matrix attachment regions (S/MARs) are used to divide chromatin into structural and functional domains.
• These components play a part in disease biology as well as the regulation of gene expression, which determines phenotype.
• Therefore, understanding these components at the genomic scale offers significant therapeutic potential.
• There have been numerous attempts to identify S/MARs in the genomes of different creatures, including the human.
• But there isn't yet a complete genome-wide atlas of human S/MARs.​

​Explanation:

• A three-dimensional filamentous RNA-protein meshwork called the nuclear matrix serves as the structural backbone for the orderly compaction of DNA .
• Due to DNA sequences that connect the chromatin to the nuclear matrix, the chromatin is arranged into loops.
• The term "scaffold/matrix attachment regions" (S/MARs) refers to these anchor sequences.
• S/MARs are known to interact with a variety of proteins, known as S/MAR binding proteins (S/MARBPs), to promote chromatin looping.
• Numerous biological functions, including DNA replication, transcription, chromatin to chromosomal transition, and DNA repair, have been shown to depend on such DNA looping.
• Interestingly, there is no sequence conservation in the S/MARs that attach these loops to the nuclear matrix.
• Nevertheless, characteristics of its secondary structure seem to remain preserved and functionally significant.

​Fig 1: MARs

  Option 1: SPR ( scaffold protein region)

• Consider the explanation above thus this option is not true

Option 2: MAR (matrix associated region)

•    Consider the explanation above thus this option is  true.

  Option 3: NAR (nuclear associated regions ) 

•    Consider the explanation above thus this option is not true

  Option 4: HAR (histone associated regions)

•    Consider the explanation above thus this option is not true​

​​ hence the correct answer is option 1

Concept:

• Internal ribosome entry sites (IRESs) are sequences within the RNA molecule that allow binding of ribosomes without scanning.
• They  play an important part in allowing certain animal viruses, notably the picornaviruses, to evade the constraints of ribosomal scanning.

Explanation:

Fig 1: IRES mediated translation

• Internal ribosomal entry sites (IRESs) are regions in the mRNAs that allow the internal initiation of translation.
• Indeed, even though the vast majority of cellular mRNAs are translated by the cap-dependent mechanisms.
•  It is now known that, in certain cases, translation can also occur in a cap-independent manner, thanks to specific mRNA sequences able to recruit the 40S ribosomal subunit in the vicinity of an AUG starting codon.
• The first IRES was identified in the genome of poliovirus (PV) and encephalomyocarditis virus (EMCV), and later in many other viruses.
• During infection, the cap-dependent translation of the cell is impaired, and the viral RNAs use the internal initiation for their gene expression.

​hence the correct answer is option 3

​

Concept:

• Selenocysteine (Sec) is not one of the standard 20 amino acids and yet it is incorporated into a few rare proteins during translation of the mRNA by the ribosome.
• This occurs both in bacteria and in eukaryotes, including humans. 

​Explanation:

Concept:

• Reverse transcriptase (RT) is a multifunctional enzyme that has RNA- and DNA-dependent DNA polymerase activity and ribonuclease H (RNase H) activity, and is responsible for the reverse transcription of retroviral single-stranded RNA into double-stranded DNA.
• The essential role that RT plays in the human immunodeficiency virus (HIV) life cycle is highlighted by the fact that multiple antiviral drugs—which can be classified into two distinct therapeutic classes—are routinely used to treat and/or prevent HIV infection.

Explanation:

Fig 1:  Reverse transcriptase mechanism of action

• The RNase H or ribonuclease activity of reverse transcriptase acts as an endonuclease that hydrolyzes the RNA strand in an RNA/DNA hybrid to generate 5′ phosphate and 3′ hydroxyl ends.
  hence the correct answer is option 2

Concept:

• Noncanonical open reading frames (ORFs) that

                 (i) regulate translational control

                 (ii) create functional microproteins, and/or

                 (iii) encode protein variations of the canonical coding sequence have been highlighted by recent advancements in tools to investigate the translatome and proteome.

Explanation:

• There are accumulating examples within the literature of noncanonical ORF-encoded proteins that possess important functional and physiological roles, but there lie thousands of putative ORFs within ‘omics’ data-bases that are left unverified and functionally uncharacterized.
• A key question in genome research is whether biologically active proteins are restricted to the ~20,000 canonical, well-annotated genes, or rather extend to the many non-canonical open reading frames (ORFs) predicted by genomic analyses.
• Non-canonical open reading frames encode functional proteins essential for cancer cell survival have been deduced in recent studies.

​