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Pavlova-9 [17]
2 years ago
12

A grasshopper jumps along a line. His first jump takes him 1 cm., the second 2 cm, etc. Each jump can take him to the right or t

o the left. Prove that after 2017 jumps, the grasshopper cannot be in the same location it started.
Biology
1 answer:
loris [4]2 years ago
5 0

Answer:

The grasshopper cannot be in the same location it started because by 2017, it would have covered a distance of 1017576.5 to either left or right. and that distance is not an integer.

Explanation:

The grasshopper jumps in integer of 1 cm to either right or left

Calculating for the total distance covered in 2017 jumps

Total Distance = (2017 x 2018)/2 = 2035153 cm

To either right or left: this implies that the total distance be divided by 2

=> 1017576.5 cm

It shows that at some point, the grasshopper will not be in a position of integer on either right or left.

Therefore, it can not get to its initial point of start.

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Acentric chromosomes, which lack centromeres, can form as a result of chromosome breakage or as a consequence of inversion with
Furkat [3]

Answer:

It wouldn't be able to migrate to the cell's poles.

Explanation:

During anaphase, microtubules attach the chromosome's centromeres and start dragging them towards the poles of the cell. In mitosis, this causes the separation of both sister chromatids for each chromosome, and each chromatid migrates to a different pole. In meiosis, first each homologous chromosome is separated by the same process, and then the sister chromatids are separated during anaphase II. If a chromosome doesn't have centromeres, microtubules are unable to attach the chromosomes, and so the separation of either sister chromatids or homologous chromosomes can't take place.

8 0
2 years ago
MicroRNA (miRNA) production occurs in several steps. Arrange the statements describing the production and function of miRNAs in
zmey [24]

Answer:

The miRNAs act as post-transcriptional silencers, as they are similar to specific mRNAs and regulate their stability and translation. They are small endogenous non-coding ribonucleic acid (RNA) molecules, with about 22 nucleotides, which act as regulators of gene expression in plants and animals, at the post-transcriptional level through the cleavage of a target messenger RNA (mRNA) or repression of translation.

In general, most miRNA genes are transcribed by RNA polymerase II in the nucleus in primary miRNAs (pri-miRNAs). Individually, a pri-miRNA can produce a single miRNA or contain groups of two or more miRNAs that are processed from a common primary transcript. These long pri-miRNA are cleaved by a complex comprising the double-stranded RNAse III enzyme (DROSHA) and its essential cofactor, the binding protein DGCR8 (DiGeorge Syndrome Critical Region 8 protein) in mammals. DROSHA contains two domains of RNAse III, each of which cleaves a strand of the RNA resulting in the precursor microRNA (pre-miRNA) with about 70 base pairs, which contains a double-stranded stretch and a single-stranded loop, forming a structure in clamp. The pre-miRNA is exported to the cytoplasm by the protein exportin-5 (XPO-5), where it is cleaved by DICER1, an RNAse III that assesses the 3 'and 5' ends of the pre-miRNA, generating a mature miRNA with about 22 nucleotides. The processing of pre-miRNA by Dicer promotes the unfolding of the RNA duplex in the form of a clamp. The position in the formation of the clamp can also influence the choice of tape.

Explanation:

7 0
2 years ago
When you cross two heterozygotes (aa), the offspring will most likely be __________?
Cerrena [4.2K]
When you cross two heterozygotes with the genotype Aa, the offspring will be:   <span>Parents: Aa  x  Aa</span>
<span>F1 generation: AA  Aa  Aa  aa</span>
<span>This means that ½ of the offspring will be heterozygous, ¼ will be homozygous (dominant) and ¼ will be homozygous (recessive).</span>
6 0
2 years ago
In the stable form of protein, the ________ is generally oriented to the interior of the protein molecule.
dimulka [17.4K]

Answer:

Hydrophobic part

Explanation:

A protein is made up of amino acids. Some amino acids such as serine and threonine are polar and hydrophilic in nature as they have polar side chains. The other amino acids such as alanine and proline are non-polar as they have nonpolar side chains.

The polar amino acids of a polypeptide chain are folded towards the inside of the protein to avoid any interaction with the surrounding watery medium. On the other hand, the polar and charged amino acids are oriented towards the surface of a folded protein. This arrangement makes the proteins stable.

8 0
2 years ago
A micrograph of dividing cells from a rat, a diploid animal, shows a cell where 21 pairs of sister chromatids are being pulled a
Elanso [62]

Answer:

<h2>b) Anaphase II of meiosis </h2>

Explanation:

1. Meiosis is the process of cell division in which one cell is divided into four daughter cell, each contains equal number of chromosome, half the number of chromosomes as compared to parental cell.

2. In meiosis I, DNA duplication occurs but the sister chromatids are not separated, only homologous pair of chromosomes are separated, so this is called reductional division.

3. In meiosis II, chromatids are pulled apart and and are separated into different chromosomes, so it is called equational division. There is no DNA duplication in meiosis II.

8 0
2 years ago
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