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3 years ago

Describe what happens to the electric field lines when two objects with unlike charges are brought near each other.

2 answers:

7 0

When two objects with unlike charges are brought near each other, their electric field lines combine, showing that the objects attract each other.

this is the correct sample response :)

Schach [20]3 years ago

5 0

Answer and explanation;

When two objects with unlike charges are brought near each other, their electric field lines combine, showing that the objects attract each other.

Oppositely charged objects will exert an attractive influence upon each other. In contrast to the attractive force between two objects with opposite charges, two objects that are of like charge will repel each other. That is, a positively charged object will exert a repulsive force upon a second positively charged object. This repulsive force will push the two objects apart.

Two electrons will tend to repel each other because both have a negative electrical charge. Two protons will also tend to repel each other because they both have a positive charge. On the other hand, electrons and protons will be attracted to each other because of their unlike charges.

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Which growing season is longest? The deciduous forest biome or the desert biome?

aliina [53]

Answer:desert biome

Explanation: because the desert go a long way and probably does not

stop for 100 miles or more. :/

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3 years ago

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Example of a human geneticdisorder caused by an alteration in

Art [367]

Answer:

Cri du chat syndrome.

Explanation:

Maintenance of chromosomal structure is important for the proper functioning of the cell. The alteration or change in the chromosome structure may leads to the various genetic problem.

The cri du chat syndrome occurs due to the missing of the short arm of the chromosome number 5. The deletion in short arm of chromosome 5 is responsible for this syndrome. The infants that suffer from this syndrome has high pitched cry that is similar to the cry of a cat.

Thus, the answer is Cri du chat syndrome.

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4 years ago

In Drosophila melanogaster, vestigial wings (vg) is recessive to normal wings (vg+), black body (b) is recessive to gray body (b

yan [13]

Correct progeny phenotype:

1779 vestigial wings, black body and purple eyes, vg b pr
1665 normal wings, a grey body and red eyes, vg+ b+ pr+
252 normal wings, an black body and purple eyes, vg+ b pr
241 vestigial wings, a gray body and red eyes, vg b+ pr+
131 normal wings, an black body and red eyes, vg +b pr+
118 vestigial wings, a gray body and purple eyes, vg b+ pr
13 vestigial wings, an black body and red eyes, vg b pr+
9 normal wings, a gray body and purple eyes, vg+ b+ pr

Answer:

The order of these genes is vg --- pr --- b
Map distances between the genes vg/pr = 12.2 MU
Map distance between the genes pr/b = 6.4 MU
Map distances between the genes vg/b = 18.6 MU

Explanation:

We know that

•Normal wings expressed by vg+ is dominant over vestigial wings, vg

•Gray body b+ is dominant over black body

•Red eyes, pr+, is dominant over purple ayes, pr

We have the number of descendants of each phenotype product of the tri-hybrid cross.

•1779 vestigial wings, black body and purple eyes vg b pr

•1665 normal wings, a grey body and red eyes vg+ b+ pr+

•252 normal wings, an black body and purple eyes vg+ b pr

•241 vestigial wings, a gray body and red eyes vg b+ pr+

•131 normal wings, an black body and red eyes vg +b pr+

• 118 vestigial wings, a gray body and purple eyes vg b+ pr

•13 vestigial wings, an black body and red eyes vg b pr+

• 9 normal wings, a gray body and purple eyes vg+ b+ pr

The total number of individuals is 4208.

In a tri-hybrid cross, it can occur that the three genes assort independently or that two of them are linked and the third not, or that the three genes are linked. In this example, in particular, the three genes are linked on the same chromosome.

Knowing that the genes are linked, we can calculate genetic distances between them. First, we need to know their order in the chromosome, and to do so, we need to compare the genotypes of the parental gametes with the ones of the double recombinants. We can recognize the parental gametes in the descendants because their phenotypes are the most frequent, while the double recombinants are the less frequent. So:

Parental)

1779 vestigial wings, black body and purple eyes vg b pr

1665 normal wings, a grey body and red eyes vg+ b+ pr+

Double recombinant)

13 vestigial wings, an black body and red eyes vg b pr+
9 normal wings, a gray body and purple eyes vg+ b+ pr

Simple recombinant)

252 normal wings, an black body and purple eyes vg+ b pr
241 vestigial wings, a gray body and red eyes vg b+ pr+
131 normal wings, an black body and red eyes vg +b pr+
118 vestigial wings, a gray body and purple eyes vg b+ pr

Comparing parental with the double recombinants we will realize that between

vg b pr (parental)

vg b pr+ (double recombinant)

and

vg+ b+ pr+ (Parental)

vg+ b+ pr (double recombinant)

They only change in the position of the alleles pr/pr+. This suggests that the position of the gene pr is in the middle of the other two genes, vg and b, because in a double recombinant only the central gene changes position in the chromatid.

So, the order of the genes is:

---- vg ---- pr -----b ----

Now we will call Region I to the area between vg and pr and Region II to the area between pr and b.

Once established the order of the genes we can calculate distances between them, and we will do it from the central gene to the genes on each side. First We will calculate the recombination frequencies, and we will do it by region. We will call P1 to the recombination frequency between vg and pr genes, and P2 to the recombination frequency between pr and b.

P1 = (R + DR) / N

P2 = (R + DR)/ N

Where: R is the number of simple recombinants in each region, DR is the number of double recombinants in each region, and N is the total number of individuals.

So:

Parental)

• 1779 vestigial wings, black body and purple eyes vg pr b

• 1665 normal wings, a grey body and red eyes vg+ pr+ b+

Double recombinant)

• 13 vg pr+ b

• 9 vg+ pr b+

Simple recombinant)

• 252 vg+ pr b

• 241 vg pr+ b+

• 131 vg+ pr+ b

• 118 vg pr b+

P1 = (R + DR) / N

P1 = (252+241+13+9)/4208

P1 = 515/4208

P1 = 0.122

P2= (R + DR) / N

P2 = (131+118+13+9)/4208

P2 = 271/4208

P2 = 0.064

Now, to calculate the recombination frequency between the two extreme genes, vg and b, we can just perform addition or a sum:

P1 + P2= Pt

0.122 + 0.064 = Pt

0.186=Pt

The genetic distance will result from multiplying that frequency by 100 and expressing it in map units (MU). One centiMorgan (cM) equals one map unit (MU).

The map unit is the distance between the pair of genes. Every 100 meiotic products, one of them results in a recombinant product. Now we must multiply each recombination frequency by 100 to get the genetic distance in map units:

GD1= P1 x 100 = 0.122 x 100 = 12.2 MU

GD2= P2 x 100 = 0.064 x 100 = 6.4 MU

GD3=Pt x 100 = 0.186 x 100 = 18.6 MU

---- vg ---------------------- pr ---------------------b ----

R1 R2

-----vg----12.2MU---------pr—

----pr--------6.4 MU----b—

-----vg ----------------18.6 MU--------------------b----

3 0

3 years ago

1 point

Alekssandra [29.7K]

<h2>Option (B) is Right Answer</h2>

Explanation:

(B) bacteria that make insulin for is right answer

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6 0

3 years ago

Hydrogen covalently bonds with oxygen to form the compound water. Oxygen becomes MORE chemically stable because

Digiron [165]

A- when the atoms covalently bond, the two hydrogens share an electron with the oxygen. As a result, they both acquire the ‘noble gas’ rank of having a full out energy shell (8 valence electrons).

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4 years ago

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