CUET 2026 Biology May 30 Shift 2 Question Paper- Download Answer Key and Solution PDF

Step 1: Recall the concept of RNA interference.
RNA interference (RNAi) is a biological process in which a specific mRNA molecule is silenced by complementary RNA molecules.
Step 2: Understand its mechanism.
Double-stranded RNA (dsRNA) produces small interfering RNAs (siRNA), which bind to the target mRNA.
Step 3: Determine the effect.
Binding of siRNA leads to degradation of mRNA or prevents its translation.
Hence,
\[ \boxed{Translation\ of\ mRNA\ is\ inhibited} \]
Therefore,
\[ \boxed{(B)} \] is the correct answer. Quick Tip: RNAi is used in biotechnology to silence specific genes. Example: \[ \boxed{RNAi\ in\ nematode-resistant\ tobacco\ plants} \] Remember: \[ \boxed{RNAi \Rightarrow Gene\ Silencing} \]

Step 1: Recall the structural genes of the Lac Operon.
The Lac Operon contains three structural genes:
\[ lacZ,\; lacY,\; lacA \]
Step 2: Recall their functions.
\[ lacZ \rightarrow \beta-galactosidase \]
\[ lacY \rightarrow Permease \]
\[ lacA \rightarrow Transacetylase \]
Step 3: Match the columns.
\[ A \rightarrow III \]
\[ B \rightarrow I \]
\[ C \rightarrow II \]
Thus,
\[ \boxed{A\!-\!III,\;B\!-\!I,\;C\!-\!II} \]
Therefore,
\[ \boxed{(B)} \] is the correct answer. Quick Tip: Lac Operon Gene Functions: \[ \boxed{lacZ \rightarrow \beta-galactosidase} \] \[ \boxed{lacY \rightarrow Permease} \] \[ \boxed{lacA \rightarrow Transacetylase} \] A very common CUET/NEET molecular genetics question.

Step 1: Recall the male reproductive hormones.
The principal male sex hormone is:
\[ \boxed{Testosterone} \] which belongs to the group of hormones called androgens.
Step 2: Identify the secreting cells.
The interstitial cells present between seminiferous tubules are called:
\[ \boxed{Leydig\ Cells} \]
These cells secrete testosterone and other androgens.
Step 3: Evaluate the options:
Sertoli cells \(\rightarrow\) Nourish sperms \(\times\)
Leydig cells \(\rightarrow\) Secrete androgens \(\checkmark\)
Spermatogonia \(\rightarrow\) Germ cells \(\times\)
Follicular cells \(\rightarrow\) Female reproductive system \(\times\)
Therefore,
\[ \boxed{Leydig\ Cells} \] are responsible for androgen secretion.
Hence,
\[ \boxed{(B)} \] is the correct answer. Quick Tip: Remember: \[ \boxed{Leydig\ Cells \rightarrow Testosterone} \] \[ \boxed{Sertoli\ Cells \rightarrow Nourishment\ of\ Sperms} \] This distinction is frequently asked in CUET Biology.

Step 1: Recall the definition of a test cross.
A test cross is used to determine the genotype of an individual showing a dominant phenotype.
Step 2: Identify the crossing partner.
The individual of unknown genotype is crossed with:
\[ \boxed{Homozygous\ Recessive\ Parent} \]
Step 3: Understand its purpose.
If the unknown individual is heterozygous, both dominant and recessive phenotypes appear in the offspring.
If it is homozygous dominant, all offspring show the dominant phenotype.
Thus,
\[ \boxed{Test\ Cross\ =\ F_1 \times Recessive\ Parent} \]
Therefore,
\[ \boxed{(C)} \] is the correct answer. Quick Tip: Important Crosses: \[ \boxed{Test\ Cross\ =\ Unknown \times Homozygous\ Recessive} \] \[ \boxed{Back Cross = F_1 \times Either\ Parent} \] Every test cross is a back cross, but every back cross is not a test cross.

Step 1: Recall the purpose of PCR.
PCR is a technique used to amplify a specific DNA segment and produce millions of copies.
Step 2: Identify the first step.
The DNA strands are first separated by heating.
\[ \boxed{Denaturation} \]
Step 3: Identify the second step.
Primers bind to complementary DNA sequences.
\[ \boxed{Annealing} \]
Step 4: Identify the final step.
Taq DNA polymerase synthesizes new DNA strands.
\[ \boxed{Extension} \]
Hence the correct sequence is:
\[ \boxed{Denaturation \rightarrow Annealing \rightarrow Extension} \]
\[ \boxed{2 \rightarrow 1 \rightarrow 3} \]
Therefore,
\[ \boxed{(B)} \] is the correct answer. Quick Tip: PCR Steps: \[ \boxed{D \rightarrow A \rightarrow E} \] \[ \boxed{Denaturation \rightarrow Annealing \rightarrow Extension} \] Remember: \[ \boxed{PCR\ uses\ Taq\ DNA\ Polymerase} \]

Step 1: Recall what Copper-T is.
Copper-T is an intrauterine contraceptive device (IUD) placed inside the uterus.
Step 2: Understand the role of copper ions.
Copper ions released from the device:
Increase phagocytosis of sperms
Reduce sperm motility
Reduce sperm fertilizing capacity
Step 3: Determine the correct mechanism.
Thus Copper-T mainly prevents fertilization by adversely affecting sperm activity.
\[ \boxed{Kills\ spermatozoa\ and\ reduces\ sperm\ motility} \]
Hence,
\[ \boxed{(B)} \] is the correct answer. Quick Tip: Examples of IUDs: \[ \boxed{Copper-T} \] \[ \boxed{Cu-7} \] \[ \boxed{Multiload-375} \] Copper ions: \[ \boxed{Decrease\ sperm\ motility\ and\ fertilizing\ capacity} \] A very important NCERT fact for CUET Biology.

Step 1: Recall the interaction between orchid and mango tree.
An orchid grows on the branches of a mango tree to obtain support and better access to sunlight.
Step 2: Determine the effect on both organisms.
\[ Orchid\ benefits \]
\[ Mango\ tree\ neither\ benefits\ nor\ is\ harmed \]
Step 3: Identify the ecological interaction.
When one species benefits and the other remains unaffected, the interaction is called:
\[ \boxed{Commensalism} \]
Therefore,
\[ \boxed{(C)} \] is the correct answer. Quick Tip: Ecological Interactions: \[ (+,+) \rightarrow Mutualism \] \[ (+,0) \rightarrow Commensalism \] \[ (+,-) \rightarrow Parasitism / Predation \] NCERT Example: \[ \boxed{Orchid\ +\ Mango\ Tree\ =\ Commensalism} \]

Step 1: Recall the population growth equation.
For exponential population growth,
\[ \frac{dN}{dt}=rN \]
where \(N\) is the population size.
Step 2: Interpret the term \(r\).
The parameter \(r\) represents the growth potential of a population under ideal environmental conditions.
\[ r = Intrinsic\ Rate\ of\ Natural\ Increase \]
Step 3: Identify the correct option.
Thus,
\[ \boxed{r=Intrinsic\ Rate\ of\ Natural\ Increase} \]
Therefore,
\[ \boxed{(C)} \] is the correct answer. Quick Tip: Population Growth Equations: Exponential Growth: \[ \boxed{\frac{dN}{dt}=rN} \] Logistic Growth: \[ \boxed{\frac{dN}{dt}=rN\left(\frac{K-N}{K}\right)} \] where \[ K=Carrying\ Capacity \] and \[ r=Intrinsic\ Rate\ of\ Natural\ Increase \]

Step 1: Recall the definition of a food chain.
A food chain represents the transfer of energy from one trophic level to another.
Step 2: Identify the trophic levels.
\[ Producer \]
\[ \downarrow \]
\[ Primary\ Consumer \]
\[ \downarrow \]
\[ Secondary\ Consumer \]
\[ \downarrow \]
\[ Tertiary\ Consumer \]
Step 3: Match with the numbering.
\[ 2 \rightarrow 3 \rightarrow 1 \rightarrow 4 \]
Hence,
\[ \boxed{2 \rightarrow 3 \rightarrow 1 \rightarrow 4} \]
Therefore,
\[ \boxed{(A)} \] is the correct answer. Quick Tip: Typical Food Chain: \[ \boxed{Grass \rightarrow Rabbit \rightarrow Fox \rightarrow Tiger} \] Producer: \[ \boxed{Grass} \] Primary Consumer: \[ \boxed{Rabbit} \] Secondary Consumer: \[ \boxed{Fox} \]

Step 1: Recall Ex-situ Conservation.
Conservation carried out outside the natural habitat is called:
\[ \boxed{Ex-situ\ Conservation} \]
Examples include botanical gardens, zoological parks and gene banks.
Step 2: Recall In-situ Conservation.
Conservation within the natural habitat is called:
\[ \boxed{In-situ\ Conservation} \]
Example:
\[ \boxed{Wildlife\ Sanctuary} \]
Step 3: Recall Gene Banks.
Gene banks preserve genetic material through:
\[ \boxed{Cryopreservation} \]
Step 4: Match the columns.
\[ A \rightarrow IV \]
\[ B \rightarrow II \]
\[ C \rightarrow III \]
\[ D \rightarrow IV \]
Thus,
\[ \boxed{A-IV,\;B-II,\;C-III,\;D-IV} \]
Therefore,
\[ \boxed{(C)} \] is the correct answer. Quick Tip: Conservation Methods: \[ \boxed{In-situ \rightarrow National\ Parks,\ Sanctuaries,\ Biosphere\ Reserves} \] \[ \boxed{Ex-situ \rightarrow Zoos,\ Botanical\ Gardens,\ Gene\ Banks} \] Remember: \[ \boxed{Cryopreservation \rightarrow Gene\ Banks} \]