KCET 2011 Biology Question Paper with Answer and Solution

(D) The correct answer is $D$.
The ascending order of the Linnaean hierarchy is: $\text{Species} \rightarrow \text{Genus} \rightarrow \text{Family} \rightarrow \text{Order} \rightarrow \text{Class} \rightarrow \text{Phylum} \rightarrow \text{Kingdom}$.
This sequence represents an arrangement from the most specific level (Species) to the most inclusive level (Kingdom),which is the standard hierarchical structure used in biological classification.

(C) The correct answer is $C$. The conversion of pyruvate to acetyl CoA is a classic example of oxidative decarboxylation.
In this reaction,pyruvate $(3C)$ undergoes both oxidation and decarboxylation (loss of $CO_2$) to form acetyl CoA $(2C)$.
Simultaneously,$NAD^+$ is reduced to $NADH + H^+$.
This reaction is catalyzed by the pyruvate dehydrogenase complex and serves as a vital link between glycolysis and the Krebs cycle.

(C) The correct answer is $C$.
During fermentation,whether it is lactic acid fermentation or alcoholic fermentation,only $2$ $ATP$ molecules are produced per molecule of glucose.
This is because fermentation is an anaerobic process,which only includes glycolysis.
Glycolysis results in a net gain of $2$ $ATP$ molecules,as the subsequent steps of fermentation do not produce any additional $ATP$.

(B) $GA$.
The rosette habit of cabbage is characterized by short internodes and a cluster of leaves.
The application of gibberellins $(GA)$ promotes internode elongation,which causes the stem to grow longer and changes the rosette growth form into an elongated habit.
This process is known as bolting.

(B) asphyxia.
Asphyxia is a medical condition characterized by a severe deficiency of oxygen in the body $(hypoxia)$ and an excess of carbon dioxide in the blood $(hypercapnia)$.
This state leads to the cessation of breathing and loss of consciousness.
$Suffocation$ is a general term for oxygen deprivation,$emphysema$ is a chronic lung disease,and $eupnea$ refers to normal,quiet breathing.

(A) The stroke volume $(SV)$ is the volume of blood pumped out by each ventricle during each cardiac cycle.
It is calculated using the formula: $SV = EDV - ESV$.
Given:
End-diastolic volume $(EDV)$ = $120 \ mL$
End-systolic volume $(ESV)$ = $50 \ mL$
Therefore,$SV = 120 \ mL - 50 \ mL = 70 \ mL$.
Thus,the correct option is $A$.

(C) The correct answer is $C$.
In a striated muscle fiber,the parts are labeled as follows:
$A$ represents the Light band ($I$-band).
$B$ represents the Sarcoplasm (cytoplasm of the muscle cell).
$C$ represents the Myofibril (contractile unit).
$D$ represents the Sarcolemma (plasma membrane of the muscle fiber).
$E$ represents the Nucleus (located at the periphery).
$F$ represents the Dark band ($A$-band).

(B) midbrain,pons,medulla oblongata.
The brainstem is composed of three main structures: the midbrain,pons,and medulla oblongata.
The brainstem forms the vital connection between the brain and the spinal cord.

(A) Estrogen.
Steroid hormones are derived from cholesterol. Examples include estrogens,androgens,and glucocorticoids.
Estrogens are involved in regulating secondary sexual characteristics in females,the menstrual cycle,and the overall health of the reproductive system.
Insulin and Glucagon are peptide hormones,while Thyroxine is an amino acid derivative (iodinated tyrosine).

(C) is the correct answer.
The neurohypophysis (posterior pituitary) stores and releases oxytocin and vasopressin.
These hormones are synthesized in the hypothalamus and transported axonally to the neurohypophysis,where they are stored and later released into the bloodstream.

(A) The correct matching is $A-2, B-3, C-4, D-1$.
• Glycogenesis $(A-2)$: The process of conversion of glucose to glycogen for storage in the liver and muscles.
• Glycosuria $(B-3)$: The presence of excess glucose in the urine,often associated with diabetes mellitus.
• Gluconeogenesis $(C-4)$: The synthesis of glucose from non-carbohydrate precursors like amino acids,glycerol,and lactate.
• Glycogenolysis $(D-1)$: The breakdown of stored glycogen into glucose to maintain blood sugar levels.

(B) - $Spirogyra$.
$Spirogyra$ is a chlorophyllous thallophyte because it belongs to the group of green algae,which possess chlorophyll pigments for the process of photosynthesis.
$Volvariella$ is a fungus (non-chlorophyllous),$Nephrolepis$ is a pteridophyte,and $Gnetum$ is a gymnosperm.

(D) $Riccia$,a liverwort within the $Bryophyta$,exhibits a dominant gametophyte phase in its life cycle.
In bryophytes,the gametophyte is the main stage,whereas the sporophyte is smaller and dependent on the gametophyte for nutrition.

(D) The correct answer is $D$.
Frogs and toads are both amphibians belonging to the order $Anura$.
They can be differentiated by several morphological features:
$1$. Frogs typically have smooth,moist skin,while toads have dry,warty skin.
$2$. Frogs have long,powerful hind legs for jumping,whereas toads have shorter legs and tend to walk or hop.
$3$. The most distinct anatomical difference is that toads possess a pair of large,prominent parotid glands behind their eyes,which secrete toxins for defense,whereas frogs lack these specific glands.

(C) The correct statement is that all vertebrates are chordates,but all chordates are not vertebrates.
This is because the phylum $Chordata$ is a larger group that includes organisms with a notochord at some stage of their life cycle.
Vertebrates are a subphylum within $Chordata$ where the notochord is replaced by a vertebral column in the adult stage.
Therefore,while every vertebrate is a chordate,not every chordate (such as $Urochordata$ or $Cephalochordata$) possesses a vertebral column.

(D) The correct answer is $D$ (Mongoose).
$1$. Mammals are classified into three groups: Prototherians (egg-laying mammals like Platypus),Metatherians (marsupials like Kangaroo),and Eutherians (placental mammals).
$2$. $A$ true placenta is a characteristic feature of Eutherians,which allows for the exchange of nutrients and waste between the mother and the fetus.
$3$. Mongoose is a Eutherian mammal and thus possesses a true placenta.
$4$. Kangaroo is a marsupial (Metatherian) with a short-lived yolk-sac placenta,and Platypus is a monotreme (Prototherian) that lays eggs.

(B) The correct answer is $B$.
An enzyme is composed of an apoenzyme (the protein part) and a coenzyme (a non-protein organic molecule).
Together, they form the holoenzyme, which is the active form of the enzyme capable of catalyzing biochemical reactions.
Therefore, the relationship is defined as: $\text{Holoenzyme} = \text{Apoenzyme} + \text{Coenzyme}$.

(C) The inheritance of $ABO$ blood groups is controlled by the gene $I$. The gene $I$ has three alleles: $I^A$,$I^B$,and $i$.
Since the children have blood groups $A$,$B$,$AB$,and $O$,the parents must be heterozygous.
For a child to have blood group $O$ (genotype $ii$),both parents must contribute an $i$ allele.
For a child to have blood group $AB$ (genotype $I^A I^B$),one parent must contribute $I^A$ and the other must contribute $I^B$.
Therefore,the genotypes of the parents must be $I^A i$ (Blood group $A$) and $I^B i$ (Blood group $B$).
This cross $(I^A i \times I^B i)$ results in offspring with genotypes $I^A I^B$ $(AB)$,$I^A i$ $(A)$,$I^B i$ $(B)$,and $ii$ $(O)$.
Thus,both parents are heterozygous for blood groups $A$ and $B$ respectively.

(A) Based on the $T.S.$ of anther diagram:
$A$ points to the central region connecting the two lobes,which is the Connective tissue.
$B$ points to the layer just inside the epidermis,which is the Endothecium.
$C$ points to the microspores or Pollen grains located within the microsporangium.
Therefore,the correct identification is $A$-Connective tissue,$B$-Endothecium,$C$-Pollen grain. The correct option is $A$.

(D) The correct matching is as follows:
$A$. Parturition: $3$. Delivery of the baby from the uterus.
$B$. Gestation: $4$. The duration between pregnancy and birth.
$C$. Ovulation: $2$. The release of the egg (ovum) from the Graafian follicle.
$D$. Implantation: $1$. The attachment of the blastocyst to the uterine endometrium.
$E$. Conception: $5$. The formation of the zygote by the fusion of the egg and sperm.
Therefore, the correct sequence is $A-3, B-4, C-2, D-1, E-5$.

(D) men are hemizygous and one defective gene is enough to make them colour blind.
More men suffer from colour blindness than women because men are hemizygous for the $X$ chromosome.
Since men have only one $X$ chromosome,a single recessive allele on this chromosome is sufficient to express the trait of colour blindness.
In contrast,women have two $X$ chromosomes,so they require two defective copies of the gene to express the condition,making it significantly less common in women.

(A) In a typical Mendelian dihybrid cross,the $F_2$ phenotypic ratio is $9:3:3:1$.
Here,the parental combinations are the phenotypes that resemble the original parents (homozygous dominant and homozygous recessive),which are represented by the $9$ (both dominant) and $1$ (both recessive) classes.
Total parental combinations = $9 + 1 = 10$.
The recombinations are the new phenotypes that differ from the parents,represented by the two $3$ classes (one dominant and one recessive trait each).
Total recombinations = $3 + 3 = 6$.
Therefore,the ratio of parental combinations to recombinations is $10:6$.

(C) The correct answer is $C$.
In the genetic code,the phenomenon where $61$ codons code for only $20$ different amino acids is known as degeneracy.
This occurs because some amino acids are specified by more than one codon.
This redundancy provides a protective mechanism against mutations,as a change in the third base of a codon (wobble position) often does not result in a change to the amino acid being incorporated.

(A) genetic material of $RNA$ viruses.
In $RNA$ viruses,genetic $RNA$ refers to the $RNA$ that encodes the virus's genetic information,allowing it to replicate and produce new viral particles.
This $RNA$ acts as the virus's genetic blueprint,serving the same function as $DNA$ does in other organisms.

(D) is the correct answer.
By the statement "Survival of the Fittest," Darwin meant that the individuals of a species that are best adapted to their environment, through favorable variations and advantageous traits, are more likely to survive and reproduce in the struggle for existence.