GSEB 2022 Biology Question Paper with Answer and Solution

(C) The correct answer is $C$. Human liver flukes (such as $Fasciola \ hepatica$ or $Clonorchis \ sinensis$) belong to the class $Trematoda$ within the phylum $Platyhelminthes$. $Plasmodium$ is a protozoan,$Nematode$ is a separate phylum (roundworms),and $Wuchereria$ is a nematode parasite.

(C) The scutellum is a modified,shield-shaped cotyledon found in the embryos of monocotyledonous seeds.
Among the given options,$A$ (Beans),$B$ (Black gram),and $D$ (Green gram) are dicotyledonous seeds,which possess two cotyledons.
$C$ (Maize) is a monocotyledonous seed,which contains a single large cotyledon known as the scutellum.
Therefore,the correct option is $C$.

(B) Currently,there are $36$ recognized biodiversity hotspots in the world. However,in the context of standard $NCERT$ textbooks,the number $34$ is often cited as the recognized count. $A$ biodiversity hotspot is a biogeographic region with a significant reservoir of biodiversity that is under threat from humans. To qualify as a biodiversity hotspot,a region must meet two strict criteria: it must contain at least $1,500$ species of vascular plants as endemics,and it must have lost at least $70$ percent of its primary native vegetation.

(D) The correct answer is $D$.
Insects are the most species-rich taxonomic group on the planet, representing more than $70\%$ of all recorded animal species.
According to global biodiversity estimates, for every $10$ animals on this planet, $7$ are insects.
Therefore, among invertebrates, insects are the most abundant class.

(B) bioreactor is a vessel in which raw materials are biologically converted into specific products by microbes,plant/animal cells,or their enzymes.
Standard bioreactors are equipped with essential components to maintain optimal growth conditions,including:
$1$. An agitator system for mixing.
$2$. An oxygen delivery system for aeration.
$3$. $A$ foam control system to manage foaming.
$4$. $A$ temperature control system.
$5$. $A$ pH control system.
$6$. Sampling ports for periodic withdrawal of the culture.
Light distribution systems are typically required for photobioreactors (used for photosynthetic organisms like algae),but they are not standard components of a general-purpose industrial bioreactor.

(A) The Polymerase Chain Reaction $(PCR)$ consists of three main steps:
$1$. Denaturation: The double-stranded $DNA$ is heated to separate it into two single strands.
$2$. Annealing: The temperature is lowered to allow the primers to bind (join) to their complementary sequences on the single-stranded template $DNA$.
$3$. Extension (Elongation): The $Taq$ polymerase enzyme synthesizes the new $DNA$ strand by adding nucleotides to the primer.
Therefore,the stage where the template joins with the primer is Annealing.

(A) palindromic $DNA$ sequence is a sequence of base pairs that reads the same on the two strands when the orientation of reading is kept the same (e.g.,$5^{\prime} \rightarrow 3^{\prime}$).
$1$. Option $A$: $5^{\prime}$-$GTCGTC$-$3^{\prime}$ and its complement $3^{\prime}$-$CAGCAG$-$5^{\prime}$. Reading the complement in $5^{\prime} \rightarrow 3^{\prime}$ direction gives $5^{\prime}$-$GACGAC$-$3^{\prime}$,which is not the same as $5^{\prime}$-$GTCGTC$-$3^{\prime}$. Thus,it is not palindromic.
$2$. Option $B$: $5^{\prime}$-$GAATTC$-$3^{\prime}$ and $3^{\prime}$-$CTTAAG$-$5^{\prime}$. Reading the complement in $5^{\prime} \rightarrow 3^{\prime}$ gives $5^{\prime}$-$GAATTC$-$3^{\prime}$,which is palindromic.
$3$. Option $C$: $5^{\prime}$-$AAATTT$-$3^{\prime}$ and $3^{\prime}$-$TTTAAA$-$5^{\prime}$. Reading the complement in $5^{\prime} \rightarrow 3^{\prime}$ gives $5^{\prime}$-$AAATTT$-$3^{\prime}$,which is palindromic.
$4$. Option $D$: $5^{\prime}$-$GACCTC$-$3^{\prime}$ and $3^{\prime}$-$CTGGAG$-$5^{\prime}$. Reading the complement in $5^{\prime} \rightarrow 3^{\prime}$ gives $5^{\prime}$-$GACCTC$-$3^{\prime}$,which is palindromic.
Therefore,option $A$ is the correct answer.

(A) In the plasmid vector $pBR322$,the restriction enzyme recognition site for $BamHI$ is located within the tetracycline resistance gene $(tet^R)$. This is significant because the insertion of foreign $DNA$ at this site inactivates the tetracycline resistance gene,allowing for the selection of recombinant colonies through insertional inactivation.

(B) In an aquatic ecosystem,phytoplankton are the primary producers that synthesize food through photosynthesis. Zooplanktons feed on these phytoplankton. Since they consume the primary producers,zooplanktons are classified as primary consumers,occupying the second trophic level $(T_2)$.

(C) In a $Marine$ ecosystem,the pyramid of biomass is inverted. This is because the biomass of phytoplankton (producers) is much lower than that of the zooplankton and small fish (primary consumers) and large fish (secondary consumers) that feed on them. The rapid turnover rate of phytoplankton allows a small standing crop to support a larger biomass of consumers.

(D) Divergent evolution occurs when related species evolve different traits due to different environmental pressures,leading to homologous structures.
$A$. Flippers of penguins and dolphins are analogous structures (convergent evolution).
$B$. Eye of octopus and mammals are analogous structures (convergent evolution).
$C$. Sweet potato (root modification) and potato (stem modification) are analogous structures (convergent evolution).
$D$. Hearts and brains of vertebrates share a common ancestral plan but have evolved into different forms in various groups,representing homologous structures which are a result of divergent evolution.

(C) In the Hardy-Weinberg principle,the equation is given by $(p + q)^2 = p^2 + 2pq + q^2 = 1$.
Here,$p$ represents the frequency of the dominant allele and $q$ represents the frequency of the recessive allele.
$p^2$ represents the frequency of homozygous dominant individuals.
$q^2$ represents the frequency of homozygous recessive individuals.
$2pq$ represents the frequency of heterozygous individuals.
Therefore,the frequency of the heterozygous gene is represented by $2pq$.

(B) The brain capacity of $Homo$ $erectus$ is estimated to be approximately $900$ $C.C.$ (cubic centimeters).
$Homo$ $habilis$ had a brain capacity between $650 - 800$ $C.C.$,while $Homo$ $sapiens$ $neanderthalensis$ had a brain capacity of about $1400$ $C.C.$

(D) In $1953$,$S$. $L$. Miller conducted an experiment to test the Oparin-Haldane hypothesis regarding the origin of life.
He created a closed system using a flask containing a mixture of gases: methane $(CH_4)$,ammonia $(NH_3)$,hydrogen $(H_2)$,and water vapor $(H_2O)$.
He provided energy for the reaction by using electric discharges in the flask,simulating lightning.
After a week,he observed the formation of amino acids in the mixture,supporting the theory of chemical evolution.

(D) In the human body, primary lymphoid organs are the sites where lymphocytes originate and/or mature and become antigen-sensitive.
These organs include the $Bone marrow$ and the $Thymus gland$.
In the $Bone marrow$, all blood cells including lymphocytes are produced.
In the $Thymus gland$, $T$-lymphocytes mature.
Therefore, both $Bone marrow$ and $Thymus gland$ are primary lymphoid organs.

(A) The diagnostic test for typhoid fever is the $Widal$ test. It is a serological test that detects the presence of antibodies against the bacterium $Salmonella$ $typhi$ in the patient's blood serum. The test works on the principle of agglutination,where the antibodies in the serum react with the $Salmonella$ antigens.

(A) Innate immunity consists of four types of barriers: physical,physiological,cellular,and cytokine barriers.
$1$. Physical barriers include skin and mucus coatings.
$2$. Physiological barriers include acid in the stomach,saliva in the mouth,and tears from eyes.
$3$. Cellular barriers include leukocytes like polymorphonuclear leukocytes ($PMNL$-neutrophils),monocytes,and natural killer cells.
$4$. Cytokine barriers involve virus-infected cells that secrete proteins called interferons,which protect non-infected cells from further viral infection.
Therefore,the correct answer is $A$.

(B) In humans,fertilisation is the process of fusion of a sperm with an ovum. This process occurs in the $Ampullary$ region of the fallopian tube. The ovum released by the ovary is transported to the $Ampullary$ region where it meets the sperm. Fertilisation can only occur if the ovum and sperms are transported simultaneously to the $Ampullary$ region. Therefore,the correct option is $B$.

(B) The placenta acts as an endocrine tissue and produces several hormones essential for pregnancy,such as human Chorionic Gonadotropin $(hCG)$,human Placental Lactogen $(hPL)$,estrogens,and progestogens.
Relaxin is also produced during the later stages of pregnancy,but it is primarily secreted by the ovary (corpus luteum) and not the placenta.
Therefore,the correct option is $B$.

(C) The process of cleavage in the zygote leads to the formation of a $16$-celled stage called the $Morula$. The $Morula$ continues to divide and transforms into the $Blastocyst$. It is the $Blastocyst$ that gets embedded or implanted into the endometrium of the uterus to initiate pregnancy.

(B) $Trichoderma$ species are free-living fungi that are very common in the root ecosystems. They are effective biocontrol agents of several plant pathogens. They act as antagonists to various soil-borne plant pathogens,thereby helping in the treatment and prevention of plant diseases. Therefore,option $B$ is the correct answer.

(A) In a Sewage Treatment Plant $(STP)$, the primary treatment involves the physical removal of particles through filtration and sedimentation.
The sediment that settles at the bottom during this process is known as primary sludge.
The supernatant liquid that remains above the primary sludge is called the primary effluent.
This primary effluent is then taken for secondary treatment (biological treatment), where it is passed into large aeration tanks.
Therefore, the primary effluent is the key substance that acts as the link between primary and secondary treatment.

(C) The correct matching is as follows:
$1$. $Streptococcus$ $pneumoniae$ produces the enzyme $streptokinase$, which is used as a 'clot buster' for removing clots from the blood vessels of patients who have undergone myocardial infarction.
$2$. $Aspergillus$ $niger$ is a fungus used for the commercial production of $citric$ $acid$.
$3$. $Clostridium$ $butyricum$ is a bacterium used for the production of $butyric$ $acid$.
$4$. $Acetobacter$ $aceti$ is a bacterium used for the production of $acetic$ $acid$.
Therefore, the correct sequence is $1-C, 2-D, 3-B, 4-A$.

(A) The total length of $DNA$ in $E. coli$ is calculated by multiplying the total number of base pairs by the distance between two consecutive base pairs.
$E. coli$ has $4.6 \times 10^6$ base pairs.
The distance between two consecutive base pairs is $0.34 \times 10^{-9} \ m$.
Length = $(4.6 \times 10^6) \times (0.34 \times 10^{-9} \ m) \approx 1.36 \times 10^{-3} \ m = 1.36 \ mm$.
Therefore,the correct option is $A$.

(A) $AUG$ is a dual-function codon.
$1$. It codes for the amino acid Methionine $(Met)$.
$2$. It acts as an initiator codon,which signals the start of protein synthesis (translation) on the mRNA strand.
Therefore,the correct option is $A$.

(A) The $DNA$ template strand is $ATG$.
First,the $mRNA$ is synthesized from the $DNA$ template through transcription. According to the base-pairing rules ($A$ pairs with $U$,$T$ pairs with $A$,$G$ pairs with $C$),the $mRNA$ sequence will be $UAC$.
Next,the $t-RNA$ anticodon pairs with the $mRNA$ codon. The $mRNA$ codon is $UAC$.
The anticodon on $t-RNA$ is complementary to the $mRNA$ codon ($U$ pairs with $A$,$A$ pairs with $U$,$C$ pairs with $G$).
Therefore,the anticodon on $t-RNA$ will be $AUG$.

(C) In the process of transcription,the $DNA$ double helix has two strands with opposite polarity ($5' \rightarrow 3'$ and $3' \rightarrow 5'$).
The promoter is located at the $5'$ end (upstream) of the coding strand,which corresponds to the $3'$ end of the template strand.
The terminator is located at the $3'$ end (downstream) of the coding strand,which corresponds to the $5'$ end of the template strand.
In the provided diagram,'$X$' is at the $3'$ end of the coding strand (which is the $5'$ end of the template strand),identifying it as the Terminator.
'$Y$' is at the $5'$ end of the coding strand (which is the $3'$ end of the template strand),identifying it as the Promoter.
Therefore,$X = \text{Terminator}$ and $Y = \text{Promoter}$.

(B) The population density at time $t+1$ is calculated by considering the initial population at time $t$ and adding the net changes due to births $(B)$,immigration $(I)$,deaths $(D)$,and emigration $(E)$.
The formula is given by: $N_{t+1} = N_t + [(B + I) - (D + E)]$.
Here,$(B + I)$ represents the total number of individuals added to the population,and $(D + E)$ represents the total number of individuals lost from the population. Therefore,the correct option is $B$.

(A) When resources are unlimited and there are no environmental resistance factors (growth-limiting factors),the population grows exponentially. This type of growth is represented by a $J$-shaped curve. In contrast,an $S$-shaped (sigmoid) curve represents logistic growth,which occurs when resources are limited and environmental resistance is present.

(D) The condition characterized by the presence of $45$ chromosomes with an $XO$ genotype (where one $X$ chromosome is missing) is known as Turner's syndrome.
In this disorder,the individual is phenotypically female but has underdeveloped ovaries and lacks secondary sexual characteristics.
Down's syndrome involves trisomy of chromosome $21$ ($47$ chromosomes).
Klinefelter's syndrome involves an $XXY$ genotype ($47$ chromosomes).
Therefore,the correct option is $D$.

(A) To determine the mode of inheritance from the pedigree chart:
$1$. Observe that an affected mother (filled circle) and an unaffected father (empty square) produce both affected sons and affected daughters.
$2$. Since the trait appears in every generation and affected individuals have at least one affected parent,it is a dominant trait.
$3$. If it were $X$-linked dominant,an affected father would pass the trait to all his daughters,which is not observed here. Also,an affected mother would pass it to all her sons,but here she has an unaffected son as well.
$4$. The pattern of inheritance where an affected mother passes the trait to both sons and daughters,and the trait does not skip generations,is characteristic of Autosomal Dominant inheritance.
$5$. Therefore,the correct option is $A$.

(B) Alpha-thalassemia is a genetic blood disorder caused by mutations in the genes responsible for the production of the alpha-globin chain of hemoglobin.
The genes $HBA1$ and $HBA2$ are responsible for encoding the alpha-globin protein.
These two genes are located on the short arm of chromosome $16$ $(16p13.3)$.
Therefore,the correct option is $B$.

(D) The provided diagram represents the structure of a dicot embryo. In this structure, the lower pointed portion is the radicle, which is covered by a protective sheath-like structure at the tip known as the root cap. Therefore, "$X$" represents the radicle and "$Y$" represents the root cap. Thus, the correct option is $D$.

(C) In angiosperms,the process of double fertilization involves the fusion of one male gamete $(n)$ with two polar nuclei $(n + n)$ or a secondary nucleus $(2n)$ present in the central cell.
This fusion results in the formation of the Primary Endosperm Nucleus $(PEN)$,which is $3n$ or triploid.
Therefore,the correct option is $C$.

(D) In flowering plants,the life cycle involves an alternation of generations between the haploid gametophyte and the diploid sporophyte.
$A$. Pollen grains are haploid $(n)$ structures produced by meiosis.
$B$. The egg cell is a haploid $(n)$ female gamete.
$C$. The male gamete is a haploid $(n)$ cell.
$D$. The zygote is formed by the fusion of the male and female gametes $(n + n)$,resulting in a diploid $(2n)$ cell.
Therefore,the zygote is the diploid structure.

(A) The residual,persistent nucellus in a seed is known as $Perisperm$.
In some seeds,such as those of black pepper and beet,the nucellus is not completely consumed during the development of the embryo and remains as a nutritive tissue surrounding the embryo.
$Pericarp$ is the wall of the fruit developed from the ovary wall.
$Coleorrhiza$ and $Coleoptile$ are protective sheaths found in monocot embryos.

(D) In angiosperms,the egg cell is a female gamete formed through meiosis,so it is haploid $(n)$.
The endosperm is formed by the fusion of one male gamete $(n)$ with two polar nuclei $(n + n)$,a process known as triple fusion,resulting in a triploid $(3n)$ structure.
Therefore,the egg cell is haploid and the endosperm is triploid.

(D) In many aquatic plants, such as $Vallisneria$ and $Hydrilla$, pollination occurs through water (hydrophily).
However, in several freshwater plants like $Water \text{ } lily$ (Nymphaea) and $Water \text{ } hyacinth$, the flowers emerge above the level of water.
In these cases, pollination is brought about by insects or wind, similar to most of the land plants.
Therefore, $Water \text{ } lily$ is the correct answer.

(A) In the provided diagram of an anatropous ovule:
$X$ points to the central structure within the nucellus,which is the embryo sac (female gametophyte).
$Y$ points to the stalk that attaches the ovule to the placenta,known as the funicle.
Therefore,the correct identification is $X = \text{Embryo sac}$ and $Y = \text{Funicle}$.