This biology question covers important biological concepts and processes. The step-by-step explanation below helps you understand the underlying mechanisms and reasoning.
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Here are the solutions to the questions: 28. Step 1: Identify the function of Organelle I from the table. The table states that Organelle I "Controls all cell activities". Step 2: Recall which cell organelle performs this function. The nucleus is responsible for controlling all cell activities, as it contains the genetic material and regulates gene expression. The correct option is C. Answer: C. nucleus* 29. Step 1: Identify the function of Organelle II from the table. The table states that Organelle II "Carries out photosynthesis". Step 2: Recall where photosynthesis occurs and in which plant parts. Photosynthesis occurs in chloroplasts, which are found in plant cells that are exposed to light and contain chlorophyll, typically in green parts like leaves. Roots and woody logs do not perform photosynthesis. The correct option is C. Answer: C. green leaf* 30. Step 1: Understand the effect of a hypotonic solution on plant cells. In a hypotonic solution, water moves into the plant cell by osmosis, causing the cell to swell. Step 2: Explain why plant cells do not burst in hypotonic solutions, unlike animal cells. Plant cells have a rigid cell wall outside their cell membrane. This cell wall provides structural support and prevents the cell from bursting when it takes in too much water, allowing it to become turgid. The correct option is C. Answer: C. the cell wall provides rigidity for them.* 31. Step 1: Understand blood group compatibility and agglutination. Agglutination occurs when the recipient's antibodies react with antigens on the donor's red blood cells. A donor with blood group B has B antigens on their red blood cells and anti-A antibodies in their plasma. Step 2: Determine which recipient blood groups would not cause agglutination with blood group B. • A patient with blood group AB has no anti-A or anti-B antibodies, so they can receive blood from group B without agglutination. • A patient with blood group B has anti-A antibodies but no anti-B antibodies, so they can receive blood from group B without agglutination. • A patient with blood group A has anti-B antibodies, which would react with B antigens, causing agglutination. • A patient with blood group O has both anti-A and anti-B antibodies, which would react with B antigens, causing agglutination. Therefore, agglutination would not occur if blood group B is transfused to a patient with blood group AB or B. The correct option is B. Answer: B. AB or B* 32. Step 1: Determine the possible gametes for each parent. Parent 1: BbFf can produce four types of gametes: BF, Bf, bF, bf. Parent 2: bbff can produce only one type of gamete: bf. Step 2: Determine the possible genotypes of the offspring. When the gametes combine, the possible offspring genotypes are: • BF (from BbFf) + bf (from bbff) → BbFf • Bf (from BbFf) + bf (from bbff) → Bbff • bF (from BbFf) + bf (from bbff) → bbFf • bf (from BbFf) + bf (from bbff) → bbff There are 4 possible distinct genotypes for the offspring. The correct option is D. Answer: D. Four* 33. Step 1: Analyze the genetic cross shown in the diagram. Parent F has a pointed abdomen, and Parent G has a rounded abdomen. All the F1 offspring have pointed abdomens. Step 2: Determine the dominant and recessive traits. When two parents with different traits produce offspring that all show one of the parental traits, that trait is dominant. In this case, pointed abdomen is dominant over rounded abdomen. Step 3: Relate the gene for pointed abdomen to the parents. Since pointed abdomen is dominant, Parent F (pointed) must be homozygous dominant (let's say HH) and Parent G (rounded) must be homozygous recessive (hh). The gene for pointed abdomen (H) is dominant. Parent G expresses the recessive trait (rounded abdomen). Therefore, the characteristic of G is recessive. The correct option is A. Answer: A. recessive in G.* 34. Step 1: Use the information from Question 33. We established that pointed abdomen is dominant (H) and rounded abdomen is recessive (h). Parent F is HH, and Parent G is hh. Step 2: Determine the genotypes of the F1 offspring. A cross between HH and hh will produce all offspring with the genotype Hh. HH × hh → Hh, Hh, Hh, Hh All offspring will be heterozygous (Hh) and will have pointed abdomens. The correct option is C. Answer: C. Hh, Hh, Hh, Hh.* 35. Step 1: Recall the diploid number of chromosomes in humans. Human somatic cells are diploid (2n) and contain 46 chromosomes. Step 2: Calculate the haploid number. The haploid number (n) is half of the diploid number. n = (2n)/(2) = (46)/(2) = 23 The correct option is D. Answer: D. 23* 36. Step 1: Determine the genotypes of the first cross (P generation). Let T represent the allele for tall (dominant) and t represent the allele for dwarf (recessive). A heterozygous tall plant (Tt) is crossed with a dwarf plant (tt). Tt × tt Step 2: Determine the genotypes of the F1 generation. The possible offspring from Tt x tt are: Tt, Tt, tt, tt So, 50% are tall (Tt) and 50% are dwarf (tt). Step 3: Determine the genotypes of the second cross (F1 x F1). "Two of the tall F1 plants were crossed." From Step 2, all tall F1 plants are Tt. So, the second cross is Tt x Tt. Step 4: Determine the phenotypic ratio of the F2 generation. From the cross Tt x Tt, the offspring genotypes are: TT, Tt, Tt, tt The phenotypic ratio is 3 Tall (TT, Tt, Tt) : 1 Dwarf (tt). Step 5: Calculate the number of dwarf plants in the F2 generation. The total number of F2 plants obtained is 120. The fraction of dwarf plants is (1)/(4). Number of dwarf plants = (1)/(4) × 120 = 30. The correct option is A. Answer: A. 30* Send me the next one 📸