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marishachu
1 month ago
9

Rutherford tracked the motion of tiny, positively charged particles shot through a thin sheet of gold foil. Some particles trave

led in a straight line and some were deflected at different angles. Which statement best describes what Rutherford concluded from the motion of the particles? Some particles traveled through empty spaces between atoms and some particles were deflected by electrons. Some particles traveled through empty parts of the atom and some particles were deflected by electrons. Some particles traveled through empty spaces between atoms and some particles were deflected by small areas of high-density positive charge in atoms. Some particles traveled through empty parts of the atom and some particles were deflected by small areas of high-density positive charge in atoms.
Biology
2 answers:
-BARSIC- [2.5K]1 month ago
7 0

Answer:

Some particles went through vacant regions of the atom, while others were redirected by concentrated clusters of positive charge within the atoms.

Explanation:

Tresset [2.2K]1 month ago
5 0

Answer:

Some particles went through vacant regions of the atom, while others were redirected by concentrated clusters of positive charge within the atoms.

Explanation:

In Rutherford's experiment, particles either passed straight through the gold foil or were deflected by the positively charged nucleus.

This response is accurate because the particles moved through vacant regions of the atom (not just spaces among atoms), with some being deflected by zones of concentrated positive charge (the nucleus).

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If Jack and Jill have a child with an Aaa genotype, during which meiotic division, and in which parent, could nondisjunction hav
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Answer:

Meiosis I in the mother, meiosis I in the father, and meiosis II in the father are all potential stages where nondisjunction may occur.

Explanation:

When a haploid sperm fertilizes a haploid egg that experiences nondisjunction, there is a chance that the resulting offspring's genotype could be Aaa.

Additionally, if an egg undergoing nondisjunction in meiosis II is fertilized by a haploid sperm, the genotype of the offspring could be AAA.

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In the case of the mother, both X chromosomes would end up in one cell, leaving the other cell without any X chromosome.

If this occurs in the father, both the X and Y chromosomes could also be present in one cell, resulting in the absence of a sex chromosome in the other. Such scenarios imply that a parent might fail to pass on a sex chromosome to their child, leading to Turner syndrome, resulting in an XO configuration.

During meiosis II, sister chromatids ordinarily separate. A nondisjunction event in this phase means that the sister chromatids would migrate to the same daughter cell instead of separating into two.

When this happens in the mother, one cell would contain two identical copies of an X chromosome, while the other would lack an X chromosome. In the father, it could happen that neither of the two identical copies of the X chromosome is present in one cell, or alternatively, both identical Y chromosomes could be found together in one cell, leaving no sex chromosomes in the other. Such instances could result in a parent not transferring a sex chromosome to their child and again, this could lead to Turner syndrome with an XO genotype.

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