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What is the relationship between Pokemon, marshmallow and comic?

2025-08-12 14:29
2025-08-12 18:36

It's hard to say for sure. Maybe there could be a comic that features Pokemon characters enjoying marshmallows, or a story where marshmallows have powers similar to Pokemon. It all depends on the imagination of the creator.

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I think it's quite possible that they have no inherent connection. But in the world of creativity, anything can happen. Maybe there's a comic where Pokemon use marshmallows as a tool or a treat to gain special abilities.

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At the middle and late stages of the second cell division, the number of embryos in a cell and the number of embryos labeled with 32P were respectively () - Answer: Due to the semi-conservative replication of DNA, in the middle stage of the second cell division, the number of microorganisms was 20, and one strand of the DNA molecules on each of the microorganisms was labeled. Therefore, the number of microorganisms labeled by 32P was 20. In the anaphase, the number of microorganisms became 40, but the number of microorganisms labeled by 32P was still 20. The answer was A. 20 and 20 in the middle stage, 40 and 20 in the anaphase. 4. ** Meiosis with marker DNA ** - For example, if the DNA molecules of a sperm cell containing a pair of homological embryos are labeled with 15N and supplied with 14N-containing materials, the ratio of 15N-labeled sperm among the four sperm produced by the cell during Meiosis is () - Answer: Since DNA replication is semi-conservative replication, a single sperm cell contains a pair of homolog chromomes. After replication, the two sister chromatids of each of the primary sperm cells contain 15N. Among the four sperm formed by Meiosis, each sperm contains 15N. Therefore, the proportion of sperm containing 15N is 100%. The answer is D. 5. ** Type of gametes produced by the sperm cell ** - For example: Droplets have eight embryos, and the type of gametes produced by the Meiosis of a sperm cell is () - Answer: A single sperm cell would produce two types of four gametes during the Meiosis. The answer was A.2. 6. ** Cell division after fusion of neurospora crassa ** - For example, the haploid cell of the neurospora crassa has seven nuclei. Two different types of neurospora crassa become diploids after fusion, and then a typical Meiosis occurs, followed by a mitosis. The final number of daughter cells and the number of embryos in each daughter cell are () - Answer: After the two haploids fuse into a diploid, the number of microorganisms will be 14. After Meiosis, four daughter cells will be formed, and each daughter cell will have seven microorganisms. After another mitosis, the number of daughter cells will become eight, and the number of microorganisms in each daughter cell will still be seven. The answer is A. Eight, seven. ** 2. Calculation of Inheritance Law ** 1. ** Genotype-ratio of offspring of multiple gene hybrids ** - For example, if sunflower hybrids with AaBbCcdd and AabbCcDd are crossed, according to the law of free gene combination, the proportion of individuals with AabBccDd in the offspring should be () - The answer was to calculate the ratio of each pair of genes to produce a specific gene type. The ratio of Aa× AAto AAis 1/2; the ratio of Bb×Bb to BBis 1/4; the ratio of Cc×Cc to CCis 1/4; and the ratio of DD×Dd to Dd is 1/2. Multiplying these proportions, 1/2×1/4 ×1/4×1/2 = 1/64, the answer is D. 1/64. 2. ** Analysis of the proportion of the crossbred offsprings ** - For example, the red fruit of tomato is dominant to the yellow fruit, and the round fruit is dominant to the long fruit, and they are freely combined. Now, the red long fruit and the yellow round fruit tomato are crossed. Theoretically, the proportion of the offspring's genetics that cannot appear is () - Answer: According to the law of free combination of genes, analyze the proportion of the genes in the offspring of different parent combinations. The ratio of 1:0 may appear when a pair of genes are crossed with a pair of genes. 1:1 may appear when a pair of genes are tested. 1:1:1:1 may appear when a pair of genes are tested. 1:2:1 is the ratio of the genes of the self-bred offspring of a pair of genes. The answer is B. 1:2:1. 3. ** Genetic pedigree calculation ** - For example, the picture on the right is the genetic pedigree of an albino family. Please estimate the probability of the couple II-2 and II-3 having an albino child () - Answer: First of all, albinism is an autosomal-type hereditary disease. Suppose the pathogenic gene is A and the normal gene is A. According to the pedigree, the II-2 gene type was 1/3AA or 2/3Aa, and the II-3 gene type was Aa. The probability of them having an albino child (aa) is 2/3×1/4 = 1/6. (The original answer here is wrong. This result is obtained according to the normal calculation process.) 4. ** Calculation of Homozyte Progeny of Pea Self-cross ** - For example, there is a green (yy) round (Rr) pea. Their relative characteristics are yellow and wrinkled. It was known that these two pairs of genes were located on two pairs of homolog genes. The peas were planted and self-pollinated, and the offspring were all planted without selection. They were self-pollinated again and harvested n seeds. It can be inferred that the number of green and round grains in the n grains is about () - Answer: Green (yy) self-bred offspring are all green, round (Rr) self-bred offspring account for 1/4. The first generation of the second generation was obtained by self-crossing. The ratio of green circles (yyRV) was 1×1/4 = 1/4, so there were about n/4 green and round grains in the n grains. The answer was C.n/4. 5. ** Calculating the ratio of the crossbred offspring of two pairs of relative traits ** - For example, the tall and short stems of rice are a pair of relative traits, while the glutinous and non-glutinous traits are a pair of relative traits. Some people let a high-stem non-glutinous rice cross with another short-stem non-glutinous rice, and the resulting offspring are as shown in the right picture (these two pairs of traits are inherited according to the law of free combination). Among the short-stem non-glutinous rice of the offspring, the rice that can be stably inherited accounts for () - Answer: First, determine the parent's gene type according to the ratio of the hybrids, and then analyze the ratio of the dwarf non-glutinous individuals. Assuming tall stem (D), short stem (d), glutinous (R), and non-glutinous (r), the parent genomes were deduced as DdRr and ddRr according to the results of the crossing. 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