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GalinKa [24]
3 years ago
8

Yanni made a chart to compare the strong and weak forces. Which best describes Yanni’s error? Only the weak force is active in t

he nuclei of atoms. The weak force keeps particles that make up neutrons together. The strong force overcomes repulsion and keeps protons together. The strong force is responsible for radioactive decay..

Biology
2 answers:
cluponka [151]3 years ago
8 0

Answer: Option (d) is the correct answer.

Explanation:

Nucleus of an atom consists of protons and neutrons. Protons are positively charged and neutrons have no charge. So, due to the like charges of protons there occurs electrostatic force of repulsion inside the nucleus of the atom.

But due to similar number of neutrons and protons a force that is able to bind both of them together is known as strong nuclear binding energy.

This force is strong enough that it is able to overcome electrostatic force of repulsion. But when there is great difference in the number of protons and neutrons then binding force is not strong enough.

Hence, the atom becomes unstable and undergoes radioactive decay. So, this means weak forces are responsible for radioactive decay.

Thus, we can conclude that the statement which best describes Yanni’s error is that the strong force is responsible for radioactive decay.

svet-max [94.6K]3 years ago
3 0
So it's t<span>he strong force that overcomes repulsion and keeps the nucleus together.</span>
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A cross is made between homozygous wild-type female Drosophila (a^+ a^+ b^+ b^+ c^+ c^+) and triple-mutant males (aa bb cc) (the
marusya05 [52]

Answer:

a is the middle gene.

Distance [b-a]= 24.7 mu

Distance [a-c]= 15.8 mu

Distance [b-a} = 40.5 mu

Explanation:

A homozygous wild-type female drosophila (a⁺b⁺c⁺/a⁺b⁺c⁺) is crossed with a homozygous recessive male (abc/abc). <u>The order of the genes here is arbitrary. </u>

The F1 is heterozygous for the three genes (a⁺b⁺c⁺/abc). The F1 females were test crossed (crossed with abc/abc males).

The F2 shows the following phenotypic ratios:

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  • 15 a b⁺ c⁺

Total = 1000

The male parent is homozygous recessive for the 3 genes, so the observed phenotypes of the offspring correspond to the gametes received from the mother.

Recombination during meiosis is a rare event, so the most abundant gametes are always the parentals:  a⁺b⁺c⁺ and abc.

The least abundant gametes, following the same logic, are the double crossovers (DCO): a⁺bc and ab⁺c⁺.

<h3><u>1st. Determine the gene order</u></h3>

Compare the parental and the DCO gametes. The allele that is switched corresponds to the middle gene. In this case, gene a is in the middle of the other two.

<h3><u>2nd Determine the single crossover gametes</u></h3>

The F1 mother that generated all 8 types of gametes had the genotype b⁺a⁺c⁺/bac (correct order of genes).

  • The single crossover (SCO) gametes resulting from recombination between genes b and a are b⁺ac and ba⁺c⁺.
  • The single crossover (SCO) gametes resulting from recombination between genes a and c are b⁺a⁺c and bac⁺.
<h3><u>3) Calculate the recombination frequencies between genes </u></h3>

Recombination frequency (RF) = #Recombinants/Total progeny

  • RF [b-a]= (102+112+18+15)/1000= 0.247
  • RF [f-br]= (66+59+18+15)/1000= 0.158
<h3><u /></h3><h3><u>4) Calculate the distance in map units </u></h3>

Distance (mu) = RF x 100

Distance [b-a]= 0.247 × 100 = 24.7 mu

Distance [a-c]= 0.158 × 100 = 15.8 mu

Distance [b-a} = 24.7 mu + 15.8 mu = 40.5 mu

<h3><u>The gene map therefore looks like: </u></h3>

b------------24.7 mu--------------------------a---------15.8 mu-----------c

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I need help !!<br> And if you can give explanation it would mean a lot :)
marusya05 [52]

Answer:

d

Explanation:

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Answer:

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Explanation:

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This structure has two portions:

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Some even propose that the borderline surface between the pharynx and larynx can be called the laryngopharynx, due to its tissue structure.

The lower portion of the pharynx is covered by a squamous stratified epithelium, the inner layers of which rest on a basal lamina. The more superficial layers provide protection to the interior of the pharynx against friction, in addition to remaining lubricated by mucous secretion at that level.

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