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During prophase I, homologous chromosomes from each parent pair up and exchange genetic material, a process known as crossing over. This process creates new combinations of genes in the resulting gametes. When two gametes unite during fertilization, the offspring will have a unique combination of DNA. Genetic recombination during fertilization takes place in the prophase I stage of meiosis ( part labelled III)

Hemophilia is a rare, genetic blood disorder that happens when your blood doesn't clot and make your bleeding slow down or stop.Most people who have hemophilia are born with it. It is inherited (passed down) from a parent to a child.It is a chromosomally linked recessive disorder, the result of a mutated gene on the X chromosome. Because those assigned male at birth only have one X chromosome, they can develop hemophilia if they inherit the altered gene. The X-linked recessive inheritance pattern explains why hemophilia is more common in males than females. Females can be carriers of the mutated gene, but they are less likely to develop the disorder themselves. In summary, hemophilia in humans controlled by recessive genes.

The Olfactory lobe is the portion of the cerebral cortex concerned with the sense of smell. It is part of the Cerebrum. It is a structurally distinct cortical region on the ventral surface of the forebrain, composed of several areas. The medulla oblongata controls vital functions like breathing, heart rate, and blood pressure. The cerebrum, the largest part of the brain, is responsible for controlling conscious thought, actions, emotions, memory, language, reasoning, and sensory perception, essentially acting as the center for higher-level cognitive functions like thinking, learning, and problem-solving; The cerebellum's primary function is to coordinate movement, maintain balance and posture, regulate muscle tone, and facilitate motor learning, essentially ensuring smooth and precise muscle control by processing information from other brain areas and sensory receptors;

Marchantia is a member of the Marchantiaceae, the Marchantia family. This family is one of many thalloid liverwort families or bryophyta. A thalloid liverwort is strap-like and often forms large colonies on the surface on which it grows. A liverwort is non-vascular green plant. Spirogyra is a green algae that is a member of the Thallophyta division. It is also known as water silk, mermaid's tresses, and blanket weed. Dryopteris, also known as wood ferns, male ferns, or buckler ferns, is a genus of ferns in the Dryopteridaceae family, of pteridophyta. Cycads are part of the order Cycadales and the division Cycadophyta, which are both groups of gymnosperms. Maize belongs to the group angiosperms. Angiosperms are plants that have a well-developed vascular system Only bryophytes(Marchantia) - I and Thallophytes (Spirogyra) - II are non- vascular, others have vascular systems. Therefore option A is the correct answer.

\(\frac {\sin A}{a} = \frac {\sin B}{b} = \frac {\sin C}{c}\) \(\implies \frac {\sin 82^0}{43.2} = \frac {\sin 56^0}{AB}\) \(\implies AB \times \sin 82^0 = 43.2 \times \sin 56^0\) \(\therefore AB = \frac {43.2 \times \sin 56^0}{\sin 82^0}\) = 36.2cm (to 3 s.f)

Consider ∆XOB and using Pythagoras theorem 13\(^2\) = 12\(^2\) + h\(^2\) ⇒ 169 = 144 + h\(^2\) ⇒ 169 - 144 =h\(^2\) ⇒ 25 = h\(^2\) ⇒ h = \(\sqrt25\) = 5cm tan \theta = \(\frac {opp}{adj}\) ⇒ tan \theta = \(\frac{12}{5}\) = 2.4 ⇒ \theta = tan\(^{-1}\)(2.4) ⇒ \theta = 67.38\(^0\) ∠AOB = 2\theta = 2 x 67.38\(^o\) = 134.76\(^o\) L = \(\frac{\theta}{360^o} \times 2\pi r\) ⇒ L = \(\frac {134.76}{360} \times 2 \times \frac {22}{7} \times 13 = \frac {77082.72}{2520}\) ∴ L = 30.6cm (to 3 s.f)

An exterior angle of a n-sided regular polygon = \(\frac{360}{n}\) For (n - 1) sided regular polygon = \(\frac{360}{n - 1}\) For (n + 2) sided regular polygon = \(\frac{360}{n + 1}\) ⇒ \(\frac{360}{n - 1} - \frac{360}{n + 2}\) = 6 9Given) ⇒ \(\frac{360(n + 2) - 360(n - 1)}{(n - 1)(n + 2)}\) ⇒ \(\frac{360n + 720 - 360n + 360}{(n - 1)(n + 2)}\) ⇒ \(\frac{1080}{(n - 1)(n + 2)} = \frac{6}{1}\) ⇒ 1080 = 6 (n - 1)(n + 2) ⇒ 180 = (n - 1)(n + 2) ⇒ 180 = n\(^2\)+ 2n - n - 2 ⇒ 180 = n\(^2\) + n - 2 ⇒ n\(^2\)b+ n - 2 - 180 = 0 ⇒ n\(^2\) + n - 182 = 0 ⇒ n\(^2\) + 14n - 13n - 182 = 0 ⇒ n (n + 14) - 13 (n + 14) = 0 ⇒ (n - 13) (n + 14) = 0 ⇒ n - 13 = 0 or n + 14 = 0 ⇒ n = 13 or n = -14 ∴ n = 13 (We can't have a negative number of side)

\(PR^2 = PQ^2 + RQ^2 - 2(PQ)(RQ)cos Q\) \(\implies cos Q = \frac{PQ^2 + RQ^2 - PR^2}{2(PQ)(RQ)}\) \(\implies cos Q = \frac{8^2 + 5^2 - 7^2}{2\times8\times5}\) \(\implies cos Q = \frac{64 + 25 - 49}{80}\) \(\implies cos Q = \frac{40}{80} = 0.5\) \(\implies Q = cos^{-1} (0.5) = 60^0\) \(\therefore x = 60^0\)

Intersecting Chords Theorem states that If two chords intersect in a circle, then the products of the measures of the segments of the chords are equal. ⇒ AE * EB = CE * ED ⇒ 6 * \(x\) = 4 * (\(x\) + 5) ⇒ 6\(x\) = 4\(x\) + 20 ⇒ 6\(x\) - 4\(x\) = 20 ⇒ 2\(x\) = 20 ∴ \(x = \frac{20}{2}\) = 10 units

∠PQR + ∠PSR = 180o (opp. angles of cyclic quad. are supplementary) ⇒ 5\(x\) - \(y\) + 10 + (-2\(x\) + 3\(y\) + 145) = 180 ⇒ 5\(x\) - \(y\) + 10 - 2\(x\) + 3\(y\) + 145 = 180 ⇒ 3\(x\) + 2\(y\) + 155 = 180 ⇒ 3\(x\) + 2\(y\) = 180 - 155 ⇒ 3\(x\) + 2\(y\) = 25 ----- (i) ∠QPS + ∠QRS = 180o (opp. angles of cyclic quad. are supplementary) ⇒ -4\(x\) - 7\(y\) + 150 + (2\(x\) + 8\(y\) + 105) = 180 ⇒ -4\(x\) - 7\(y\) + 75 + 2\(x\) + 8\(y\) + 180 = 180 ⇒ -2\(x\) + \(y\) + 255 = 180 ⇒ -2\(x\) + y = 180 - 255 ⇒ -2\(x\) + \(y\) = -75 ------- (ii) ⇒ \(y\) = -75 + 2\(x\) -------- (iii) Substitute (-75 + 2\(x\)) for \(y\) in equation (i) ⇒ 3\(x\) + 2(-75 + 2\(x\)) = 25 ⇒ 3\(x\) - 150 + 4\(x\) = 25 ⇒ 7\(x\) = 25 + 150 ⇒ 7\(x\) = 175 ⇒ \(x = \frac{175}{7} = 25\) From equation (iii) ⇒ \(y\) = -75 + 2(25) = -75 + 50 ⇒ \(y\) = -25 ∴ \(x\) + \(y\) = 25 + (-25) = 0

On the \(y\)-axis, each box is \(\frac{1 - 0}{5} = \frac{1}{5}\) = 0.2unit On the \(x\)-axis, each box is \(\frac{90 - 0}{6} = \frac{90}{6} = 15^o\) ⇒ \(\theta_1 = 180^o + (2.5\times15^o) = 180^o + 37.5^o = 217.5^o ≃ 218^o \)(2 and half boxes were counted to the right of 180\(^o\)) ⇒ \(\theta_2 = 270^o + (3.5\times15^o) = 270^o + 52.5^o = 322.5^o ≃ 323^o \)(3 and half boxes were counted to the right of 270\(^o\)) ∴ \(\theta = 218^o, 323^o\)

|AP| = |PB| = \(x\) (The perpendicular to a chord bisects the chord if drawn from the center of the circle.) From ∆OPB Using Pythagoras theorem ⇒ \(13^2 = 12^2 + x^2\) ⇒ \(169 = 144 + x^2\) ⇒ \(169 - 144 = x^2\) ⇒ \(x^2 = 25\) ⇒ \(x = \sqrt25 = 5 cm\) ∴ Length of the chord |AB| = \(x + x = 5 + 5 = 10 cm\)

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\(402_y = 102_ten\) ⇒ 4 x y\(^2\) + 0 x y\(^1\) + 2 x y\(^0\) = 102 ⇒ 4y\(^2\) + 0 + 2 x 1 = 102 ⇒ 4y\(^2\) + 2 = 102 ⇒ 4y\(^2\) = 102 - 2 ⇒ 4y\(^2\) = 100 ⇒ y\(^2 = \frac {100}{4}\) ⇒ y\(^2\) = 25 ∴ y = \sqrt25 = 5

16\(^{0.16}\) \times 16\(^{0.04}\) \times 2\(^{0.2}\) =2\(^{4(0.16)}\) \times 2\(^{4(0.04)}\) \times 2\(^{0.2}\) =2\(^{0.64}\) \times 2\(^{0.16}\) \times 2\(^{0.2}\) =2\(^{0.64\times0.16\times0.2}\) =2\(^1\) 2