Board Paper of Class 12 2022 Biology (Theory) Set 057/1/4 - Solutions

Ans. (a)

People suffering from STDs do not go for timely detection and proper treatment due to the less significant symptoms of the STDs. Usually symptoms in early stages are itching, burning sensation, inflammation in genitals, etc. Many STDs are said to be socially stigmatized, therefore people are judged or condemned for being infected with STDs, for example in case of person suffering from AIDS.

Ans. (d)

Vasectomy is a sterilisation procedure advised for males. In vasectomy, a small part of the vas deferens is removed or tied so as to block gamete transport, therefore preventing conception. In case of females, the surgical method employed for contraception is termed as Tubectomy which involves cutting and tying of fallopian tube so as to inhibit gamete transfer.

Ans. (d)

Each testis consist of 250 compartments called as testicular lobules and each lobule contains 1-3 highly coiled seminiferous tubules surrounded by connective tissue in which numerous cells called interstitial cells or Leydig cells occur. These cells produce male sex hormones named androgens like testosterone. Each seminiferous tubule is lined on its inside by two types of cells called male germ cells and Sertoli cells or sustentacular cells or nurse cells. Sertoli cells are nutritive in function. The germinal epithelial cells produce sperms by spermatogenesis.

Ans. (c)

Phenylketonuria is caused by mutation in the gene coding the enzyme phenylalanine hydroxylase. The affected individuals lack an enzyme (phenylalanine hydroxylase) that catalyzes the conversion of amino acid phenylalanine into tyrosine. The affected individuals show mental retardation, and hair and skin pigmentation.

Ans. (c)

Mammary glands or Breasts are modified sweat glands. These are paired structure consisting of glandular tissue, fibrous tissue and variable amount of fat. Each mammary gland consists of 15-20 mammary lobes that forms the glandular tissue and contains cluster of milk secreting alveoli which open into tubules; they join to form mammary ducts. These ducts further join to form mammary ampulla which is connected to the lactiferous duct through which milk is sucked out.

Ans. (b)

In figure (i), the plant has multiple carpels which are fused together representing multicarpellary syncarpous condition. In case of figure (ii), the plant has separate carpels (not fused together), therefore, it is termed as multicarpellary apocarpous.

Ans. (c)

The IUDs are available as non-medicated IUDs (e.g., Lippes loop), copper releasing IUDs (Cu7, CuT, etc.) and hormone releasing IUDs (e.g., LNG-20). The copper releasing IUDs suppress sperm motility and the fertilising capacity of sperms. However, the hormone releasing IUDs make the uterus unsuitable for implantation and also cervix hostile to the sperm.

Ans. (c)

An infection caused by an amoeba is called amoebiasis. It spreads through eating or drinking contaminated food or water. The symptoms of the disease include diarrhea, cramps, stomach pain, etc.

Ans. (b)

Oogonia undergo meiosis I and form primary oocytes. The primary oocytes temporarily remain in prophase I until the first menstrual cycle. After that, each month, the primary oocytes resume the meiosis process. This type of cell division where cells remain in some phase of the cycle is called a suspended cell division.

Ans. (c)

• The zygote or oospore is divided by a transverse wall into suspensor cell and an embryonal cell.
• The suspensor cell, which lies towards the micropylar end, is divided by transverse divisions to constitute 6-10 celled suspensor.
• The upper cell of the suspensor filament towards the micropylar end is called haustorial cell, whereas the cell lying above the embryo cell is called hypophysis.
• The haustorial cell enlarges in size and attaches the suspensor to the tip of the embryo sac.
• The embryonal cell is divided by second longitudinal division at right angle to the first and then by transverse division to form an octant or an eight celled embryo.
• Out of these eight cells, the lower four cells of the octant away from the suspensor give rise to the plumule and the two cotyledons, while the above four cells of the octant near the suspensor form the hypocotyl and the stele of radicle.

Ans.(a)

The continued self-pollination of plants leads to inbreeding depression. To avoid and discourage self-pollination, there are some outbreeding devices, such as, in some species, pollen release and stigma receptivity are not synchronised. Another way is to place the stigma in a different position so that the pollen cannot come in contact with it. Also, through genetic mechanisms, inbreeding can be avoided. Here, the pollen is prevented from fertilising the ovules by inhibiting pollen germination or pollen tube growth in the pistil.

Ans. (c)

n eukaryotes, this organisation is much more complex. There is a set of positively charged, basic proteins called ‘histones’. Histones are rich in basic amino acid, lysine and arginine. Both the amino acid residues carry positive charges in their side chains. The structure is organised to form a unit of eight molecules known as ‘histone octamer’. The negatively charged DNA is wrapped around the positively charged histone-octamer to form a structure called ‘nucleosome’.

Ans. (b)

Let’s make a cross between the parents (Aa × aa) given in the question and find out the percentage of the offspring.

From this cross, it can be concluded that 50% of the offspring have an axial flower position while the other 50% have a terminal flower position.

Ans. (b)

According to the question, the genotype of both the parents must be heterozygous (Rr) for the trait. If the parents are homozygous for the trait, then all of their children should also have that disease. But as per the question, only their son is suffering from the disease.

From knowing the genotype of both the parents, we can find the genotype of their children by making a cross between them.

From the cross, their son is either RR or Rr, and their daughter is rr because she is not suffering from the rolling of the tongue disorder. Therefore, she will not have any dominant allele for the rolling of the tongue in her genotype. As their son is suffering from this disease, therefore, at least one of the genes should have a dominant allele.

Ans. (c)

In the first generation, the mother has a disease and the father is normal. In the second generation, both the daughter and son have the disease, indicating that the disease was passed on to the daughter or son by their mother; thus, the trait is not linked to sex. Now, the diseased daughter in the second generation is heterozygous dominant, whereas the person she marries is normal. In the third generation, the disease is only passed to their daughter, and their sons are normal. From this, it is indicated that the disease is an autosomal recessive disease.

Ans. (d)

As per the question, the father is heterozygous for blood group A, thus his genotype is I^Ai. Similarly, the mother is heterozygous for blood group B, thus her genotype is I^Bi.

Now, the possibility of the blood groups in the offsprings will be:

Ans. (b)

It is an autosomal recessive blood disorder. The defect occurs due to the mutation or deletion of the genes controlling the formation of the globin chain of haemoglobin. An imbalanced synthesis of globin chains in haemoglobin causes anemia. Thalassemia is of two types, depending upon which chain of the haemoglobin molecule is affected. When $\mathrm{\alpha }$ globin chain is affected, it is called $\mathrm{\alpha }$ thalassemia. It is controlled by two closely linked genes, HBA1 and HBA2, on chromosome 16. Whereas, when the production of $\mathrm{\beta }$ globin chain is affected, it is called $\mathrm{\beta }$ Tthalassemia. It is controlled by a single gene, HBB, on chromosome 11.

Ans. (a)

In the question, a nucleotide sequence of coding strand is given. But, as we know, the mRNA is transcribed from the template strand. The coding strand does not code for anything.

Given,
Coding strand = 5^'- TACGCCG -3^'
Template strand = 3^' – ATGCGGC –5^'
Therefore, the sequence of bases on the mRNA transcribed will be
Template strand = 3^' – ATGCGGC –5^'
mRNA strand = 5^'- UACGCCG -3^' (In case of RNA, thymine is replaced by uracil.)

Ans. (c)

The number of base pairs in a DNA molecule is 160, as given in the question.

Therefore, the total number of bases = 160 ×2 = 320 bases (because one pair has two nitrogenous bases).

According to Chargaff's rules, the amount of adenine is equal to the amount of thymine, and the amount of guanine is equal to the amount of cytosine.

So, adenine = 20%, as given in the question; therefore, thymine = 20% and rest of the sequence consist of G and C that is 60% (30% each of the nitrogenous bases).

Now,
$$\begin{gathered} \mathrm{Cytosine} \mathrm{bases} =320 \times \frac{30}{100} \\ =96 \mathrm{base} \end{gathered}$$

Ans. (a)

In the question, base sequence of a template strand is given, that is, 5’-AGGTTTAACG -3'.

As we know, mRNA is transcribed from the template base or has a sequence base complementary to that of the template strand. Therefore, the RNA sequence will be

RNA strand = 3’- UCCAAUUGC -3’ (In case of RNA, thymine is replaced by uracil.)

Ans. (b)

Lactose is the inducer in lac operon; when lactose is present/added to the cell, it binds to the repressor and changes the conformation of the repressor, thereby making it inactive. Such a repressor does not bind to the operator and, hence, RNA polymerase has access to the structural genes. The transcription precedes and lactose is metabolised.

Ans. (c)

In 1958, Taylor and his colleagues performed an experiment involving the use of radioactive thymidine. They performed their experiment on Vicia faba (faba beans). From their experiment, they proved that the chromosomes inside the DNA also replicate semi-conservatively.

Ans. (d)

The presence of a hydroxyl group at the 2^nd carbon in RNA makes the RNA more reactive and less stable. Therefore, RNA is not an ideal genetic material. Only in a few species is the RNA found as a genetic material like viruses.

Ans. (c)

Mendel conducted the hybridisation experiments on garden pea(Pisum Sativum). He chose the pea plant due to the presence of contrasting characters like tall or dwarf (the stem’s height), round and wrinkled (the seed’s shape), etc. One of the traits in this character is a dominant trait, while the other is a recessive trait.

In this question, the incorrect pair is yellow and green (pod’s colour) because green is a dominant trait while yellow is a recessive trait.

Ans. (c)

The temperature of the body at rest is known as basal body temperature (BBT). By looking at the graph given in the question, it can be concluded that the temperature during menstruation, i.e., the first day of the cycle, is low. Similarly, BBT at the time of ovulation is also reduced. The rise in temperature is the result of the release of progesterone hormone. Women can use charting average basal body temperatures over the length of a menstrual period so as to determine the time of ovulation.

Ans. (c)

A couple should have the knowledge of time of ovulation as it is the most fertile stage of menstrual cycle. During the ovulatory phase, the egg is released and is ready for fertilisation.

Ans. (a)

The second half of the menstrual cycle is called secretory phase or luteal phase which is characterized by formation of corpus luteum from the ruptured follicle after ovulation. Ovulation occurred under the effect of LH hormone and corpus luteum secretes progesterone.

Ans. (d)

LH and FSH are increased in the middle of the cycle. The rapid increase in LH leads to the rupturing of Graafian follicle and, thereby, the release of ovum takes place.

Ans. (b)

The corpus luteum is formed in the mammalian ovary from a ruptured Graafian follicle after ovulation. It degenerates in the absence of fertilisation as the level of gonadotropins (FSH and LH) drops.

Ans. (b)

The menstrual cycle has three phases:

• Menstrual phase: Degeneration of endometrium and bleeding (1 to 4 days)
• Proliferative phase: Regeneration of endometrium (5 to 13 days)
• Luteal phase: This phase lasts for 12 to 15 days. Ovulation occurrs around day 14, also it leads to formation of corpus luteum from the ruptured follicle. Corpus luteum secretes progesterone in this phase (days 15 to 28).

Ans. (b)

Ans. (a)

According to the karyotype given, all autosomes are present in pairs except the one sex chromosome. Thus, one Y chromosome is missing, therefore, the individual has 45 with XO set of chromosomes. This disorder is called Turner’s syndrome. It occurs due to the absence of one X chromosome. In such a case, females are sterile as ovaries are rudimentary and they also lack other secondary sexual features.

Ans. (c)

Incomplete linkage: Genes located on the same chromosome have a tendency to segregate due to cross-over. As a result, recombinant progeny are produced besides the parental type. In independent assortment, the number of recombinant individuals is lower than expected. In each case, two parental types and two recombinant types are each 25% in the case of independent assortment. However, in the case of linkage, each of the two parental types is more than 25%, whereas recombinant types are less than 25%.

Examples: Morgan and his students discovered that linked genes have different recombinations, some of which are more tightly linked than others. For example, in Drosophila, crossing yellow-bodied (Y) and white-eyed (W) females with brown-bodied (Y+) and red-eyed (W+) males resulted in F1, which was brown bodied and red eyed. Morgan discovered that the two genes did not segregate independently of each other when F1 progeny were intercrossed, and that the F2 ratio departed greatly from the expected 9: 3: 3: 1 ratio. He discovered that 98.7% of the samples were parental and only 1.3% were recombinants. (ii) In a second Drosophila cross between white-eyed and miniature-winged females with wild type or red-eyed normal-winged males, all of the F1 were red eyed and normal winged. After then, an F1 female fly was crossed with a white-eyed, miniature-winged male. 62.8% of the population progeny were of parental types while 37.2% were recombinants.

Ans. (c)

Transformation experiments were initially conducted by F. Griffith in 1928.

• He injected a mixture of two strains of Pneumococcus (Diplococcus pneumoniae) into the mice. The strains were S III (virulent form) and strain RII (non-virulent).
• The S III type bacteria when injected into the mice cause pneumonia and ultimately lead to death of mice while R II type bacteria when injected didn’t resulted in death of the mice as it didn’t cause the disease.
• Then heat killed S III bacteria was taken and injected into the mice. They did not cause the disease and thus mice lived.
• Heat killed S III and R II (non-virulent) bacteria, when they were injected into the mice, causing pneumonia. Finally, death occurred.

Ans. (a)

The DNA strands open up to a certain point forming a replication fork. Replication fork is the Y-shaped structure formed as a result of small opening of DNA helix for replication. Since DNA polymerase can polymerise the nucleotides only in 5’-3’ direction, in the replication fork, it polymerises the nucleotides into short stretches. Later, these short stretches, Okazaki fragments, are joined by the action of DNA ligases into the lagging strand.

Ans. (c)

The figure showing the diagrammatic representation of translation: