MikaStars39/nano-eval huggingface dataset:
Since the primary oocytes of a certain organism contain 16 chromosomes, therefore?

Options:
A. During metaphase of the second division in meiosis, paired centromeres align on the equatorial plate, symmetrically connected to the spindle fibers.
B. During telophase of the second meiotic division, homologous chromosomes are restored, creating 16 identical nuclei.
C. During prophase of the second meiotic division, chromosome pairs undergo synapsis, creating tetrads for separation.
D. There are 16 dyads in the prophase of the second division of meiosis.
E. During the anaphase of mitosis, all 16 chromatids move toward opposite poles, identical to parental cells.
F. There are 16 tetrads in the prophase of the first division of meiosis.
G. During metaphase of the first meiotic division, homologous chromosomes randomly align, but no chromatids separate.
H. During the anaphase of the second meiotic division, 8 chromatids migrate to one pole, while another 8 migrate to the opposite pole.
I. During metaphase of the first meiotic division, eight dyads form spindle connections, awaiting chromatid separation.
J. During the anaphase of the first meiotic division, 16 chromatids migrate to opposite poles, forming balanced gametes.
Answer: H
Who were the first to walk upright on two legs?

Options:
A. Homo erectus
B. Cro-Magnons
C. Homo habilis
D. Paranthropus
E. Homo floresiensis
F. Australopithecus
G. Capable Person
H. Ardipithecus
I. Homo sapiens
J. Neanderthals
Answer: F
An organ primarily composed of glandular epithelium is called a gland. In the human body, exocrine gland cells commonly employ a secretion method in which secretory granules are surrounded by a limiting membrane. During release, the limiting membrane of the secretory granules fuses with the plasma membrane, thereby discharging the secretion without damaging the integrity of the glandular cells. What is described above?

Options:
A. Membrane fusion secretion
B. Limited release secretion
C. Integrated secretion
D. Preservation-release secretion
E. Non-destructive secretion
F. Constrained secretion
G. Partial secretion
H. Safe fusion secretion
I. Local secretion
J. Total Secretion
Answer: I
In corn, the allele **b** results in a white endosperm color, while **b⁺** corresponds to the wild-type purple color. Another gene, **w** (*waxy*), leads to waxy endosperm, while **wx⁺** represents the wild-type starchy phenotype. A third gene, **cn**, causes pink-colored cotyledons, whereas **cn⁺** represents the wild-type yellow cotyledons. Female heterozygotes of these three genes were test-crossed, yielding 1,000 progeny with the following phenotypes and their respective counts: 5 wild-type; 6 with white, waxy endosperm and pink cotyledons; 69 with waxy endosperm and pink cotyledons; 67 with white endosperm; 382 with pink cotyledons; 379 with white, waxy endosperm; 48 with waxy endosperm; and 44 with white and pink cotyledons. In the progeny phenotypes, unless otherwise indicated, all traits are in their wild-type form.
What is the genetic distance between **wx** and **cn**?

Options:
A. 10.3
B. 13.6
C. 22.8
D. 11.5
E. 20.1
F. 17.4
G. 9.2
H. 14.7
I. 15.2
J. 8.9
Answer: H
Which structure is the source of the rays in the root?

Options:
A. Epidermal Cell Region
B. Root Tip Meristematic
C. Intercellular Pith Strand
D. The original cambium
E. Secondary Xylem Origin
F. Edges of Protoxylem
G. Pericycle Layers Source
H. Phloem Ray Diffusion
I. Vascular Cambium
J. Primary Meristem
Answer: I
The birth weight of most human newborns is generally stable between 3 to 4 kilograms, and the mortality rates are higher for babies who are either underweight or overweight. What type of natural selection pattern does this result from?

Options:
A. Transitional Selection
B. Split Selection
C. Balancing Selection
D. Targeted Selection
E. Disruptive Selection
F. Random selection
G. Directional Selection
H. Kinetic Selection
I. Stable Selection
J. Convergent Selection
Answer: I
In the process of root absorption of inorganic ions, which of the following factors plays a decisive role?

Options:
A. The Respiration Rate of Root Systems
B. The Density of Stomata on Leaf Surfaces
C. The chemical potential difference across the apical epidermal cells
D. The Temperature of Epidermal Cellular Layers
E. The Photosynthetic Activity of Root Systems
F. The Distribution of Sunlight on Root Surfaces
G. The Rate of Plant Transpiration
H. The Water Absorption Rate by Root Hairs
I. The Osmotic Pressure in Soil Surrounding Roots
J. The concentration difference of inorganic ions inside and outside the root apex epidermal cells
Answer: A
A certain type of bird that lives 50 kilometers from the coast arrives at the seaside 50 minutes later each day to forage. What does this belong to?

Options:
A. Feeding Interval
B. Irregular.
C. Diurnal Pattern
D. Migratory Behavior
E. Lunar Rhythm
F. Seasonal Cycle
G. Tidal Rhythm
H. Solar Tracking
I. Weather Influence
J. Circadian Rhythm
Answer: G
In corn, the number of double recessive types among the offspring of the AB/ab × AB/ab cross accounts for 16% of all progeny. What is the genetic map distance between these two genes?

Options:
A. 50
B. 45
C. 30
D. 10
E. 15
F. 20
G. 35
H. 40
I. 60
J. 25
Answer: F
Arthropods possess an open circulatory system. Which of the following statements accurately describes their blood-filled cavities?

Options:
A. A coelom filled with blood at birth
B. A cavity with confined blood and specialized vessels
C. An organ-specific compartment enclosed by membranes
D. A cavity filled with blood where the primary body cavity and secondary body cavity are combined
E. A series of tubes transporting blood in a loop
F. Perivisceral compartments filled with clear fluid instead of blood
G. Pericardial cavity filled with blood
H. An enclosed chamber devoid of blood flow
I. A structure combining muscles and blood for circulation
J. Interstitial spaces filled with blood
Answer: D
The bones of an animal's head are connected to the cervical vertebrae via the occipital bone. Humans have two occipital condyles. Which type of animal, like humans, also has two occipital condyles?

Options:
A. Mammals
B. Mollusks
C. Reptiles
D. Arachnids
E. Birds
F. Fish
G. Insects
H. Bony fish
I. Crustaceans
J. Amphibians
Answer: J
What are the main differences between the seeds of gymnosperms and angiosperms?

Options:
A. Distribution Range of Seeds
B. The number of embryos varies.
C. The structure of the endosperm is uniform.
D. Presence or absence of seed coat
E. Size of the Embryo
F. Embryo Nutrient Source Differences
G. The function of the endosperm is the same.
H. Presence or absence of endosperm
I. The shapes of seeds vary.
J. The colors of seeds are different.
Answer: F
Which fatty acid has the highest degree of unsaturation?

Options:
A. Palm oil
B. Olive Oil
C. Tropical Marine Fish Oil
D. Corn oil
E. Cold-Water Sea Fish Oil
F. Peanut oil
G. Butter
H. Soybean oil
I. Sunflower Seed Oil
J. Walnut Oil
Answer: E
The human haploid genome refers to all the genetic information present on how many chromosomes?

Options:
A. 21 autosomes + X + Y
B. 22 autosomes + Y chromosome
C. 24 Chromosome Sets
D. 22 autosomes
E. 20 Autosomes
F. 23 autosomes
G. 22 autosomes + X + Z
H. 22 autosomes + X chromosome
I. 23 pairs of autosomes + X
J. 22 autosomes + X + Y
Answer: J
Predators have a significant impact on prey populations, whereas parasites have little effect on host populations. Why is this?

Options:
A. Parasites slowly weaken life, while predators swiftly prey on their targets.
B. Parasites do not kill their hosts, whereas predators kill their prey.
C. Parasites do not kill large prey, whereas predators kill small prey.
D. Parasites kill their hosts, while predators only kill the old, weak, sick, or injured prey.
E. Parasites inhibit host reproduction, whereas predators promote prey reproduction.
F. Parasites gain symbiotic relationships, while predators gain from the departure of their prey.
G. Parasites only secrete toxins, while predators kill their prey.
H. Parasites parasitize multiple hosts, while predators prey on a single species.
I. Parasites rely on their hosts for survival, whereas predators are entirely dependent on hunting.
J. Parasites benefit the growth of the host, while predators contribute to population development.
Answer: B
What is the relationship between the quality of vision among the three types of animals: fish, birds, and mammals?

Options:
A. Bird > Fish = Mammal
B. Birds > Fish > Mammals
C. Fish > Mammals > Birds
D. Fish > Birds > Mammals
E. Mammals > Fish > Birds
F. Fish = Bird > Mammal
G. Mammals > Fish = Birds
H. Fish > Birds = Mammals
I. Mammals > Birds > Fish
J. Birds > Mammals > Fish
Answer: J
Two DNA molecules (① and ②) have the same number of base pairs (1000 bp), but their base compositions are different. ① contains 44% G+C, while ② contains 66% G+C. How many T are there in ① and ②, respectively?

Options:
A. 380, 620
B. 600, 320
C. 520, 360
D. 570, 330
E. 480, 240
F. 450, 550
G. 500, 400
H. 240, 480
I. 340, 560
J. 560, 340
Answer: J
What is the mRNA produced by eukaryotes through transcription?

Options:
A. Cis-trans isomer.
B. Structural enhancer.
C. Non-coding RNA transcript.
D. Insertion sequence
E. Multiple isomers
F. Alternative looping factor.
G. Gene sequence regulator.
H. Transcription bubble unit.
I. Intron
J. Splice variant protein.
Answer: A
Regarding voltage-gated ion channels, which of the following descriptions is correct?

Options:
A. This channel remains permanently active, regardless of membrane potential.
B. These types of channels are activated by depolarization of the cell membrane.
C. This channel primarily regulates osmotic balance within the cell.
D. This channel is the foundation for generating action potentials in excitable cells.
E. This channel operates exclusively in non-excitable cells and tissues.
F. This channel allows direct intracellular diffusion of small molecules.
G. This type of channel can transport ions against the electrochemical gradient.
H. This channel is directly activated by the binding of specific ligands.
I. This channel facilitates passive ion transport through gap junctions.
J. This type of channel is ATPase.
Answer: D
The father of a boy with hemophilia, Xiao Zhang, once worked at a nuclear power plant for several years. Neither Xiao Zhang nor his wife's family had any history of hemophilia. Another man, Xiao Li, who also worked at the same nuclear power plant for several years, had a child with achondroplasia. Similarly, there was no history of this condition in Xiao Li's or his wife's family. As a result, both Xiao Zhang and Xiao Li filed lawsuits against their employer. As a lawyer, whose case would you prefer to take on? (Note: Hemophilia is an X-linked recessive inherited condition; achondroplasia is an autosomal dominant inherited condition.)

Options:
A. Neither Xiao Zhang nor Xiao Li has a strong case to present in court.
B. Xiao Zhang's condition supports litigation more than Xiao Li's does.
C. The likelihood of either person winning the case is very small; it would be wise not to accept the commission.
D. Both are fine, and the likelihood of winning the case is very high.
E. The probability of Xiao Zhang winning the case is identical to Xiao Li's.
F. Xiao Zhang's case is stronger than Xiao Li's in this situation.
G. Xiao Zhang has a high likelihood of winning the case, whereas Xiao Li does not.
H. Xiao Li's case may be valid, but Xiao Zhang's is far more compelling.
I. The likelihood of Xiao Li winning the lawsuit is high, whereas for Xiao Zhang, it is not the case.
J. Both individuals are equally likely to lose their respective cases.
Answer: I
Cofactors are extremely important for enzyme activity. What is typically used as a cofactor for carboxylase? And what serves as a cofactor for decarboxylase?

Options:
A. Biotin; Nicotinamide Adenine Dinucleotide (NAD)
B. Niacinamide; Tetrahydrofolate
C. Pyridoxal phosphate (PLP); Pantothenic acid D
D. Biotin; Riboflavin Phosphate (FMN)
E. Biotin; Cytosine Triphosphate Pyrophosphate Thiamine (CTPP)
F. Biotin; Adenosine Triphosphate (ATP)
G. Biotin; Flavin Adenine Dinucleotide (FAD)
H. Biotin; Guanosine Triphosphate (GTP)
I. Biotin; Uridine Monophosphate (UMP)
J. Lipoic acid: Cobalamin (Vitamin B12)
Answer: E
Which type of vitamin, when consumed in excess, accumulates less in the body?

Options:
A. Vitamin H
B. Vitamin K
C. Vitamin C
D. Vitamin L
E. Vitamin F
F. Vitamin B
G. Vitamin D
H. Vitamin A
I. Vitamin P
J. Vitamin E
Answer: F
In plant tissue culture, theoretically, which hormone formula is conducive to bud formation?

Options:
A. 6-BA:IAA=1:4
B. 6-BA:IAA=2:1
C. 6-BA:IAA=1:2
D. 6-BA:IAA=5:1
E. 6-BA:IAA=3:2
F. 6-BA:IAA=1:1
G. 6-BA:IAA=2:2
H. 6-BA:IAA=1:3
I. 6-BA:IAA=3:1
J. 6-BA:IAA=4:1
Answer: I
In a certain rainforest system, there are four species of rodents with average individual biomasses of 20, 30, 40, and 50 grams, respectively. Under the condition that all other factors remain the same, which species has the highest relative abundance?

Options:
A. A species with a biomass of 40 grams
B. A species with a biomass of 35 grams
C. A species with a biomass of 60 grams
D. A species with a biomass of 25 grams
E. A species with a biomass of 20 grams
F. A species with a biomass of 45 grams
G. Species with a biomass of 50 grams
H. Species with a biomass of 70 grams
I. A species with a biomass of 30 grams
J. Species with a biomass of 80 grams
Answer: E
The enzyme that catalyzes peptide bond formation is called peptidyl transferase because it is capable of?

Options:
A. Transfer the growing peptide chain to a new amino acid.
B. Transferring amino acids from tRNA to an elongating peptide chain
C. Moving the peptide chain directly into the nucleus.
D. Attaching the peptide chain to DNA for replication purposes.
E. Breaking down the peptide chain for recycling materials.
F. Transferring the growing peptide chain to the ribosomal subunit.
G. Transfer the peptide chain from one amino acid to another amino acid.
H. Transfer the peptide chain from one ribosome to another charged tRNA.
I. Transferring the peptide chain to a free-floating amino acid.
J. Joining two ribosomes to facilitate peptide synthesis.
Answer: A
The distance between genes A and B is known to be 20 centimorgans, and the recombination value measured from the hybrid offspring is 18%. This indicates that the double crossover value between them is approximately?

Options:
A. 8%
B. 7%
C. 4%
D. 2%
E. Unable to determine
F. 3%
G. 1%
H. 10%
I. 6%
J. 5%
Answer: G
During the development of the sea pancake, it undergoes a larval stage known as the veliger. What is a veliger?

Options:
A. Bilateral symmetry
B. Segmented body larva phase
C. Radially symmetrical larva
D. Radial Symmetry
E. Symmetrical on both sides
F. Transitional polyp structure
G. Spirally coiled shell stage
H. Sessile adult stage structure
I. Free-swimming juvenile phase
J. Pentaradial symmetry
Answer: E
Before the germination of barley seeds, the embryo is removed, and the starch within the seed cannot be hydrolyzed. What can be applied to most effectively restore the seed's ability to hydrolyze starch?

Options:
A. Auxins
B. Prolamin
C. Ethylene
D. Photosynthase
E. Gibberellin
F. Amylase
G. DNA enzyme
H. Lipidase
I. Polysaccharide
J. Carbohydrates
Answer: E
What type of ovary does the sunflower's fruit develop from?

Options:
A. With 1 carpel, 1 chamber, and 1 ovule, a superior ovary
B. An inferior ovary with 1 carpel, 2 locules, and 1 ovule
C. An inferior ovary with 2 carpels, 1 locule, and 1 ovule
D. An inferior ovary with 1 carpel, 1 locule, and no ovules
E. An superior ovary with 3 carpels, 1 locule, and 2 ovules
F. A hypogynous ovary with 2 carpels, 1 locule, and 2 ovules
G. An superior ovary with 2 carpels, 1 locule, and 1 ovule
H. A superior ovary with 1 carpel, 1 locule, and 2 ovules
I. A hypogynous ovary with 1 carpel, 1 locule, and 1 ovule.
J. An inferior ovary with 2 carpels, 2 locules, and 1 ovule
Answer: C
The following are five types of animals: a. Tuna; b. Sturgeon; c. Australian lungfish; d. Silver chimaera; e. Bowfin. In the order of the names listed above, which option corresponds to their correct classifications?

Options:
A. Teleosts: osteichthyans (bony fish with cartilage-covered scales); Sarcopterygii: cartilaginous fish; Holostei: fully ossified fish
B. Cartilaginous fish; True bony fish; Lobe-finned fish; Cartilaginous ganoid fish; Holosteans
C. Cartilaginous and bony fish species: cartilaginous fish; holostean fish; lobe-finned fish; teleost fish
D. Teleosts: osteichthyans (bony fish with no scales); Sarcopterygii: primitive cartilaginous fish; Holostei: ossified ganoid-likes
E. Cartilaginous fish: semi-teleosts; Sarcopterygii: ossified cartilage; Holostei: advanced skeletal structure
F. Osteichthyans: cartilaginous predators; Sarcopterygii: armored ganoid; Holostei: simple-finned bony species
G. Cartilaginous fish: advanced teleosts; Sarcopterygii: semi-cartilaginous species; Holostei: mixed-scale structure
H. Holostei: Sarcopterygii: Chondrostei; Chondrichthyes: Teleostei
I. Osteichthyans: cartilaginous ganoid fish; Sarcopterygii: bony fish; Holostei: fully ossified cartilage
J. Teleosts: jawless fishes with cartilaginous skeletons; Sarcopterygii: hard-scaled fish; Holostei: ganoid-bodied cartilage
Answer: A
What are the main forms of transport for oxygen and carbon dioxide in the blood?

Options:
A. Physical dissolution; carbaminohemoglobin
B. Oxyhemoglobin; Plasma glucose
C. Dissolved oxygen; Nitrogen bubbles
D. Oxyhemoglobin; Carbon monoxide
E. Bicarbonate; Water vapor
F. Oxyhemoglobin: Physical Dissolution
G. Oxyhemoglobin: Bicarbonate
H. Myoglobin; Hemoglobin
I. Oxyhemoglobin; Carbaminohemoglobin
J. Plasma proteins; Carbaminohemoglobin
Answer: G
In a certain organism, the following metabolic pathway exists, and loci G1 and G2 are located on different linkage groups. A test cross is performed with a double heterozygote for the recessive mutations g1g1 and g2g2. What proportion of the offspring can produce substance C?

Options:
A. 3/8
B. 5/16
C. 1/4
D. 9/16
E. 7/16
F. 3/4
G. 1/16
H. 1/8
I. 1/2
J. 3/16
Answer: C
Which of the following statements about ATP synthesis in animal cells is correct?

Options:
A. ATP synthesis does not require energy from glucose metabolism.
B. ATP synthesis occurs exclusively in the nucleus.
C. ATP can be synthesized both in mitochondria and the cytoplasm.
D. ATP is synthesized only during protein translation.
E. ATP is synthesized in lysosomes during cellular respiration.
F. ATP synthesis is a function of the endoplasmic reticulum.
G. ATP is synthesized only in the chloroplasts of animal cells.
H. ATP is synthesized in the cytoplasm.
I. ATP is synthesized in the Golgi apparatus.
J. ATP is synthesized in the mitochondria.
Answer: C
If NADH(H+) and FADH2 are calculated as 2.5 and 1.5 ATP respectively, how many ATP molecules are released when one molecule of pyruvate is completely oxidized into CO2 and H2O?

Options:
A. 16
B. 10
C. 17
D. 12
E. 12. 5
F. 13
G. 14
H. 14. 5
I. 15
J. 11. 5
Answer: I
The island is densely and evenly covered with cormorant nests. What type of behavior ensures the establishment and maintenance of their territories?

Options:
A. Friendly Behavior
B. Foraging behavior
C. Performative Acts
D. Defensive Behavior
E. Aggressive Behavior
F. Migration Behavior
G. Parental Care Behavior
H. Mating Behavior
I. Cooperative Behavior
J. Altruistic Behavior
Answer: E
Primarily distributed along the eastern coast of the Eurasian continent, summers are hot and rainy, while winters are cold with little rainfall. The annual precipitation ranges between 1000 and 1500mm. The forest structure is simple and not very tall, mainly composed of species from the Fagaceae, Lauraceae, and Theaceae families. What type of forest is this?

Options:
A. Subarctic Coniferous Forests
B. Tropical Savanna
C. Forest-Grassland Transition Zone
D. Tropical Monsoon Forest
E. Monsoon Evergreen Forest
F. Temperate Grassland
G. Northern Coniferous Forest
H. Subtropical Evergreen Broadleaf Forest
I. Tropical Rainforest
J. Temperate Deciduous Broadleaf Forest
Answer: H
What are the factors of α diversity of species in biodiversity in general?

Options:
A. Relative abundance and evenness of species
B. Species Diversity and Distribution
C. Relative Diversity and Abundance of Species
D. Species Evenness and Heterogeneity
E. Relative abundance and abundance of species
F. Relative abundance and distribution of species
G. Relative Species Abundance and Diversity
H. Range and density of species distribution
I. Species Diversity and Integration
J. Species Richness and Evenness
Answer: A
What are the two driving forces behind population and species evolution?

Options:
A. Genetic Mutations and Natural Selection
B. Gene Flow and Genetic Mutation
C. Genetic Variation and Natural Selection
D. Genetic Variation and Biological Competition
E. Geographical Differentiation and Directional Selection
F. Genetic Drift and Natural Selection
G. Geographic Isolation and Natural Selection
H. Gene Mutation and Gene Flow
I. Mutation Accumulation and Population Selection
J. Genetic Drift and Trait Differentiation
Answer: F
Given that the concentration of the anti-competitive inhibitor [I] = $\mathrm{K_i}$ and the substrate concentration [S] = 5$\mathrm{K_m}$, what is the reaction rate $v$ calculated according to the Michaelis-Menten equation?

Options:
A. $\mathrm{\frac{3V_{max}}{16}}$
B. $\mathrm{\frac{V_{max}}{4}}$
C. $\mathrm{\frac{9V_{max}}{32}}$
D. $\mathrm{\frac{2V_{max}}{8}}$
E. $\mathrm{\frac{7V_{max}}{16}}$
F. $\mathrm{\frac{3V_{max}}{8}}$
G. $\mathrm{\frac{4V_{max}}{16}}$
H. $\mathrm{\frac{5V_{max}}{16}}$
I. $\mathrm{\frac{11V_{max}}{32}}$
J. $\mathrm{\frac{5V_{max}}{8}}$
Answer: H
What is the reason for reduced urine output during excessive sweating?

Options:
A. Plasma crystalloid osmotic pressure decreases, inhibiting antidiuretic hormone secretion.
B. Reduced blood volume leads to an increase in glomerular filtration.
C. The decrease in plasma crystalloid osmotic pressure triggers the secretion of antidiuretic hormone.
D. Increased plasma osmolality leads to the secretion of antidiuretic hormone.
E. Reduction in plasma sodium ions triggers the secretion of antidiuretic hormone.
F. The increase in blood volume leads to a decrease in glomerular filtration.
G. Sympathetic nerve inhibition leads to a reduction in antidiuretic hormone.
H. Plasma protein concentration increases, leading to enhanced secretion of antidiuretic hormone.
I. Sympathetic nervous system excitation induces the secretion of antidiuretic hormone.
J. Reduced blood volume leads to decreased glomerular filtration.
Answer: D
When using the variance-to-mean ratio to detect biological distribution patterns, which of the following ratios indicates a uniform distribution?

Options:
A. $\mathrm{s^2}$/m=0
B. $\mathrm{s^2}$/m = less than 0
C. $\mathrm{s^2}$/m = just slightly greater than 1
D. $\mathrm{s^2}$/m = much greater than 1
E. $\mathrm{s^2}$/m = exactly equal to 1
F. $\mathrm{s^2}$/m approaches 0.5
G. $\mathrm{s^2}$/m is slightly less than 1
H. $\mathrm{s^2}$/m=1
I. $\mathrm{s^2}$/m = greater than 2
J. $\mathrm{s^2}$/m is significantly greater than 1.
Answer: A
The successional process of "bare rock → lichen → moss → herbaceous plants → shrubs → trees" can be categorized as?

Options:
A. Secondary Succession
B. Heterotrophic Succession
C. Trophic Cascade Cycle
D. Primary Succession
E. Cyclic Succession
F. Heterotrophic succession
G. Autotrophic Succession
H. Catastrophic Succession
I. Quick substitution
Answer: G
What are the developmental processes of Orthoptera and Lepidoptera insects?

Options:
A. Direct Metamorphosis, Ectometabolism
B. Hemimetabolous metamorphosis, gradual metamorphosis
C. Proto-Metamorphosis, Hypermetabolism
D. Gradual Metamorphosis, Hemimetabolism
E. Instant Metamorphosis, Endopterygotes
F. Ametamorphosis, incomplete metamorphosis
G. Incomplete Metamorphosis, Neometabolism
H. Complete metamorphosis, incomplete metamorphosis
I. Abrupt Metamorphosis, Protopod
J. Complete Metamorphosis, Ametabolism
Answer: D
A female fruit fly with a genotype of AaBbCcDd can produce 16 types of gametes, with the types and proportions as follows:
ABCD: 18% aBCD: 18% Abcd: 18% abcd: 18% ABcd: 2% aBcd: 2% AbCD: 2% abCD: 2% ABCd: 4% aBCd: 4% AbcD: 4% abcD: 4% ABcD: 1% abcD: 1% ABCd: 1% abCd: 1%
The relationship between the four genes is:

Options:
A. AC/ac BD/bd
B. Ac/ac BD/bc
C. AB Cd ab cd
D. A/a BCD/bcd
E. A/B CD cd/bc
F. ABC/abc D/d
G. AB/ab C/D
H. AB/ab CD/cd
I. Aa Bb Cc Dd
J. A/a BC/bc Dd/dd
Answer: D
In the bacterial growth curve, at which point are the total bacterial count and the viable bacterial count nearly equal?

Options:
A. Grace Period
B. Logarithmic phase
C. Stable Period
D. Juvenile stage
E. Exponential decline phase
F. Inoculation phase
G. Period of Death
H. Mid-stationary phase
I. Early stationary phase
J. Quiescent growth phase
Answer: B
Microevolution refers to evolutionary changes at the level of individuals and populations within a species. What is the basic unit of microevolution?

Options:
A. Population
B. Individual Genotype
C. Cellular Process
D. Genetic Mutation
E. Genetic Variation
F. Specific Ecosystem
G. Environmental Shift
H. Biological Individual
I. Species as a Whole
J. Individual Phenotype
Answer: A
Which DNA molecule with the following different base composition ratios has the highest Tm value?

Options:
A. C+T=30%
B. G+C=35%
C. G+T=50%
D. G+C=20%
E. A+G=10%
F. A+T=15%
G. C+C=40%
H. A+T=80%
I. C+C=25%
J. A+C=25%
Answer: F
The scales of carp and snakes play an important protective role for the body. What is correct regarding the position and origin of these scales?

Options:
A. The scales of both carp and snakes are located within the epidermis and are derived from the dermis.
B. The scales of the carp and the snake are located beneath the epidermis and beneath the epidermis, respectively, and are derived from the epidermis and the dermis, respectively.
C. The scales of the carp and the snake are located outside the epidermis and beneath the epidermis, respectively, and are derived from the epidermis and the dermis, respectively.
D. The scales of both carp and snakes are located outside the epidermis and are derived from the dermis.
E. The scales of both the carp and the snake are located beneath the epidermis but are derived from the epidermis.
F. The scales of carp and snakes are located outside the epidermis and outside the dermis, respectively, and are derived from the epidermis and the dermis, respectively.
G. The scales of both carp and snakes are located beneath the epidermis and are derived from the dermis.
H. The scales of carp and snakes are located beneath the epidermis and outside the epidermis, respectively, and are derived from the dermis and the epidermis, respectively.
I. The scales of carp and snakes are located outside the epidermis and outside the epidermis, respectively, and are derived from the dermis and the dermis, respectively.
J. The scales of both the carp and the snake are located on the innermost layer of their bodies and are derivatives of the epidermis.
Answer: H
The mean arterial pressure refers to the average value of arterial blood pressure at every moment during one cardiac cycle. So, what should its value be closer to?

Options:
A. The overall blood pressure peak
B. Arterial pressure at ventricular systole
C. The systolic and diastolic pressure product
D. Pulse pressure average
E. Diastolic pressure
F. A midpoint of blood pressure fluctuations
G. Systolic Blood Pressure
H. Systolic pressure and diastolic pressure sum
I. The difference between systolic pressure and diastolic pressure
J. Maximum systolic pressure
Answer: E
The number of unsaturated bonds in the fatty acids of the cell membrane lipids: the _____ the number, the lower the solidification temperature, and the stronger the cold resistance.

Options:
A. constant
B. limited
C. equal
D. unchanged
E. singular
F. minimal
G. lesser
H. reduced
I. higher
J. lowered
Answer: I
If global carbon dioxide levels increase and climate warming continues for 100 years, predict the "fate" of plants distributed at altitudes above 4,000 meters. In other words, what will happen to these plants after 100 years?

Options:
A. Adaptation will allow survival without any distribution change.
B. The upper limit of the distribution will shift downward.
C. These plants will thrive better at lower altitudes.
D. Perish due to the inability to adapt to change
E. The range of distribution will remain unchanged.
F. These plants will undergo complete extinction worldwide.
G. Climate change will not impact these high-altitude species.
H. The upper limit of the distribution will shift upward.
I. The range of these plants will shrink in both directions.
J. The plants will migrate to lowland ecosystems instead.
Answer: H
Research has found that dibenzocyclooctadiene lignans are considered characteristic chemical components of plants in the genus Schisandra, known for their hepatoprotective effects, antioxidant properties, anti-tumor activities, and anti-HIV activities. According to the Cronquist classification system, to which subclass under the phylum of dicotyledonous plants does this family belong?

Options:
A. Caryophyllidae
B. Magnoliids
C. Fabidae
D. Asteridae
E. Rosidae
F. Commelinidae
G. Ranunculids
H. Hamamelidae Subclass
I. Dilleniidae
J. Subclass Pentapetalae
Answer: B
Which group of organisms belongs to prokaryotes?

Options:
A. Viruses and Mycoplasma
B. Actinomycetes and Molds
C. Protozoa and Archaea
D. Bacteria and Candida
E. Fungi and Protozoa
F. Algae and Bacteria
G. Viruses and Actinomycetes
H. Cyanobacteria and Fungi
I. Mycoplasma and Algae
J. Archaea and Molds
Answer: D
Divide a willow twig into four segments. Let A and B represent the morphological upper and lower ends of the twig, respectively, and hang them in a humid environment. After several days, which type of twig will not sprout buds at its morphological upper end?

Options:
A. B side facing up, A side facing down
B. Place the willow twigs horizontally after cutting them into segments.
C. Hang the twig with end B exposed to sunlight.
D. Position the twig with side A submerged in water.
E. Coat the A-side with rooting powder and seal.
F. Apply 1AA lanolin to the B-side.
G. Add 2AA growth hormone to the A-side.
H. Cover the twigs with dry cloth horizontally.
I. End A facing up, end B facing down.
J. Apply water lanolin to the A-side.
Answer: F
What similarities exist between chordates and advanced invertebrates?

Options:
A. Hemocoel, asymmetrical body plan, radial symmetry
B. Notochord, deuterostome, and true coelom
C. Deuterostome, trilaminar germ layers, true coelom
D. Closed circulatory system, exoskeleton, segmentation
E. Spiral cleavage, pseudocoelom, chitinous body covering
F. External fertilization, hydrostatic skeleton, open circulation
G. Cartilage skeleton, radial body plan, non-segmented body
H. Digestive system, radial cleavage, metamorphic growth
I. Neural tube, three germ layers, bilateral symmetry
J. Pharyngeal gill slits, bilateral symmetry, true coelom
Answer: C
After processing for one hour, adding iodine solution will cause the contents of which test tube to turn blue?

Options:
A. After thoroughly mixing rice soup with a certain amount of cooled boiled water, place the test tube in water maintained at 37°C.
B. Combine mashed potatoes with some digestive enzymes, then maintain the test tube in water at 25°C.
C. Mix glucose solution with the same amount of hot water, then leave the test tube at 50°C.
D. Stir oatmeal with a certain amount of herbal infusion, placing the test tube in ice-cold water.
E. Blend protein shake with a measured amount of lukewarm water, and keep the test tube at 40°C.
F. Blend wheat flour with a calculated amount of artificial saliva, keeping the test tube at 30°C.
G. After mixing the essential amino acids required by the human body with a certain amount of saliva, place the test tube in water maintained at 37°C.
H. Rice soup is thoroughly mixed with a certain amount of saliva and the test tube is kept in water at 37°C.
I. Mix starch with a certain amount of saliva, then place the test tube in water at room temperature.
J. After being mixed with a certain amount of saliva, the lipid droplets are kept in a test tube in water at 37°C.
Answer: A
The dynamics of biological communities should encompass at least three aspects: aside from the internal dynamics of the community, what else is involved in the evolution of biological communities on Earth?

Options:
A. Dispersal of Communities
B. Habitat Partitioning Mechanisms
C. Interference
D. Mutation within Population
E. Succession of Communities
F. Seasonal Changes
G. Competitive Relationships
H. Population Migration Patterns
I. Food Web Alterations
J. Annual Changes in Biological Communities
Answer: E
What is required in the process of developing the megaspore into the embryo sac during the sexual reproduction of angiosperms?

Options:
A. One mitosis produces six nuclei.
B. The second meiotic division forms six nuclei.
C. Three rounds of meiosis result in the formation of four nuclei.
D. Three rounds of mitosis result in the formation of 8 nuclei.
E. One meiosis and two mitoses result in the formation of 8 nuclei.
F. Two rounds of mitosis result in the formation of four nuclei.
G. Meiosis II results in the formation of 8 nuclei.
H. One meiotic division results in the formation of two nuclei.
I. Four mitotic divisions result in the formation of eight nuclei.
J. Meiosis forms four nuclei.
Answer: D
Which of the following ferns has an aerial stem with a distinct division between nodes and internodes?

Options:
A. Cibotium barometz
B. Selaginella
C. Floatingfern
D. Quillwort
E. Clubmoss
F. Mangqi
G. Scouring Rush
H. Maidenhair Fern
I. Phoenix Tail Fern
J. Staghorn Fern
Answer: G
Prolonged malnutrition can cause edema. What is the main reason for this?

Options:
A. Plasma protein concentration is too low
B. The osmotic pressure of plasma crystalloid remains unchanged.
C. The plasma protein concentration remains unchanged.
D. Plasma protein concentration is too high
E. The colloid osmotic pressure of plasma is too high.
F. Plasma crystalloid osmotic pressure is too high
G. Plasma crystalloid osmotic pressure too low
H. The plasma protein concentration is initially high and then decreases.
I. The colloid osmotic pressure of plasma remains constant.
J. Plasma colloid osmotic pressure too low
Answer: J
A student who enjoys fishing once caught two scaleless grass carp from a lake. She wanted to determine whether this unique phenotype had a genetic basis. When she crossed the scaleless fish with homozygous wild-type fish, the $\mathrm{F_1}$ generation exhibited two phenotypes: 50% were scaled fish, while the other 50% had a single row of scales on each side. Why?

Options:
A. The gene that determines the characteristic of scales is certainly not the only one.
B. The inheritance of scale traits does not follow Mendel's laws.
C. A single genetic trait determines all subsequent phenotypes.
D. Having scales is a dominant trait.
E. All offspring should exhibit the recessive scale trait.
F. A single gene mutation explains all observed phenomena.
G. The trait of scales is controlled by a single gene and exhibits complete dominance.
H. The gene determining the trait of scales may be only one.
I. The presence of scales is a recessive trait.
J. The proportion of offspring shows that the dominant trait completely suppresses the recessive trait.
Answer: A
Microtubules are an important type of cytoskeletal structure. Among the following microtubule-related cellular structures, which group exhibits a similar structural pattern of microtubules?

Options:
A. Flagella and Spindle Apparatus
B. Elementary Particles and Intermediate Particles
C. Cilia and Centrioles
D. Flagella and Basal Bodies
E. Cilia and Intermediate Granules
F. Centrosome and Spindle Apparatus
G. Elementary Particles and Centrosomes
H. Intermediate Particles and Spindle Apparatus
I. Cilia and Basal Bodies
J. Flagella and Centrioles
Answer: B
Which of the following descriptions about the chemiosmotic theory is correct?

Options:
A. $\mathrm{H^+}$ is primarily present in the space outside the mitochondria, while $\mathrm{OH^-}$ is absent near the membrane.
B. $\mathrm{OH^-}$ is actively transported and accumulates outside the mitochondria, while $\mathrm{H^+}$ diffuses and distributes evenly.
C. $\mathrm{OH^-}$ and $\mathrm{H^+}$ are both concentrated outside the mitochondrial inner membrane.
D. $\mathrm{OH^-}$ accumulates outside the mitochondrial inner membrane, while $\mathrm{H^+}$ accumulates inside the membrane.
E. Both $\mathrm{OH^-}$ and $\mathrm{H^+}$ are concentrated on the inner mitochondrial membrane.
F. The concentration of $\mathrm{H^+}$ is the same on the inner membrane and the outer membrane, while $\mathrm{OH^-}$ is higher outside the inner membrane.
G. $\mathrm{H^+}$ accumulates on the inner side of the mitochondrial membrane, while $\mathrm{OH^-}$ accumulates on the outer side of the membrane.
H. $\mathrm{H^+}$ accumulates outside the inner membrane of the mitochondria, while $\mathrm{OH^-}$ accumulates inside the membrane.
I. $\mathrm{H^+}$ and $\mathrm{OH^-}$ are evenly distributed on both sides of the inner and outer membranes.
J. $\mathrm{OH^-}$ accumulates in the middle of the mitochondrial inner membrane, while $\mathrm{H^+}$ accumulates at both ends of the membrane.
Answer: H
A married couple had a son (named "OK"), and little "OK" not only is colorblind, but also has an extra sex chromosome compared to other boys: XY. Medical examinations show that "OK"'s parents have normal vision and no hereditary or other diseases. How did little "OK" acquire this genetic condition?

Options:
A. The problem lies with the "OK" father. During the second meiotic division of spermatocytes, the XX chromosomes failed to separate and instead both entered the same daughter cell.
B. The problem lies with the "OK" father. During the first meiotic division of spermatogonia, the XX chromosomes failed to separate and both entered the same daughter cell.
C. The issue lies with the "OK" mother. During the second meiotic division of the oocyte, the X and Y chromosomes failed to separate and entered the same daughter cell together.
D. The issue lies with the "OK" mother. During the second meiotic division of the oocyte, the XX chromosomes failed to separate and both ended up in the same daughter cell.
E. The problem lies with the "OK" mother. During the normal division of the oocyte, the YY chromosomes simultaneously enter the same daughter cell.
F. The issue lies with the "OK" father. During the second meiotic division of spermatocytes, the XY chromosomes did not separate and directly entered the same daughter cell.
G. The problem lies with the "OK" father. During the first meiotic division of the primary spermatocyte, the X and Y chromosomes failed to separate and directly entered the same daughter cell.
H. The issue lies with the "OK" father. During normal spermatocyte division, the X and Y chromosomes simultaneously enter the same daughter cell.
I. The issue lies with the "OK" mother. During the first meiotic division of the oocyte, the XY chromosomes failed to separate and entered the same daughter cell together.
J. The issue lies with the "OK" mother. During the first meiotic division of the oocyte, the XX chromosomes failed to separate and directly entered the same daughter cell.
Answer: D
Add the two oligonucleotide fragments 5'-ACCACGTAACGGA-3' and 5'-GTTAC-3' together with DNA polymerase into a reaction mixture containing dATP, dGTP, dCTP, and dTTP. Predict the ratio of the incorporated bases in the final reaction product.

Options:
A. 3G:1A:1T
B. 2G:1A
C. 3G:2T
D. 3G:2C:1T
E. 1G:1T
F. 3G:3T:2C
G. 2G:3T:1C
H. 4G:2T
I. 2C:1T
J. 5T:4G:3C:1A
Answer: C
In intact isolated mitochondria, in the presence of oxidizable substrates, which substance can increase electron transfer and oxygen uptake?

Options:
A. $\mathrm{FADH_2}$
B. Succinate
C. Coenzyme Q
D. Enzymes of the TCA Cycle
E. ADP
F. Citrate
G. Cytochrome c
H. Pyruvate
I. NADH
J. ATP
Answer: E
To establish the ethogram of the saiga antelope, researchers observed and recorded the behaviors of a group of antelopes one by one at fixed time intervals. What type of sampling method in ethological research does this represent?

Options:
A. Event-based Sampling
B. Time-point Sampling
C. Interval Sampling
D. Scanning Sampling
E. Target Sampling
F. Random Sampling
G. Behavior Sampling
H. Group-centered Sampling
I. Focus Sampling
J. Sequential Sampling
Answer: D
What are the changes in heart rate, cardiac output, blood pressure, and respiration caused by pressing on the carotid sinus and carotid body of the test rabbit?

Options:
A. Heart rate stabilizes, cardiac output rises, blood pressure increases, and respiration speeds up
B. Increased heart rate, reduced cardiac output, lowered blood pressure, and deeper breathing
C. Slower heart rate, reduced cardiac output, lowered blood pressure, deeper and faster breathing
D. Increased heart rate, elevated cardiac output, raised blood pressure, and slowed breathing.
E. Reduced heart rate, higher cardiac output, unchanged blood pressure, and slowed breathing
F. Increased heart rate, higher cardiac output, elevated blood pressure, and deeper, faster breathing
G. Rapid heart rate, decreased cardiac output, elevated blood pressure, and shallow breathing
H. Increased heart rate, decreased cardiac output, elevated blood pressure, and faster breathing
I. Heart rate slows down, cardiac output decreases, blood pressure lowers, and respiration slows down.
J. Heart rate decreases, cardiac output stabilizes, blood pressure rises, and breathing remains steady
Answer: C
What is the umbrella-shaped fruiting body of a mushroom also called?

Options:
A. Hyphal cluster
B. Subconstellation
C. Zi Nang Guo
D. Canopy shell
E. Mycelium top
F. Clustered fruit
G. Seed layer
H. Fungal bloom
I. Spore bundle
J. Cap structure
Answer: F
The thoracic cavity is a progressive characteristic in the evolutionary process of animals. In which class did it first appear?

Options:
A. Reptiles
B. Fish
C. Crustaceans
D. Mammals
E. Arachnids
F. Insects
G. Birds
H. Mollusks
I. Amphibians
Answer: A
What type of division occurs when the unicellular pollen further develops into a generative cell and a vegetative cell?

Options:
A. Asymmetric Mitosis
B. Asymmetric Meiosis
C. Asymmetric Nuclear Division
D. Symmetric Binary Fission
E. Asymmetric Cytokinesis
F. Asymmetric Binary Fission
G. Symmetric Meiosis
H. Symmetrical Mitosis
I. Symmetric Nuclear Division
J. Symmetric Cytokinesis
Answer: A
Which of the following descriptions regarding the distinguishing characteristics of Bryopsida (mosses) and Hepaticopsida (liverworts) is correct?

Options:
A. The sporangium of mosses in the Hepaticopsida lacks a calyptra, peristome teeth, and annular structure, whereas the sporangium of mosses in the Bryopsida possesses a calyptra, peristome teeth, and annular structure.
B. The sporophyte of bryophytes is composed only of the sporangium and the foot, while the sporophyte of mosses is composed of the sporangium, seta, and foot.
C. The archegonia in mosses are borne on specialized stalks in Bryopsida but are embedded in thalloid structures in Hepaticopsida.
D. The rhizoids of the gametophyte in bryophytes of the Hepaticae class are unicellular, while the rhizoids in the Musci class are composed of a single row of cells and are unbranched.
E. The gametophytes of Hepaticopsida always develop from a protonema, while gametophytes of Bryopsida directly arise from a spore.
F. The leaves or leaf-like structures of gametophytes in Bryophyta lack a midrib; however, the vast majority of leaves in Musci possess a costa.
G. Bryophytes within Hepaticopsida exhibit alternating heteromorphic generations, while Bryopsida exhibits strictly homomorphic generations.
H. The reproduction of Hepaticopsida primarily depends on gemmae cups, and Bryopsida lacks structures for vegetative reproduction.
I. Spore dispersal in Hepaticopsida is facilitated by elaters, while spore dispersal in Bryopsida depends on hygroscopic peristome teeth.
J. Hepaticopsida produces a sporophyte with photosynthetic capabilities, unlike the non-photosynthetic sporophyte of Bryopsida.
Answer: A
In the primary structure of the root, which tissue serves a protective function?

Options:
A. Vascular cambium
B. Inner cortex
C. Leaf mesophyll
D. Endothelium
E. Epidermis
F. Root hairs
G. Pericycle
H. Middle cortex
I. Outer cortex
J. Secondary xylem
Answer: I
During the development of the Pinus tabulaeformis (Chinese pine), which of the following cells have a haploid number of chromosomes?

Options:
A. Germ cell
B. Epidermal cell
C. Granule cell
D. Hypodermal cell
E. Endosperm cell
F. Guard cell
Answer: E
In the late stages of seed development, which proteins are associated with the dehydration process?

Options:
A. Lignin-modified proteins aid desiccation resistance
B. Late embryonic defense proteins evolve for protection
C. Early embryonic development triggers several proteins
D. ABA Binding Protein
E. GTP-binding protein
F. Calmodulin
G. Dormancy-related proteins activate during late stages
H. Development-specific dehydration-linked proteins form
I. Late embryonic development enriched with proteins
J. Hormone-regulated proteins dominate seed dehydration
Answer: I
The balance of plant hormones plays a regulatory role in maintaining and breaking bud dormancy. What plays the main role in this process?

Options:
A. ABA and Eth
B. CTK and GA
C. Aux and ABA
D. IAA and CTK
E. GA and CK
F. TAA and GA
G. Eth and TAA
H. CTK and Eth
I. ABA and GA
J. GA and Eth
Answer: I
When 20 mol% $YF_3$ is added to $CaF_2$ to form a solid solution, the experimentally measured density of the solid solution is 3.64 g/cm³, and the lattice parameter at this time is $a=0.55$ nm. (The atomic masses are as follows: Y: 88.9, Ca: 40.0, F: 19.0.) Calculate and determine what type of solid solution this is ().

Options:
A. Formation of a Covalent Solid Solution.
B. Formation of a Defect-Induced Solid Solution.
C. Formation of a Polymorphic Solid Solution.
D. Formation of an Amorphous Solid Solution.
E. Formation of an Interstitial Solid Solution.
F. Formation of a Phase-Separated Mixture.
G. Formation of a Metallic Solid Solution.
H. Formation of an Ionic Solid Solution.
I. Formation of a Non-Crystalline Solid Solution.
J. Formation of a Substitutional Solid Solution.
Answer: E
Your statistical mechanics class has twenty-five students. What is the probability that at least two classmates have the same birthday?

Options:
A. 0.5837
B. 0.7214
C. 0.4587
D. 0.4712
E. 0.6894
F. 0.5128
G. 0.6105
H. 0.4925
I. 0.5343
J. 0.6421
Answer: A
You play a slot machine in Las Vegas. For every \$1 coin you insert, there are three outcomes: (1) you lose \$1, (2) you win \$1, so your profit is \$0, (3) you win \$5, so your profit is \$4.Suppose you find that your average expected profit over many trials is \$0. (What an optimist!) What is the maximum entropy distribution for the probabilities $p_{1}$, $p_{2}$ and $p_{3}$ of observing each of these three outcomes?

Options:
A. $$
p_{1}=0.500
$$
$$
p_{2}=0.350
$$
$$
p_{3}=0.150
$$
B. $$
p_{1}=0.510
$$
$$
p_{2}=0.370
$$
$$
p_{3}=0.120
$$
C. $$
p_{1}=0.450
$$
$$
p_{2}=0.400
$$
$$
p_{3}=0.150
$$
D. $$
p_{1}=0.495
$$
$$
p_{2}=0.380
$$
$$
p_{3}=0.125
$$
E. $$
p_{1}=0.520
$$
$$
p_{2}=0.360
$$
$$
p_{3}=0.120
$$
F. $$
p_{1}=0.480
$$
$$
p_{2}=0.390
$$
$$
p_{3}=0.130
$$
G. $$
p_{1}=0.490
$$
$$
p_{2}=0.385
$$
$$
p_{3}=0.125
$$
H. $$
p_{1}=0.505
$$
$$
p_{2}=0.375
$$
$$
p_{3}=0.120
$$
I. $$
p_{1}=0.498
$$
$$
p_{2}=0.377
$$
$$
p_{3}=0.125
$$
J. $$
p_{1}=0.470
$$
$$
p_{2}=0.420
$$
$$
p_{3}=0.110
$$
Answer: I
Experiments show that the retractive force $f$ of polymeric elastomers as a function of tem- perature $T$ and expansion $L$ is approximately given by $f ( T, L )=a T ( L-L_{0} )$ where $a$ and $L_{0}$ are constants.
Use Maxwell relations to determine the entropy and enthalpy, $S ( L )$ at constant $T$ and $p$ ,$S ( L )$ is ().

Options:
A. $$
S ( L )=\frac{-a ( L_{0}-L )^{2}} {2}
$$
B. $$
S ( L )=\frac{-a ( L-L_{0} )^{2}} {2}
$$
C. $$
S ( L )=a ( L-L_{0} )^{2}
$$
D. $$
S ( L )=\frac{-a ( L-L_{0} )^{2} T} {2T}
$$
E. $$
S ( L )=\frac{-a ( L+L_{0} )^{2}} {2}
$$
F. $$
S ( L )=\frac{-a T ( L-L_{0} )^{2}} {2}
$$
G. $$
S ( L )=\frac{-a ( L-L_{0} )} {2}
$$
H. $$
S ( L )=\frac{-a ( L-L_{0} )^{3}} {3}
$$
I. $$
S ( L )=\frac{a ( L-L_{0} )^{2} T} {2}
$$
J. $$
S ( L )=\frac{a ( L-L_{0} )^{2}} {2}
$$
Answer: B