MikaStars39/nano-eval huggingface dataset:
Sorghum is a widely cultivated plant in China, scientifically known as *Sorghum bicolor* (L.) Moench. Which of the following statements is correct?

Options:
A. Sorghum is scientifically recognized as *Zea mays*.
B. Sorghum belongs to the genus *bicolor*.
C. The revised name of sorghum is *Triticum aestivum*.
D. Sorghum is classified under the genus *Oryza*.
E. The botanical name of sorghum has been revised.
F. Sorghum's official name was never altered or modified.
G. The scientific name of sorghum is *Sorghum vulgare*.
H. The name of sorghum is bicolor.
I. Sorghum belongs to the genus *Sorghum Moench*.
J. Sorghum is mostly cultivated for ornamental purposes only.
Answer: E
In biodiversity measurement indices, different diversity indices (α, β, γ, δ) are suitable for indicating species diversity at different spatial scales. What does β diversity usually measure?

Options:
A. Within the biogeographic region
B. Landscape connectivity
C. Habitat transition
D. Landscape Interval
E. Ecosystem structure
F. Genetic variation
G. Resource allocation
H. Individual abundance
I. Species population
J. Community dominance
Answer: C
The feather color in a type of parrot is controlled by two pairs of non-allelic genes. The dominant allele Y guides the production of yellow pigment, while the dominant allele B guides the production of blue pigment. When yellow and blue pigments coexist, the feathers appear green; in the absence of pigment, the feathers appear white. A green parrot is crossed with a yellow parrot, and the phenotypic ratio of the offspring is 12 green: 4 blue: 13 yellow: 3 white. What are the genotypes of these two parrots?

Options:
A. YyBb and YyBb
B. Yybb and yyBB
C. Yybb and YyBB
D. YyBb and yyBb
E. YyBB and yybb
F. YyBb and YyBB
G. YyBb and Yybb
H. YyBb and YYBb
I. YYBB and YyBb
Answer: G
Which of the following cellular structural components is non-polar?

Options:
A. Mitochondrial cristae
B. Plasma membrane protein
C. Endoplasmic reticulum
D. Phospholipid Layer
E. Intermediate filament
F. Microtubule
G. Microfilament
H. Actin filament
I. Carbohydrate Chain
J. Stress fibers
Answer: E
Collect blood samples from malaria patients to prepare smears. Under the microscope, the nucleus of the trophozoite is observed, separated from the external environment outside the red blood cell by the basal red cell membrane?

Options:
A. Six layers
B. No layers
C. Seven layers
D. Four layers
E. Three layers
F. Single layer
G. Two layers
H. First Floor
I. Five layers
J. Multiple layers
Answer: G
In both somatic and reproductive cells of wild-type maize, starch is present. However, when the Wx gene is inactivated (mutated to wx), the wx wx homozygous cells do not contain starch, resulting in wrinkled fruit. The best method to determine the frequency of Wx mutation to wx in homozygous wild-type plants is

Options:
A. Perform a sectioning of the corn leaves and calculate the proportion of cells without starch grains.
B. Assess the absence of flower color change as an indicator of Wx gene mutation frequency.
C. Stain the pollen with iodine solution and count the proportion of unstained pollen.
D. Perform pollen hydration tests and determine the proportion of pollen grains that burst.
E. Collect hundreds of corn ears and count the proportion of shriveled kernels.
F. Stain roots of the plant with iodine solution and calculate the percentage of unstained root cells.
G. Use electrophoresis to separate kernel proteins and count the bands representing starch presence.
H. Stain the corn kernel surfaces with iodine solution and calculate the proportion of stained areas.
I. Extract the genome of each rice plant from the field separately, perform PCR amplification on the gene, and then sequence it to calculate the proportion of mutated sequences.
J. Survey the ear of the maize plant for mutations by analyzing visually mutated kernels.
Answer: C
What methods are commonly used to identify the contents of plant cells?

Options:
A. Using Iodine Solution to Identify Proteins and Fats
B. Using Sudan III to Identify Fats and Starch
C. Use Sudan III to identify proteins and starch.
D. Using Sudan III to Identify Proteins and Fats
E. Use iodine solution to identify proteins and starch.
F. Use Benedict's solution to identify proteins and starch.
G. Using methylene blue to identify proteins and starch.
H. Using iodine solution to identify fats and starch.
I. Using Iodine Solution to Identify Fats and Proteins.
J. Use iodine solution for proteins and carbohydrates.
Answer: E
A certain warm-temperature plant exhibits the phenomenon of "heavy top, light bottom" due to improper development of its ten growth shoots - with excessive growth in the aerial part. Which of the following treatment methods can alleviate this phenomenon?

Options:
A. Increase the supply of water significantly.
B. Apply more nitrogen (N) fertilizer appropriately and reduce the application of phosphorus (P) and potassium (K) fertilizer.
C. Avoid any fertilization and let the plant self-adjust naturally.
D. Appropriately increase the temperature of the surrounding environment.
E. Maintain high humidity to encourage even shoot development.
F. Apply more phosphorus (P) and potassium (K) fertilizer instead.
G. Minimize the temperature to slow plant growth rates.
H. Appropriately reduce the supply of water.
I. Increase the light intensity of the surrounding environment.
J. Decrease the light intensity to balance growth distribution.
Answer: H
The recipient is known to have blood type B. In the cross-matching test, there is no agglutination on the major side, but agglutination occurs on the minor side. What is the donor's blood type?

Options:
A. AB type
B. Type B-
C. Type A+
D. Type B
E. Type AB
F. Type B+
G. Type A
H. Type O
I. Type A-
J. Type AB-
Answer: H
Lehrman designed an experiment involving turtledoves, dividing the subjects into three groups: (1) exclusively female birds, housed individually; (2) female birds paired with male birds but without nesting materials provided; (3) female birds paired with male birds with nesting materials provided. Each female turtledove in the experimental groups was then given a clutch of eggs to observe their incubation behavior. The results showed the following: in group (1), none of the female turtledoves incubated the eggs; in group (2), the proportion of female turtledoves incubating the eggs began to increase on days 6, 7, and 8, but not all females incubated; in group (3), by day 8, all the female turtledoves had started incubating the eggs. Based on this experiment, what can we infer?

Options:
A. Female doves incubate eggs only in solitary environments without external stimuli.
B. Male partners can induce all brooding behaviors of female vocalization.
C. Nesting materials alone are sufficient for female doves to incubate eggs without male presence.
D. Male partners prevent female doves from displaying incubation behavior entirely.
E. Females paired with mates but lacking nesting materials begin incubation on day one.
F. The materials used for nest building can elicit the complete incubation behavior of the female bird.
G. The absence of male mates amplifies female reliance on artificial stimuli for incubation.
H. The male mate and nesting materials are prerequisites for the female dove to incubate eggs.
I. The experimental design is unreasonable, and no conclusions can be drawn.
J. Social isolation promotes incubation behavior more effectively than group interactions.
Answer: H
If 4% of individuals in a balanced population exhibit a recessive trait controlled by a single gene, what is the probability that the offspring of two individuals who do not exhibit this trait will show the recessive trait?

Options:
A. 8×$\mathrm{10^{-3}}$
B. 5.12×$\mathrm{10^{-2}}$
C. 6.4×$\mathrm{10^{-3}}$
D. 1.024×$\mathrm{10^{-1}}$
E. 1.28×$\mathrm{10^{-1}}$
F. 1.6×$\mathrm{10^{-2}}$
G. 2.56×$\mathrm{10^{-2}}$
H. 3.84×$\mathrm{10^{-2}}$
I. 4×$\mathrm{10^{-2}}$
J. 7.68×$\mathrm{10^{-2}}$
Answer: G
What is the mycelium of fresh wood ear mushrooms harvested from the wood used for cultivating them?

Options:
A. Bicore filamentous body
B. Multinucleate filamentous structures
C. Non-septate filamentous structure
D. Septate fibrous layers
E. Unicellular membranous network
F. Binucleate branching tissue
G. Multinucleate hyphal projections
H. Mononucleate tubular strands
I. Nonlinear filament growth
J. Mononucleosome filament
Answer: A
On one side of the domestic pigeon, the aortic arch degenerates, and on one side of the female domestic pigeon, the ovary and oviduct also degenerate. What are these degenerated organs?

Options:
A. The right aortic arch and no reproductive structures
B. The left uterine artery and the right ovary and fallopian tube
C. The left aortic arch but no reproductive structures
D. Left uterine artery, left ovary, and fallopian tube.
E. The left aortic arch and the left ovary and fallopian tube
F. The right aortic arch and the left ovary and fallopian tube
G. The right aortic arch and the right ovary and fallopian tube
H. Right aortic arch and right ovary and fallopian tube
I. The left aortic arch and the right ovary and fallopian tube
Answer: I
What is meant by a draft genome of a certain species?

Options:
A. Genetic Map
B. Conclusions derived from single-gene sequencing
C. High-Density Linkage Map
D. Analysis limited to non-coding RNA regions
E. Determination of results obtained from 4-5 times the genome
F. Sequence results from whole-genome duplication
G. Results obtained from 10x genome sequencing
H. Physics Diagram
I. Partial unsequenced fragments of the genome
J. Outcomes of a low-coverage genetic analysis
Answer: E
The bodies of arthropods are distinctly segmented, with several main variations: ① divided into head, thorax, and abdomen; ② divided into cephalothorax and abdomen; ③ divided into head and trunk. Based on this characteristic alone, what conclusions can be drawn?

Options:
A. It is possible to distinguish between the classes Myriapoda, Insecta, and Arachnida, but it is not possible to differentiate between the classes Entognatha and Crustacea.
B. It is possible to distinguish between Arachnida and Chelicerata, but it is not possible to differentiate between the classes Insecta, Myriapoda, and Crustacea.
C. It is possible to differentiate between the classes Insecta and Crustacea, but it is not possible to distinguish between the classes Arachnida, Myriapoda, and Chelicerata.
D. It is possible to distinguish between Insecta and Chelicerata, but it is not possible to differentiate between the classes Arachnida, Myriapoda, and Crustacea.
E. It is possible to differentiate between Arachnida and Crustacea, but it is not possible to distinguish between the classes Myriapoda, Chelicerata, and Insecta.
F. It is possible to distinguish between Crustacea and Myriapoda, but it is not possible to differentiate between the classes Arachnida, Chelicerata, and Insecta.
G. It is possible to differentiate between the classes Myriapoda and Arachnida, but it is not possible to distinguish between the classes Insecta, Chelicerata, and Crustacea.
H. It is possible to differentiate between the classes Myriapoda and Insecta, but it is not possible to distinguish between the classes Chelicerata, Arachnida, and Crustacea.
I. It is possible to distinguish between Insecta and Arachnida, but it is difficult to differentiate Myriapoda, Onychophora, and Crustacea.
J. It is possible to differentiate between Crustacea, Insecta, and Arachnida, but it is not possible to distinguish between Onychophora and Myriapoda.
Answer: H
What pigment is present in all oxygenic photosynthetic organisms?

Options:
A. Chlorophyll a, Chlorophyll b
B. Chlorophyll-d, phycobilins
C. Chlorophyll-d, anthocyanins
D. Chlorophyll-a, carotenoids
E. Chlorophyll-b, phycocyanin
F. Chlorophyll-b, betacyanins
G. Chlorophyll a, xanthophyll
H. Chlorophyll a, fucoxanthin
I. Chlorophyll-c, carotenoids
J. Chlorophyll-a, phycobiliproteins
Answer: D
Plastocyanin is synthesized in the cytoplasm and then transported to the thylakoid lumen within the chloroplast to perform its electron transfer function. How many transit peptides does the plastocyanin precursor synthesized in the cytoplasm contain?

Options:
A. 4 unit
B. 5 unit
C. 3 unit
D. 1 unit
E. 9 unit
F. 8 unit
G. 0 unit
H. 2 unit
I. 6 unit
J. 7 unit
Answer: H
What primarily causes acid rain?

Options:
A. $\mathrm{NO_3}$,$\mathrm{SO_3}$
B. $\mathrm{H_2S}$,$\mathrm{CFCs}$
C. $\mathrm{NO_2}$,$\mathrm{SO_2}$
D. $\mathrm{CO_2}$,$\mathrm{SO_2}$
E. $\mathrm{CH_4}$,$\mathrm{CO_2}$
F. $\mathrm{CO}$,$\mathrm{HCl}$
G. $\mathrm{O_3}$,$\mathrm{NO}$
H. $\mathrm{CO_2}$,$\mathrm{NO_2}$
I. $\mathrm{CO_2}$,$\mathrm{NO_2}$,$\mathrm{SO_2}$
J. $\mathrm{NH_3}$,$\mathrm{H_2O}$
Answer: C
There is a mutant yeast strain lacking an enzyme in the tricarboxylic acid cycle. It can only grow when α-ketoglutarate is added to the medium. Which enzyme is this yeast lacking?

Options:
A. Malate dehydrogenase
B. Pyruvate carboxylase enzyme
C. Isocitrate dehydrogenase
D. Pyruvate dehydrogenase
E. Citrate isomerase
F. α-ketoglutarate dehydrogenase
G. Citrate synthase enzyme
H. Aconitase enzyme
I. Fumarase enzyme
J. Succinate dehydrogenase
Answer: C
The placenta is formed by the combination of the maternal uterine lining and which of the following parts of the fetus?

Options:
A. Yolk Sac and Placenta Tissue
B. Chorion and Amnion
C. Chorion
D. Chorion and Allantois
E. Amnion and Umbilical Cord
F. Allantois
G. Umbilical Cord and Chorion
H. Amnion and Yolk Sac
I. Amnion and Placental Disc
J. Amnion
Answer: D
The base sequence of the template DNA is as follows: 5'-TGCAGT-3'. What is the base sequence of the RNA transcribed from it?

Options:
A. 5'-ACTGCA-3'
B. 5'-UGCAGU-3'
C. 5'-ACGCGU-3'
D. 5'-ACGUCA-3'
E. 5'-ACUGCA-3'
F. 5'-ACUACA-3'
G. 5'-ACGUCG-3'
H. 5'-ACUCCA-3'
I. 5'-AGUCUG-3'
J. 5'-ACGTCA-3'
Answer: E
After being peeled, apples turn brown due to the action of which substance?

Options:
A. Oxidative reductase
B. Peroxide catalase
C. Phenol oxidase
D. Catechol oxidase
E. Cytochrome oxidase
F. Tyrosinase activity
G. Lactate dehydrogenase
H. Cytochrome oxidase inhibition
I. Polyphenol synthase
J. Alternative Oxidase
Answer: C
Flukes (trematodes), which belong to parasitic flatworms, exhibit a parasitic lifestyle, and many species have complex life cycles requiring more than two hosts. Which of the flukes have a life cycle that involves more than two hosts?

Options:
A. Clonorchis sinensis, Schistosoma mansoni, Diphyllobothrium latum
B. Clonorchis sinensis, Schistosoma haematobium, Fasciola gigantica
C. Schistosoma mansoni, Paragonimus westermani, Fasciola hepatica
D. Clonorchis sinensis, Diphyllobothrium latum, Fasciola hepatica
E. Fasciolopsis buski, Clonorchis sinensis, Taenia solium
F. Clonorchis sinensis, Schistosoma japonicum, Fasciola hepatica
G. Clonorchis sinensis, Schistosoma intercalatum, Opisthorchis felineus
H. Paragonimus westermani, Schistosoma japonicum, Opisthorchis viverrini
I. Emerald Branch Fluke, Liver Fluke, Pork Tapeworm
J. Clonorchis sinensis, paragonimiasis, Fasciola hepatica
Answer: F
During the volcanic eruption disaster on the Indonesian island of Sumatra, a person was trapped on an island where, apart from an ample supply of drinkable water, it was very difficult to find edible food. Fortunately, he had 3 kilograms of grain and a 1-kilogram female duck with him. To maximize his survival time while awaiting rescue, what should he do?

Options:
A. Cook the grains immediately and avoid relying on the duck.
B. First eat the mother duck, then eat the grains.
C. Feed the ducks with 3 kilograms of grain, eat the duck eggs, and then eat the duck.
D. Eat 2 kilograms of grain, use 1 kilogram of grain to feed ducks, eat duck eggs, and then eat the duck.
E. Feed all the grains to the duck and eat only the duck eggs.
F. Eat the grains gradually and never eat the mother duck.
G. Keep the grains for yourself and release the mother duck into the wild.
H. Feed 1 kilogram of grain to the duck, save the rest, and live on duck eggs.
I. Eat the food first, then eat the mother duck.
J. Trade the duck for another source of food and consume the grains sparingly.
Answer: B
Cytological examination of mitosis in a somatic cell of a biological individual revealed the presence of a chromosomal bridge during the anaphase, indicating which type of chromosome this biological individual might contain?

Options:
A. Inversion within the arm
B. Unequal chromosomal split
C. Chromosomal duplication
D. Mutual Transposition
E. The apex is missing.
F. Centrosome misalignment
G. Inverted Position Between Arms
H. Extra centromere presence
I. Chromatid lagging bridge
J. Telomere shortening problem
Answer: E
Due to changes in environmental conditions, a branch of ancient forest-dwelling wolves was forced to live on the ground, eventually evolving into humans through a lengthy developmental process. During this evolutionary process, there were two transitional types of ancient apes, which were sequentially what?

Options:
A. Ramapithecus and Australopithecus
B. Anthropoids and Hominoids
C. Australopithecus and Homosapiens
D. Australopithecus, Ramapithecus
E. Homo sapiens and Anthropoids
F. Ramapithecus and Neanderthals
G. Homo habilis and Neanderthals
H. Ramapithecus, Anthropoids
I. Australopithecus, Anthropoid Apes
J. Hominoids and Homo erectus
Answer: A
What is correctly described about the shared characteristics between Escherichia coli and mouse cells?

Options:
A. Do not contain linear DNA and ribosomes.
B. Both lack ribosomes and DNA.
C. Both exclusively contain linear DNA and chloroplasts.
D. All contain circular DNA and ribosomes.
E. Both contain circular DNA and linear DNA
F. Neither have linear DNA or mitochondria.
G. Neither possess circular DNA or ribosomes.
H. All have nuclei but lack circular DNA.
I. All contain nuclei and nucleoli.
J. All contain linear DNA and mitochondria.
Answer: D
Which of the following statements about population spatial distribution patterns is correct?

Options:
A. For cluster sampling, the number of individuals in each small grid is constant across all regions.
B. Cluster sampling shows that the frequency of individuals in each small grid follows a normal distribution.
C. For cluster sampling, the frequency of individuals in each small cell is evenly distributed across all grids.
D. For uniform distribution sampling, the number of individuals in each small cell is equal, so the ratio of variance to the mean is 1.
E. For uniform distribution sampling, the frequency of individuals in each small cell matches a bell-shaped curve.
F. For random stratified sampling, the frequency of individuals in each small cell follows a binomial distribution.
G. For cluster sampling, the frequency of very few individuals and very many individuals appearing in each small cell is relatively low.
H. For uniform distribution sampling, the frequency of individuals in each small cell follows a skewed distribution.
I. For random stratified sampling, the frequency of individuals in each small cell follows a Poisson distribution.
J. For random stratified sampling, the presence of individuals in each small cell is uniform across all areas.
Answer: I
Lamarck systematically proposed the theory of biological evolution before Darwin. Darwin's theory of evolution inherited some of Lamarck's ideas in certain aspects but also differed significantly. Which of the following represents Darwin's viewpoint that differs from Lamarck's?

Options:
A. Evolution is driven solely by divine creation.
B. Traits developed by effort are passed to offspring.
C. Variations arise to meet specific environmental needs.
D. All living organisms originate from a common ancestor.
E. Biological organisms follow the principle of use and disuse.
F. Species evolve through spontaneous generation.
G. Organisms adapt traits based on personal desires.
H. Acquired traits can be inherited.
I. Complexity in species grows by progressive perfection.
J. All living beings are constantly changing.
Answer: D
What important role does moderate low temperature play in the aging process during plant growth, and what are its effects?

Options:
A. Does not affect aging
B. Enhancing Reproduction
C. Delaying Senescence
D. Promoting Aging
E. Accelerating Growth
F. Disrupting Metabolism
G. Suppressing Germination
H. Uncertain
I. Inducing Dormancy
J. Inhibiting Aging
Answer: J
The partial pressure of $\mathrm{CO_2}$ is highest in human body fluids in:

Options:
A. Plasma membrane
B. Blood
C. Alveolar air
D. Lymph
E. Synovial fluid
F. Interstitial fluid
G. Intracellular fluid
H. Cerebrospinal fluid
I. Arterial blood
J. Bronchial fluid
Answer: G
Research the blood flow in several parts of the human circulatory system. Which of the following locations has the highest blood flow per minute?

Options:
A. The renal artery as it branches off from the abdominal aorta.
B. The brachial artery located in the upper arm.
C. The right pulmonary artery at the point where it branches into the left and right pulmonary arteries.
D. The left coronary artery that just branches off from the aorta
E. The left leg artery at the point where the abdominal aorta branches into the left and right leg arteries.
F. The left subclavian vein before it joins the superior vena cava.
G. The hepatic vein before entering the inferior vena cava
H. The left jugular vein as it drains blood from the head.
I. The superior mesenteric artery as it supplies blood to the intestines.
J. The ascending aorta just after it leaves the left ventricle of the heart.
Answer: C
Identify which of the following groups are in genetic equilibrium:
(1) 100% AA
(2) 100% Aa
(3) 1% AA, 98% Aa, 1% aa
(4) 32% AA, 64% Aa, 4% aa
(5) 4% AA, 32% Aa, 64% aa

Options:
A. (1) and (3)
B. (2), (4), and (5)
C. (1) and (5)
D. (2) and (4)
E. (1), (3), and (4)
F. (1) and (4)
G. (1), (2), and (3)
H. (2) and (3)
I. (2) and (5)
J. (3) and (5)
Answer: C
In this food chain: grass → mouse → eagle, if the eagles are exterminated, how will the population of field mice change?

Options:
A. Chaotic fluctuation forever
B. Immediate disappearance
C. Rapid stabilization
D. Sharp decline, then stabilize
E. Rapid ascent
F. Gradual decline
G. Continuous rapid growth
H. Endless cyclical fluctuations
I. First rise, then fall.
J. Maintain the original state
Answer: I
Teeth are commonly found in vertebrates, and what do mammals, which have developed to a high level, classify their teeth as?

Options:
A. Serrated Molars
B. Grazing Teeth
C. Dental Armor
D. Non-standard Gear Teeth
E. Homodont teeth
F. Canine Structure
G. Extra teeth
H. Tusk-like Protrusions
I. Lateral incisor
J. Conical Dentition
Answer: D
In which stage does meiosis occur in higher plants?

Options:
A. After the formation of gametes
B. Before gametogenesis
C. After fertilization occurs
D. Before cell division
E. After spore release
F. After the development of the ovule
G. After the seed germinates
H. Before Spore Formation
I. Before pollen tube formation
J. After spore formation
Answer: H
The genes located on autosomes include two pairs that assort independently (Aa and Bb) and three pairs that are completely linked (EFG/ efg). What is the proportion of dominant homozygous gametes produced by this organism?

Options:
A. 1/16
B. 1/8
C. 1/10
D. 1/64
E. 1/32
F. 1/4
G. 1/2
H. 1/14
I. 1/12
J. 1/6
Answer: B
Some people believe that land development is limited by water. What is the factor limiting the development of marine life?

Options:
A. Deep Pressure
B. Surface Wind
C. Sunlight Adaptation
D. Acidity of Water Bodies
E. Nutrient
F. Ocean Temperature
G. Oxygen
H. Minerals
I. Ocean Current Intensity
J. High salt content
Answer: E
According to the classification of storage substances, what are the seeds of more than 90% of plants?

Options:
A. Salt content
B. Fibrous
C. Protein's
D. Moisture quality
E. Sugary
F. Vitamin's
G. Starch
H. Enzymatic
I. mineral
J. Oily
Answer: J
What are the distinct characteristics of ferns?

Options:
A. Young curled leaves, compound leaves, and flower stems
B. Rolled young leaves, simple leaves, and rhizomes
C. Young coiled leaves, compound leaves, and rhizomes
D. Unfurling young leaves, compound leaves, and rhizomes
E. Tender expanded leaves, compound leaves, and grape stems
F. Fist-curled young leaves, simple leaves, and tubers
G. Young curled leaves, compound leaves, and tubers
H. The young curled leaves, compound leaves, and stem nodes
I. Young expanding leaves, single leaves, and tubers
J. Expanded young leaves, simple leaves, and rhizomes
Answer: C
In the following series of biochemical reactions occurring in the human body: phenylalanine (A) → tyrosine (B) → $\mathrm{CO_2}$ + $\mathrm{H_2O}$, the absence of enzyme A leads to phenylketonuria (PKU), and the absence of enzyme B leads to alkaptonuria (AKU). Both diseases are not sex-linked. If a PKU patient marries an AKU patient, what will be the phenotypes of their children?

Options:
A. Their daughter is half normal, half affected by PKU.
B. Half of their children have PKU and AKU, while the other half are normal.
C. Half of their children have PKU, and the other half have AKU.
D. Their children all have PKU (Phenylketonuria) and AKU (Alkaptonuria).
E. Their son is half normal and half has AKU.
F. Their son is half normal and half has PKU (Phenylketonuria).
G. Half of their children have AKU, while the other half are normal.
H. Half of their children have PKU, while the other half are normal.
I. Their children all get sick.
J. Their children are all normal.
Answer: J
BFA (Brefeldin A) is a commonly used protein transport inhibitor that can prevent the transport of proteins from the endoplasmic reticulum to the Golgi apparatus. When cells are treated with this drug, what other cellular structures will undergo changes in the short term?

Options:
A. Peroxisome, vesicle, plasma membrane
B. Lysosome, vesicle, plasma membrane
C. Lysosome, vesicle, chloroplast
D. All organelles and membrane structures within the cell
E. Endosome, vesicle, plasma membrane
F. Lysosome, vesicle, mitochondria
G. Lysosome, ribosome, plasma membrane
H. Lysosome, peroxisome, plasma membrane
I. Vesicles, mitochondria, plasma membrane
J. Lysosome, vesicle, endoplasmic reticulum
Answer: B
The skin of a certain animal is composed of a multi-layered epidermis and dermis. The epidermis contains numerous unicellular mucous glands, and the skin is closely connected to the muscles, with very little subcutaneous tissue. Which of the following animals best matches the description above?

Options:
A. Salamander
B. Snake
C. Lancelet
D. Lizard
E. Turtle
F. Prawn
G. Fish
H. Spotted Dove
I. Frog
J. Starfish
Answer: G
Which of the following groups constitutes a set of homologous organs?

Options:
A. The Feathers of Birds and the Hair of Mammals
B. The Antennae of Insects and the Wings of Birds
C. The Endostyle of the Lancelet and the Human Parathyroid Gland
D. The Gills of the Carp and the Lungs of the Frog
E. The Claws of Crabs and the Nails of Primates
F. The three Weir's ossicles of fish and the three auditory ossicles of mammals
G. The Eyes of Octopuses and the Eyes of Mammals
H. The Beaks of Squids and the Mouths of Mammals
I. The Tongue-and-Jaw Bone of Sharks and the Stapes Bone of Humans
J. The Fins of Sharks and the Legs of Humans
Answer: I
Red-green color blindness is determined by an X-linked recessive allele. Assuming a population in Hardy-Weinberg equilibrium, where 1 in 10 males is colorblind, what would happen?

Options:
A. The frequency of the color blindness gene is approximately 0.3.
B. The proportion of color blindness in females is much lower than in males, but the specific ratio cannot be determined from the given conditions in the question.
C. The proportion of color blindness in females is equivalent to that in males in a Hardy-Weinberg population.
D. The ratio of males to females with color blindness will always be the same regardless of population changes.
E. The frequency of the color blindness gene is exactly 10 times higher in males than in females.
F. The proportion of color blindness in women is approximately 1/20 that of men.
G. The incidence of color blindness in women depends entirely on their father's genetic makeup.
H. The proportion of color blindness in women is approximately 1/10 that of men.
I. The frequency of the color blindness gene is independent of whether the population is in Hardy-Weinberg equilibrium.
J. If colorblind men only have daughters, the frequency of the colorblind gene will decrease rapidly.
Answer: H
When studying the properties of a protein, we generally first need to isolate and purify the protein, as well as determine its molecular weight. Which of the following experimental techniques can serve both the purpose of isolation and purification and the determination of its molecular weight?

Options:
A. Size exclusion electrophoresis
B. Ion Exchange Chromatography
C. Paper Chromatography
D. Gel filtration chromatography
E. Mass spectrometry ionization
F. Differential centrifugation
G. Circular dichroism spectroscopy
H. Affinity Chromatography
I. Co-immunoprecipitation assay
J. Western blotting technique
Answer: D
In a normal cardiac cycle, the duration from the closure to the opening of the atrioventricular valves is equivalent to what?

Options:
A. The sum of atrial systole and ventricular systole
B. The sum of ventricular diastole and rapid ejection phase
C. The sum of ventricular diastole and atrial systole
D. The sum of ventricular systole and isovolumetric relaxation period
E. The sum of ventricular diastole and isovolumic contraction phases
F. The sum of atrial diastole and isovolumetric relaxation phase
G. The sum of rapid ejection and reduced filling phases
H. The sum of atrial diastole and isovolumetric contraction period
I. The sum of isovolumetric contraction and rapid ejection period
J. The sum of diastolic phase and slow ejection phase of the ventricle
Answer: D
The water potential of a living plant cell is composed of osmotic potential (Ψs) and pressure potential (Ψp). Changes in the volume of a plant cell can cause alterations in turgor pressure, where $e$ represents the magnitude of the change in pressure potential induced by a change in the volume per unit of the cell. It is known that the rigidity of the cell walls of two plant cells differs significantly: the cell wall of Cell 1 is highly rigid, while the cell wall of Cell 2 is less rigid. When plant cells lose water and the volume changes by one unit, how does the change in Ψp differ between these two cells?

Options:
A. Ψp(l)= -Ψp(2)
B. Ψp(l)>Ψp(2)
C. Ψp(l)≠Ψp(2)
D. Ψp(1)<<Ψp(2)
E. Ψp(l)<Ψp(2)
F. The conditions are insufficient to make a determination.
G. Ψp(l)+Ψp(2)=0
H. Ψp(l)=0.5*Ψp(2)
I. Ψp (l) =Ψp(2)
J. Ψp(l)<<Ψp(2)
Answer: B
Large baleen whales primarily feed on small crustaceans. Which of the following processes aligns with this biological phenomenon?

Options:
A. Promote secondary consumers to balance aquatic food web structures
B. Increase energy transfer efficiency through higher metabolic demand
C. The feeding method of baleen whales is suited for preying on crustaceans.
D. The feeding habits of baleen whales enhance the species diversity of crustaceans.
E. Signal mutualism to enhance survival of predator-prey partnerships
F. Support herbivorous prey populations for maintaining biodiversity
G. Resource Competition Between Baleen Whales and Crustaceans
H. Extend the food chain to maximize resource utilization in ecosystems
I. Limit predation pressure to stabilize trophic level populations
J. Shorten the food chain to reduce energy consumption between trophic levels
Answer: J
Using chemical mutagens to induce mutations in fruit flies (P), which of the following statements is correct?

Options:
A. Every F1 offspring will carry mutations if mutagenesis happens during meiosis.
B. Mutagenesis before fertilization ensures all F2 contain identical mutations.
C. Mutagenesis during DNA replication erases prior inherited mutations.
D. Mutagen exposure ensures dominant traits appear in all generations.
E. Mutagenesis should be performed before meiosis, not after meiosis.
F. No matter when mutagenesis occurs, half of the F2 generated from the F1 crosses contain the same mutation.
G. When the first-generation offspring of mutagenized fruit flies are crossbred, there is a 1/4 probability of producing homozygous mutants.
H. Post-fertilization mutagenesis produces uniform homozygous mutants directly.
I. Mutagenesis during gamete formation prevents mutation inheritance.
J. Mutagenesis after meiosis leads to the formation of chimeras in F1.
Answer: J
In the floral structure of the Chinese redbud (Cercis chinensis Bunge), which petal is located on the outermost side?

Options:
A. Lotus petal
B. Leaflet petal
C. Helmet petals
D. Spear petal
E. Dragon bone petal
F. Sword petal
G. Petal or flag petal
H. Shield petal
I. Crown petal
J. Winglet
Answer: E
Where on the ridge of the pennate diatom can one observe which surface?

Options:
A. Annular surface
B. Ridge crest margin
C. Interior ridge section
D. Apex ridge boundary
E. Outer edge contour
F. Upper shell surface
G. Bottom shell surface
H. Side view
I. Lower shell surface
J. Peripheral shell layer
Answer: F
What kind of environment do plants with stomata mainly distributed on the upper epidermis of their leaves usually live in?

Options:
A. Sunrise
B. Wetland Marshes
C. Aquatic Life
D. Drought Resistance
E. Arctic Tundra
F. Mountain Slopes
G. Forest Canopies
H. Yinsheng
I. Desert Habitat
J. Dry Farmlands
Answer: C
The perianth segments of orchids are divided into two whorls, inner and outer, with three segments in each whorl. One of the segments in the inner whorl is specialized; what is it called in terms of morphology and taxonomy?

Options:
A. Pistil and Stamen
B. Petal sheath
C. Floral bracts
D. Ovary walls
E. Stigma lobes
F. Lower lip
G. Sepal margins
H. Column of united filaments
I. Lips
J. Apex tube
Answer: I
The shell of the mussel is formed by the secretion of the mantle. Among them, the periostracum and the prismatic layer of the shell are secreted by the edge of the mantle, while the nacreous layer is secreted by the entire mantle. Therefore, what changes occur as the animal grows?

Options:
A. The thickness of all shell layers increases evenly with no cessation.
B. The shell layers fuse entirely, halting independence in growth stages.
C. The thickness of the stratum corneum and prismatic layer does not change, while the nacreous layer continues to thicken.
D. The thickness of the stratum corneum, prismatic layer, and nacreous layer no longer changes.
E. The stratum corneum, prismatic layer, and nacreous layer continuously thicken
F. The nacreous layer forms ridges, whereas other layers erode entirely.
G. The boisterous outer layers grow slower, while the nacreous layer dissolves.
H. The mantle stops secreting layers, causing shell growth to cease.
I. The stratum corneum and prismatic layer continuously thicken, while the nacreous layer remains unchanged.
J. The prismatic layer and nacreous layer shrink, while the outermost layer grows.
Answer: C
Infect *E. coli* simultaneously with two types of bacteriophages, a-b- and a+b+, and then plate the resulting lysate onto nutrient agar plates. Count the number of plaques formed. The results are as follows: a+b+ = 7125, a+b- = 555, a-b+ = 490, a-b- = 6675. Based on this, determine the recombination frequency between gene loci a and b.

Options:
A. 12.8%
B. 18%
C. 6.4%
D. 9.2%
E. 20%
F. 23%
G. 7%
H. 4.5%
I. 14%
J. 11%
Answer: G
Which of the following statements about the balanced lethal system is correct?

Options:
A. The balanced lethal system represents a permanently homozygous genotype strain.
B. One of the characteristics of a balanced lethal system is that two dominant lethal genes are located on the same chromosome of a homologous chromosome pair, exhibiting mutual exclusivity.
C. One of the characteristics of a balanced lethal system is that two dominant lethal genes are located on different chromosomes of a homologous chromosome pair, ensuring effective recombination.
D. One of the characteristics of a balanced lethal system is that only a single recessive lethal gene is present on a homologous chromosome pair, leading to non-exclusive behavior.
E. One of the characteristics of a balanced lethal system is that two recessive lethal genes are located on different chromosomes of a homologous chromosome pair, exhibiting mutual exclusivity.
F. One of the characteristics of a balanced lethal system is that two recessive lethal genes coexist on a single chromosome without influencing recombination.
G. One of the characteristics of a balanced lethal system is: two dominant lethal genes are located on different chromosomes of a homologous chromosome pair, with no effective recombination occurring between the lethal genes.
H. The balanced lethal system is used to maintain special strains that preserve dominant homozygous lethal genes.
I. One of the characteristics of a balanced lethal system is that two recessive lethal genes are located on the same chromosome of a homologous chromosome pair, exhibiting mutual exclusivity.
J. One of the characteristics of a balanced lethal system is that one dominant and one recessive lethal gene are located on the same homologous chromosome, exhibiting exclusivity.
Answer: E
What stage are most of the cells within dormant seeds in?

Options:
A. State of Preservation
B. Phase G2
C. Epoch of Resilience
D. Phase G1
E. S phase
F. Period of Enlightenment
G. Moment of Silence
H. Stage M
I. Cycle of Restfulness
J. Duration of Dormancy
Answer: F
What life history strategies are typically adopted by plants living in extremely arid desert regions?

Options:
A. r-strategy
B. Assisted Propagation
C. Perennial Succession
D. Rapid Phototropism
E. Symbiotic Parasitism
F. Opportunistic Dormancy
G. Coerced Endurance
H. Competitive Strategies
I. Prolonged Anthesis
J. Weed Control Strategies
Answer: G
The growth and development of animals require a certain temperature. Which of the following statements is correct?

Options:
A. The developmental rate of animals always accelerates with the increase in environmental temperature.
B. The starting point of insect development is exactly the same across all temperature ranges.
C. The effective accumulated temperature for insect development is calculated by multiplying the duration of the developmental period by the average temperature during that period, and then summing the results.
D. The external temperature directly determines the body temperature of animals, thereby influencing their growth and development.
E. Temperature changes in the external environment primarily regulate insect reproduction rather than their growth and development.
F. When the external temperature falls below a certain threshold, insects cease growth and development. This threshold temperature is referred to as the developmental zero temperature.
G. The growth and development rate of animals is directly related to the changes in daylight duration and is not affected by temperature.
H. Changes in ambient temperature only affect the activity frequency of insects and do not impact their developmental cycle.
I. The external temperature determines the behavioral patterns of animals but is unrelated to their growth and development.
J. When the external temperature falls below a certain level, animals enter a state of dormancy rather than ceasing growth and development.
Answer: F
Which of the following ecosystems has the lowest productivity?

Options:
A. Tropical Rainforest Ecosystem
B. Saltwater Lake Ecosystem
C. Coastal Mangrove Ecosystem
D. Deep-Sea Ecosystems
E. Alpine Desert Ecosystem
F. Arctic Tundra Ecosystem
G. Polar Ice Cap Ecosystem
H. Subalpine meadow ecosystem
I. Temperate Broadleaf Forest Ecosystem
J. Desert Ecosystem
Answer: J
A certain population is uniformly distributed in an area; what can generally be inferred from this?

Options:
A. Individuals in this population aggregate under environmental guidance.
B. The individuals of this population are distributed according to the uniform distribution of resources.
C. Individuals in this population exhibit mutual repulsion.
D. There is no influence between individuals in this population.
E. The individuals of this population exhibit an aggregated distribution due to resource scarcity.
F. Individuals in this population are completely randomly distributed.
G. The distribution of individuals in this population depends on the predators.
H. Individuals in this population are randomly distributed to avoid competition.
I. The individuals within this population exhibit a high degree of cooperative distribution.
J. Individuals within this population are mutually attracted to one another.
Answer: C
What phenomenon will occur when 5 mL of ethanol solution at a concentration of 0.3 mol/L is added to 1 mL of mouse erythrocyte suspension?

Options:
A. Hemolysis
B. No change.
C. The cell nucleus is damaged.
D. The cell swells but does not undergo hemolysis.
E. The solution turns cloudy.
F. The cell membrane detaches but remains viable.
G. Cell shrinkage
H. The cell membrane is completely dissolved.
I. Extracellular fluid becomes isotonic
J. Complete lysis of red blood cells
Answer: A
Where is the higher central control for birds' song learning located?

Options:
A. Forebrain
B. External Brain
C. Back of the head
D. Edge Brain
E. Primal Brain
F. Midbrain
G. Diencephalon
H. Cerebellum
I. Deep Brain
J. Brain
Answer: J
The anther wall consists of four layers from the outside to the inside. What are their correct names and arrangement?

Options:
A. Epidermis - Middle Layer - Drug Chamber Inner Wall - Fiber Layer
B. Epidermis - Middle Layer - Drug Chamber Inner Wall - Velamen Layer
C. Epidermis - One-chamber inner wall - One sealing layer - Middle layer
D. Epidermis - Intermediate Layer - Fibrous Layer - Compact Layer
E. Epidermis → Single-layered wall of the drug chamber → Fiber layer → Tapetum layer
F. Epidermis - Fiber Layer - Middle Layer - Tapetum Layer
G. Epidermis - one-chambered internal wall - middle layer - endothelium layer
H. Epidermis - One-Cell Layer - Middle Layer - Velamen Layer
I. Epidermis - Middle Layer - Medicine Chamber Inner Wall - Cuticle Layer
J. Epidermis - Middle Layer - Tapetum Layer - Inner Wall of the Anther Locule
Answer: G
Which of the following statements about communities is correct?

Options:
A. In ecotone areas, the number of individuals and biomass tend to decrease.
B. In ecotone areas, species richness and diversity remain unchanged over time.
C. Community species diversity increases with the progression of community succession; therefore, the climax community stage has the highest diversity.
D. In ecotone areas, species composition is identical to adjacent habitats.
E. Predation inevitably enhances community diversity
F. In ecotone areas, environmental stability prevents species from thriving.
G. In ecotone areas, population density is lower than in central habitats.
H. During the process of ecological succession, the ratio of total productivity to total respiration (P/R) consistently remains greater than 1, and as succession reaches the climax stage, P/R approaches or equals 1.
I. In ecotone areas, ecological interactions are less significant than elsewhere.
J. In ecotone areas, the number of species and population density tend to increase.
Answer: J
The thylacines, marsupial predators, marsupial gliders, numbat, marsupial moles, and wallabies of Australia are very similar in form and habits to wolves, leopards, flying squirrels, anteaters, moles, and rats of the Eurasian continent, apart from the difference of having pouches. What type of evolutionary pattern is this phenomenon a typical example of?

Options:
A. Radiation Evolution
B. Divergent Evolution
C. Coevolution
D. Parallel Evolution
E. Convergent Evolution
F. Parallel Speciation
G. Adaptive Radiation
H. Morphological Convergence
Answer: D
Porifera, Rotifera, and Diploblastic animals belong to which groups?

Options:
A. Mesozoa, Diploblasts, Protostomes
B. Coelomates, Parazoa, Diploblasts
C. Pseudocoelomates, Mesozoa, Deuterostomes
D. Protostomes, Diploblasts, Coelomates
E. Parazoa, Pseudocoelomates, Mesozoa
F. Lateral animals, mesozoans, pseudocoelomates
G. Pseudocoelomates, Mesozoans, Placozoans
H. Parazoa, Coelomates, Protostomes
I. Mesoaquatic animals, pseudocoelomate animals, laterally symmetrical animals
J. Diploblasts, Pseudocoelomates, Protostomes
Answer: E
The color of the seed coat in wheat is determined by multiple copy genes. The more dominant alleles (R), the redder the seed coat; the more recessive alleles (r), the whiter the seed coat. When true-breeding red-seed-coat wheat is crossed with true-breeding white-seed-coat wheat, the ratio of red seed coat to white seed coat in the F2 generation is 63:1. How many R gene loci segregate in this cross?

Options:
A. 5 copies
B. 2 copies
C. 4 copies
D. 5 genes
E. 5 units
F. 2 loci
G. 3 units
H. 4 units
I. 6 loci
J. 2 units
Answer: G
What proteins are associated with the dehydration process during the late stage of seed development?

Options:
A. Heat Shock Protein
B. GTP-binding protein
C. ABA Binding Protein
D. Endosperm Storage Protein
E. Late-stage embryonic development enriched proteins
F. Calmodulin
G. Germination Trigger Protein
H. Late Maturation Protein
I. Seed Desiccation Protein
J. Abscisic Acid Carrier Protein
Answer: C
What characteristics are unique to cancer cells?

Options:
A. Cells undergo frequent division.
B. Cell volume reduction
C. Organizational infiltration
D. Cells do not require energy.
E. Completely Avoid Apoptosis
F. Unlimited generation of lipids
G. Angiogenesis
H. The genome remains unchanged.
I. Does not rely on signal transduction
J. Gene Mutation
Answer: C
The bases of chromatin DNA can be methylated. What is the function of DNA methylation?

Options:
A. Certain genes can be deactivated, while others can be activated.
B. Turning off certain genes
C. Influencing RNA translation rates in cytoplasm.
D. Activation of certain genes
E. Gene repair mechanisms during cell replication.
F. Facilitating DNA replication in the nucleus.
G. Encoding specific instructions for protein assembly.
H. Impacting cell signaling pathways in response.
I. Unrelated to gene regulation
J. Chromosome alignment during mitosis and meiosis.
Answer: A
Which aspect is the same between pulmonary circulation and systemic circulation?

Options:
A. Carbon dioxide diffusion
B. Aortic Expandability
C. Venous return rate
D. Blood vessel elasticity
E. Systolic Blood Pressure
F. Stroke volume
G. Peripheral resistance
H. Pulmonary artery function
I. Oxygen levels in veins
J. Flow direction consistency
Answer: F
Liverworts are commonly used materials for studying bryophytes; they are dioecious, with sex determined by sex chromosomes and belonging to the XY type. Which part of the liverwort's cells contains both X and Y chromosomes?

Options:
A. Nutritional Cell
B. Bud
C. Capsule base
D. Spore
E. Antheridial cells
F. Gametangial layer
G. Archegonial tissue
H. Stalk of thallus
I. Pedicel cells
J. Rhizoid structures
Answer: I
If the genetic compositions of the two nuclei that fuse to produce the plant zygote are labeled as A and B, and another fertilization product in the same embryo sac is labeled as ABB, what must the tube nucleus of the corresponding pollen tube be labeled as?

Options:
A. AA
B. BB
C. BAB
D. BBB
E. A
F. AB
G. BA
H. ABA
I. None of the above is correct
J. B
Answer: E
The eyes of a certain strain of fruit flies exhibit four colors: wild type, orange 1, orange 2, and pink. The following displays the true inheritance patterns resulting from crosses between these strains. If F1 offsprings from an orange 1 x orange 2 cross are crossed, what proportion of wild type would occur in the F2 generation?
Cross
Wild type x Orange 1
Wild type x Orange 2
Orange 1 x Orange 2
Orange 2 x Pink
F1 (Orange 1)(Orange 2) x Pink
F1 Results:
All wild type
All wild type
All wild type
All orange 2
1/4 wild type, 1/4 orange 1, 1/4 orange 2, 1/4 pink

Options:
A. 9/16
B. 2/3
C. 5/16
D. 3/8
E. 11/16
F. 1/2
G. 3/4
H. 7/16
I. 5/8
J. 1/3
Answer: A
The earliest jawed fish was discovered in strata dating back 415 million years and belongs to the ancestral placoderm lineage of cartilaginous coloration. This species is considered an intermediate link in the evolutionary chain between cartilaginous fish and bony fish. Why is that?

Options:
A. Possessed operculum without skeletal articulation
B. The mandibular arch appeared.
C. Retained fin spines without jointed bases
D. Lacked ossified scales covering the dermis
E. Exhibited early forms of external gill slits
F. Featured primitive cartilage without dermal fusion
G. Hard-ossified jawbone originating from the dermis
H. Demonstrated unmineralized jaw structure overall
I. Originated in the Silurian period
J. Having a developed cartilaginous cranium
Answer: G