Trắc Nghiệm Di Truyền Quần Thể 1

The correct answer is "tăng tỉ lệ thể đồng hợp, giảm tỉ lệ thể dị hợp." This means that in a population, self-fertilization increases the proportion of homozygous individuals and decreases the proportion of heterozygous individuals. This is because self-fertilization leads to offspring that have two identical copies of the same alleles, resulting in homozygosity. As a result, the genetic diversity within the population decreases.

The given answer represents a population in genetic equilibrium. This is indicated by the presence of both homozygous dominant (AA) and homozygous recessive (aa) individuals, as well as heterozygous (Aa) individuals in the population. In a population in genetic equilibrium, the allele frequencies remain constant from generation to generation, indicating that no evolutionary forces such as mutation, migration, genetic drift, or natural selection are acting on the population.

The correct answer is "quần thể tự phối." This is because in a population that undergoes self-fertilization, the number of heterozygous individuals decreases over time, while the number of homozygous individuals increases. This is due to the fact that self-fertilization leads to a higher likelihood of offspring inheriting two copies of the same allele, resulting in homozygosity. Therefore, the population becomes more homozygous over time, indicating that self-fertilization is the most likely scenario in this case.

The given answer 0,16AA : 0,48Aa : 0,36aa indicates a state of genetic equilibrium. This means that the frequencies of the different genotypes (AA, Aa, aa) in the population remain constant over time. In this case, the frequencies of the genotypes AA, Aa, and aa are 0.16, 0.48, and 0.36 respectively, and these frequencies do not change. This suggests that there is no selection pressure or genetic drift acting on the population, and the population is in a stable state with respect to its genetic composition.

The given answer 0.55 A: 0.45 a is the relative frequency of the alleles in the population at the time of the census. It indicates that 55% of the alleles in the population are of type A (AA and Aa) and 45% of the alleles are of type a (aa). This suggests that the population has a higher frequency of the dominant allele A compared to the recessive allele a.

The given information states that in a population at a certain time, the genetic structure is 0.7 AA, 0.1 Aa, and 0.2 aa. This means that 70% of the population has the genotype AA, 10% has the genotype Aa, and 20% has the genotype aa.

To determine the frequency of the alleles in the population, we can consider the alleles present in each genotype. The genotype AA has two A alleles, the genotype Aa has one A allele, and the genotype aa has two a alleles.

Calculating the allele frequencies, we find that the A allele frequency is (2 * 0.7 + 0.1) / 2 = 0.75, and the a allele frequency is (2 * 0.2 + 0.1) / 2 = 0.25.

Therefore, the correct answer is 0.75A; 0.25a.

The correct answer is "tập hợp tất cả các alen của tất cả các gen có trong quần thể ở một thời điểm nhất định". This answer is supported by the definition of "vốn gen của quần thể" which refers to the total number of alleles of all genes in a population at a specific time. Therefore, the correct answer accurately describes the concept of "vốn gen của quần thể".

The correct answer is "giảm dần tỉ lệ dị hợp tử, tăng dần tỉ lệ đồng hợp tử." This means that the population's genetic structure undergoes a decrease in the proportion of heterozygous individuals and an increase in the proportion of homozygous individuals over generations. This suggests that there is a decrease in genetic variation within the population and an increase in the prevalence of specific genotypes.

The given genotype frequencies in the population are 0.3AA, 0.5Aa, and 0.2aa. To calculate the relative allele frequencies, we can consider that each individual has two alleles. Since the genotype frequencies are given, we can divide the frequency of each genotype by the number of alleles per individual (2) to get the allele frequencies.

For the A allele, there are 0.3AA individuals and 0.5Aa individuals, which gives a total of (0.3 x 2) + 0.5 = 1.1 A alleles. Dividing this by the total number of alleles (2 x total population size), we get 1.1 / (2 x 1) = 0.55.

For the a allele, there are 0.5Aa individuals and 0.2aa individuals, which gives a total of (0.5 x 2) + (0.2 x 2) = 1.4 a alleles. Dividing this by the total number of alleles, we get 1.4 / (2 x 1) = 0.7.

Therefore, the relative allele frequencies are A = 0.55 and a = 0.45.

The given answer, p2+q2+r2+2pq+2qr+2rp, represents the Hardy-Weinberg equilibrium equation for a gene with three alleles. This equation calculates the expected genotype frequencies in a population under certain assumptions, including random mating and no evolutionary forces at work. The equation includes terms for the frequencies of each allele squared (p2, q2, r2) and the frequencies of each possible combination of alleles (2pq, 2qr, 2rp). This equation ensures that the frequencies of the genotypes in the population remain stable over generations if the assumptions of the Hardy-Weinberg equilibrium are met.

In a balanced genetic population, the frequency of alleles and the proportion of genotypes can be inferred. This means that by studying the population, we can determine the prevalence of different alleles and the ratios of different genotypes within the population. This information provides insights into the genetic composition and diversity of the population.

The given correct answer is 0.64 AA + 0.32 Aa + 0.04 aa. This is the correct answer because it reflects the relative frequencies of the genotypes in the population. The relative frequency of the AA genotype is 0.8 * 0.8 = 0.64, the relative frequency of the Aa genotype is 2 * 0.8 * 0.2 = 0.32, and the relative frequency of the aa genotype is 0.2 * 0.2 = 0.04. Therefore, the correct answer is 0.64 AA + 0.32 Aa + 0.04 aa.

In a self-fertilizing population, the genotypic composition of the population tends to separate into pure lines with different genotypes. This means that individuals with similar genotypes will mate with each other, leading to the formation of pure lines that have distinct genotypes. This process is known as genetic differentiation or genetic divergence. It results in the population becoming more homogeneous within each pure line, but more diverse overall as different pure lines emerge.

The correct answer is 48,4375%. This can be calculated using the Hardy-Weinberg equilibrium equation, which states that in a population in which mating is random and no evolutionary forces are acting, the frequencies of alleles will remain constant from generation to generation. In this case, the frequency of the A allele is 0.5, and the frequency of the a allele is also 0.5. When two Aa individuals mate, there is a 0.25 probability of producing an AA offspring, which corresponds to a frequency of 0.25. Therefore, after 5 generations, the frequency of AA individuals would be 0.5^5 = 0.03125, or 3.125%. Since AA individuals have two A alleles, the frequency of the A allele in the population would be 2 * 0.03125 = 0.0625, or 6.25%. Therefore, the frequency of AA individuals in the population after 5 generations would be 6.25%.

After 3 generations of self-fertilization, the frequency of the AA genotype will be 41.875%, the frequency of the Aa genotype will be 6.25%, and the frequency of the aa genotype will be 51.875%.

In a self-fertilizing population, the genetic composition tends to differentiate into pure lines with different genotypes. This means that individuals with similar genotypes will breed with each other, leading to the formation of pure lines with distinct genotypes. This process results in the maintenance of a high level of genetic diversity and richness within the population.

The correct answer is 41.875% AA : 6.25% Aa : 51.875% aa. This is because the given question states that the initial frequencies of the genotypes in the population are 20% AA, 50% Aa, and 30% aa. After three generations of self-fertilization, the frequencies of the genotypes will change due to Mendelian inheritance patterns. By calculating the expected frequencies of each genotype using the laws of segregation and independent assortment, it is determined that the frequencies will be 41.875% AA, 6.25% Aa, and 51.875% aa.

The given answer indicates the genetic structure of the population over three generations of self-fertilization. It shows that in the first generation, 57% of the population has the genotype AA, 6% has the genotype Aa, and 37% has the genotype aa. This distribution suggests that the population is not in Hardy-Weinberg equilibrium, as the frequencies of the genotypes are not in the expected proportions of 25% AA, 50% Aa, and 25% aa.

The relative frequency of alleles does not change in successive generations in a self-fertilizing population. This is because self-fertilization preserves the genetic composition of the population, and there is no introduction of new alleles from outside sources. Therefore, the relative frequencies of alleles remain constant over time.

The correct answer is 0.425 AA + 0.15 Aa + 0.425 aa = 1. This is the correct answer because it follows the principle of Mendelian inheritance and the laws of genetic probability. In the initial population, the genotype frequencies were given as 0.2 AA, 0.6 Aa, and 0.2 aa. After two generations of random mating, the genotype frequencies have changed, but the sum of the frequencies still equals 1. Therefore, the correct answer is the option that satisfies this condition, which is 0.425 AA + 0.15 Aa + 0.425 aa = 1.

The number of possible genotypes can be calculated by multiplying the number of alleles for each gene. In this case, there are 2 alleles for gene I, 3 alleles for gene II, and 4 alleles for gene III. Therefore, the total number of possible genotypes is 2 x 3 x 4 = 24.

In a population with random mating, if a gene has 4 alleles, the number of different genotypic combinations will be 10. This is because the number of possible genotypes can be calculated using the formula n(n+1)/2, where n is the number of alleles. In this case, n = 4, so the number of genotypic combinations is 4(4+1)/2 = 10.

In the given population, there are 2 alleles A and a, and 4% of the population has the genotype aa. This means that the frequency of allele a is 0.04. Since the frequency of alleles in a population must add up to 1, the frequency of allele A is 1 - 0.04 = 0.96. To find the relative frequencies of alleles A and a, we can divide their frequencies by the total frequency of alleles in the population. So, the relative frequency of allele A is 0.96 / (0.96 + 0.04) = 0.96, and the relative frequency of allele a is 0.04 / (0.96 + 0.04) = 0.04. Simplifying these fractions gives us the ratio 0.96 : 0.04, which is equivalent to 0.8A : 0.2a. Therefore, the correct answer is 0.8A : 0.2a.

The statement "số lượng cá thể đực bằng số lượng cá thể cái" is incorrect when talking about the conditions for the Hardy-Weinberg equilibrium. In the Hardy-Weinberg equilibrium, the population must have a large size, equal frequencies of mating between males and females, no gene flow, no mutations, and no natural selection. However, the ratio of males to females does not need to be equal in order for the equilibrium to be maintained.

After 3 generations of self-fertilization, the genotype frequencies in the population will reach a Hardy-Weinberg equilibrium. The initial genotype frequencies are 0.2AA, 0.5Aa, and 0.3aa. In each generation, the genotype frequencies will remain the same, but the allele frequencies will change due to random mating. The allele frequencies can be calculated by multiplying the frequency of each genotype by the proportion of each allele it contains. After 3 generations, the allele frequencies will stabilize. The frequency of the recessive allele (a) can be calculated by adding the frequency of aa and half the frequency of Aa. In this case, it is (0.3 + 0.5/2) = 0.55. The frequency of the dominant allele (A) can be calculated by subtracting the recessive allele frequency from 1, so it is 1 - 0.55 = 0.45. The frequency of the homozygous recessive genotype (aa) is the square of the recessive allele frequency, so it is 0.55^2 = 0.3025. The frequency of the homozygous dominant genotype (AA) is the square of the dominant allele frequency, so it is 0.45^2 = 0.2025. The frequency of the heterozygous genotype (Aa) can be calculated by subtracting the frequencies of the homozygous genotypes from 1, so it is 1 - 0.3025 - 0.2025 = 0.495. Therefore, the proportion of the homozygous recessive genotype in the population after 3 generations is 0.3025/1 = 30.25%, which is approximately 30.25%. Thus, the correct answer is 51.875%.

The statement "tần số kiểu gen đồng hợp ngày càng tăng và kiểu gen dị hợp ngày càng giảm" is incorrect when talking about a random mating population. In a random mating population, the frequency of homozygous genotypes should remain stable over time, while the frequency of heterozygous genotypes should also remain stable. This is because random mating ensures that there is no selection for or against specific genotypes, leading to a stable distribution of genotypes in the population.

The given question asks how many generations of self-pollinating population are needed for the homozygous genotype to make up 0.95 of the population. The equation provided shows the genotype frequencies in the population, with BB representing homozygous dominant, Bb representing heterozygous, and bb representing homozygous recessive. Since the homozygous genotype makes up 0.3 of the population, it will take multiple generations for it to increase to 0.95. Therefore, the correct answer is 3, indicating that three generations of self-pollination are needed for the homozygous genotype to reach 0.95 of the population.

The relative frequency of allele A is 0.4 and the relative frequency of allele a is 0.6. This can be determined by dividing the number of individuals with the dominant black fur phenotype (32) by the total number of individuals (32 + 96 + 72 = 200), which gives a relative frequency of 0.16 for allele A. Similarly, dividing the number of individuals with the white fur phenotype (72) by the total number of individuals gives a relative frequency of 0.36 for allele a. Subtracting these relative frequencies from 1 gives the relative frequencies of 0.4 for allele A and 0.6 for allele a.

In a self-fertilizing population, the relative frequency of alleles remains constant over generations, meaning that the proportions of different alleles do not change. However, the frequency of genotypes can still change due to random mating and genetic recombination, leading to variations in the frequency of different genotypes over generations. Therefore, the correct answer is that the relative frequency of alleles remains constant, but the frequency of genotypes always changes over generations.

The given statement "Quần thể là một tập hợp ngẫu nhiên và nhất thời các cá thể" is incorrect. A population is not a random and temporary collection of individuals. Instead, it refers to a group of individuals of the same species that live in the same area and can interbreed. Populations have characteristic and stable gene pools, and they represent a historical and shared developmental community.

The correct answer is "duy trì được sự đa dạng di truyền của quần thể". This means that the population maintains the genetic diversity over generations.

There are 3 alleles in the first gene and 4 alleles in the second gene, located on two different homologous chromosomes. To determine the number of different genotypes for both genes in the population, we multiply the number of alleles for each gene together. Therefore, the number of different genotypes for both genes is 3 x 4 = 12. However, since the question asks for the number of different heterozygous genotypes, we divide this number by 2, resulting in 12 / 2 = 6 heterozygous genotypes for each gene. Since there are two genes, the total number of different heterozygous genotypes for both genes is 6 x 6 = 36.

In a self-fertilizing population, the rate of homozygosity increases with each generation. This means that the proportion of individuals with two identical alleles for a particular gene increases over time. The other options are incorrect because they do not accurately describe the characteristics of self-fertilizing populations. The rate of heterozygosity decreases by half with each generation, the genetic diversity decreases gradually over generations, and the advantage of hybridization decreases over generations.

In a large population with random mating and free gene flow, it is difficult to find two identical individuals because there is a high degree of genetic variation. This is due to multiple alleles for each gene and a large number of possible combinations of genetic variations. The large number of possible combinations of genetic variations leads to a high level of genetic diversity within the population, making it unlikely to find two identical individuals.

The correct answer is 24 and 18. This can be determined by multiplying the number of alleles for each gene together. For gene I, there are 2 alleles. For gene II, there are 3 alleles. And for gene III, there are 4 alleles. Multiplying these together gives us 2 * 3 * 4 = 24. Therefore, there are 24 different homozygous genotypes in the population. To find the number of heterozygous genotypes, we subtract the number of homozygous genotypes from the total number of possible genotypes, which is 2 * 2 * 2 * 3 * 3 * 4 = 144. So, there are 144 - 24 = 120 heterozygous genotypes.

The correct answer is "tần số các kiểu gen" because it refers to the frequency of different types of genes in a population at a certain time. This is an important characteristic to consider when studying the genetic equilibrium of a population.

Each gene has 2 alleles and they are located on a pair of homologous chromosomes. The maximum number of gene types in a population is determined by the number of possible combinations of alleles. Since each gene has 2 alleles, there can be a maximum of 2^10 (2 raised to the power of 10) different combinations of alleles, which is equal to 1024. Therefore, the correct answer is 10.

The question is asking for the frequency of a specific trait in the offspring when the body carries 4 pairs of heterozygous genes. The correct answer is 7/64, which means that out of a total of 64 possible combinations of genes, 7 of them will result in the dominant trait. This can be calculated using the Punnett square method, where the dominant allele has a 1/2 chance of being passed on in each pair of genes.

If the frequency of alleles in both sexes is equal, the population will reach genetic equilibrium after only one generation of random mating. This means that the genetic composition of the population will no longer change over time, and the population will be in a state of genetic equilibrium.

The given answer (0.26 AA: 0.1 Aa: 0.64 aa) represents the approximate genotype frequencies in the population at the time of census. It suggests that 26% of the population carries the homozygous dominant genotype (AA), 10% carries the heterozygous genotype (Aa), and 64% carries the homozygous recessive genotype (aa). These frequencies are estimated based on the observed counts of each genotype in the population.

The number of possible genotypes in a population can be calculated by multiplying the number of alleles for each gene. In this case, there are 2 alleles for gene I, 3 alleles for gene II, 4 alleles for gene III, and 5 alleles for gene IV. Therefore, the total number of possible genotypes is 2 x 3 x 4 x 5 = 120.

The explanation for the correct answer "toàn cây thấp" is that the question states that in a population of plants, the tall plants are completely dominant over the short plants. This means that the tall plants will always be more numerous than the short plants. Therefore, the population will consist entirely of short plants.

The number of possible genotypes can be determined by multiplying the number of alleles for each gene. In this case, the first gene on the X chromosome has 2 alleles, and the second gene on the autosomes has 3 alleles. Therefore, the total number of possible genotypes is 2 * 3 = 6. However, since there are two genes being considered, the total number of possible genotypes becomes 6 * 6 = 36. Therefore, the correct answer is 36.

The given question is asking for the probability of two individuals who are carriers of the gene for Bệnh bạch tạng (a type of blood cancer) having a child with the disease. The question states that the frequency of carriers in the population is 1%. Since the gene is on an autosomal chromosome, the probability of two carriers having a child with the disease can be calculated using the Hardy-Weinberg equation. The equation is p^2 + 2pq + q^2 = 1, where p is the frequency of the dominant allele (normal gene) and q is the frequency of the recessive allele (disease gene). In this case, p = 0.99 (1 - 1%) and q = 0.01 (1%). Plugging these values into the equation, we get (0.99)^2 + 2(0.99)(0.01) + (0.01)^2 = 0.9801 + 0.0198 + 0.0001 = 0.999. Therefore, the probability of two carriers having a child with the disease is 0.0001/0.999 ≈ 0.0025%, which matches the given answer.

The relative frequency of a gene type is the ratio of the number of individuals carrying that gene type to the total number of individuals in the population. This means that the answer "các thể chứa kiểu gen đó trong tổng số các cá thể của quần thể" (the individuals containing that gene type in the total number of individuals in the population) best describes the relative frequency of a gene type.

The correct answer is "Kiểu gen đồng hợp". In a population with genetic equilibrium, the frequency of alleles can be determined when the number of individuals with different genotypes is known.

In the given question, the frequencies of the genotypes in the parental generation are given as 0.16AA, 0.52Aa, and 0.32aa. These frequencies add up to 1, indicating that they represent the entire population. In the F1 generation, the frequencies of the genotypes can be calculated by multiplying the frequencies of the corresponding genotypes in the parental generation.

For the genotype Aa, the frequency is calculated as 0.52 (frequency of Aa in parental generation) multiplied by 0.52 (probability of Aa genotype in the offspring of Aa parents) which equals 0.2704.

Therefore, the frequency of the Aa genotype in the F1 generation is 0.2704, which is equivalent to 0.3364 when rounded to four decimal places.

If the allele frequency is different between two populations, it indicates that the gene pool is not in equilibrium. In order for the population to reach genetic equilibrium, it would take at least 2 generations of random mating. This is because in each generation, the allele frequencies will gradually approach a stable state through random mating. Therefore, after 2 generations of random mating, the population will likely reach genetic equilibrium.

The relative frequency of a gene (allele frequency) is the percentage of gametes carrying that allele in the population. It represents the proportion of individuals in the population that carry that allele. This is the most appropriate explanation for the given correct answer.

The given sequence appears to be a list of numbers. The pattern in the sequence is not immediately clear, as there is no obvious rule or progression. Each number in the sequence is different, and there is no discernible pattern in the differences between the numbers. Therefore, without further information or context, it is not possible to determine the exact reason why 156 is the correct answer.