Two identical sister chromatids are held together at whatregion on a duplicated chromosome?
a Chromatin
b Centromere
C Centriole
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Answers

Answer 1
Answer:
  • After DNA duplication, the chromosome becomes composed of two identical structures, called sister chromatids, which are joined at the centromere area.

Answer 2
Answer:

Final answer:

Sister chromatids are held together at the centromere on a duplicated chromosome. The centromere serves as the point of attachment. Chromatin and centrioles do not serve this function. The correct option is b) Centromere.

Explanation:

Two identical sister chromatids are held together at the centromere on a duplicated chromosome. The centromere serves as the point of attachment for sister chromatids. During cell division, the spindle fibers attach themselves to the centromere to pull apart the sister chromatids into two separate cells. This process helps ensure that each new cell receives an identical and complete set of chromosomes. Neither chromatin nor centriole holds sister chromatids together. Chromatin refers to the material chromosomes are made up of, which includes DNA and protein, while centrioles are involved in cell division. In summary, the precise answer to your question is option b) Centromere.

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Related Questions

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Benzene is a starting material in the synthesis of nylon fibers and polystyrene (styrofoam). Its specific heat capacity is 1.74 J/g·°C. If 16.7 kJ of energy is absorbed by a 225-g sample of benzene at 20.0°C, what is its final temperature?

Answers

Answer: The final temperature of the sample is 62.66°C

Explanation:

To calculate the amount of heat absorbed, we use the equation:

Q=mc\Delta T

where,

Q = heat absorbed = 16.7 kJ = 16700 J   (Conversion factor:  1 kJ = 1000 J)

m = Mass of the sample = 225 g

c = specific heat capacity of sample = 1.74J/g.^oC

\Delta T = change in temperature = T_2-T_1=(T_2-20.0)

Putting values in above equation, we get:

16700=225g* 1.74J/g.^oC* (T_2-20)^oC\n\nT_2=62.66^oC

Hence, the final temperature of the sample is 62.66°C

Which isomer would you expect to undergo E2 elimination faster, trans-1-bromo-4-tert-butylcyclohexane or cis-1-bromo-4-tert-butylcyclohexane? Draw each molecule in its more stable chair conformation, and explain your answer.

Answers

Answer: Cis-1-bromo-4-tert-butylcyclohexane would undergo faster elimination reaction.

Explanation:

The two primary requirements for an E-2 elimination reaction are:

1.There must be availability of β-hydrogens that is presence of hydrogen on the carbon next to the leaving group.

2.The hydrogen and leaving group must have a anti-periplanar position .

Any substrate which would follow the above two requirements can give elimination reactions.

For the structure of trans-1-bromo-4-tert-butylcyclohexane and cis-1-bromo-4-tert-butylcyclohexane  to be stable it  must have the tert-butyl group in the equatorial position as it is a bulky group and at equatorial position it would not repel other groups. If it is kept on the axial position it would undergo 1,3-diaxial interaction and would destabilize the system and that structure would be unstable.

Kindly find the structures of trans-1-bromo-4-tert-butylcyclohexane and cis-1-bromo-4-tert-butylcyclohexane in attachment.

The cis- 1-bromo-4-tert-butylcyclohexane has the leaving group and β hydrogens in anti-periplanar position so they can give the E2 elimination reactions easily.

The trans-1-bromo-4-tert-butylcyclohexane  does not have the leaving group and βhydrogen in anti periplanar position so they would not give elimination reaction easily.

so only the cis-1-bromo-4-tert butyl cyclohexane would give elimination reaction.

Final answer:

Trans-1-bromo-4-tert-butylcyclohexane is expected to undergo E2 elimination faster than cis-1-bromo-4-tert-butylcyclohexane due to less steric hindrance.

Explanation:

In determining the rate of E2 elimination, the trans-1-bromo-4-tert-butylcyclohexane would undergo E2 elimination faster than the cis-1-bromo-4-tert-butylcyclohexane. This is due to the larger degree of steric hindrance in the case of the cis isomer.

In trans-1-bromo-4-tert-butylcyclohexane, the bromine is at the equatorial position while the tert-butyl group is axial. It forms a structure that allows the compound to experience less steric hindrance with bromine in a more favorable position for leaving.

In comparison, cis-1-bromo-4-tert-butylcyclohexane has a bromine and tert-butyl group both at equatorial positions. This causes steric hindrance, and in turn, slows down the E2 elimination rate. Despite the more stable conformation, the bromine is not well-oriented for a leaving group in E2 elimination.

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Given the following equation: 8 Fe + S8 ---> 8 FeSHow many moles of iron are needed to react with 16.0 moles of sulfur?
A FULL STEP EXPLANATION WILL BE APPRECIATED

Answers

The number of moles of iron needed to react with 16.0 moles of sulfur is 128.0 moles.

Calculation of number of moles of iron

Given equation, 8Fe + S8 = 8 FeS

Moles of sulfur = 16.0

To react 1 mole of sulfur, we need 8 moles of Fe

So, for 16.0 moles of sulfur we need

16.0* 8 = 128.0 moles

Thus, to react with 16.0 moles of Sulfur, 128.0 moles of Fe is needed.

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Answer:

To react with 16.0 moles of sulfur we need 128.0 moles of iron (Fe).

Explanation:

Step 1: Data given

Number of moles Sulfur = 16.0 moles

Step 2: The balanced equation

8 Fe + S8 → 8 FeS

Step 3: Calculate mole Fe

For 8 moles Fe we need 1 mol S8 to produce 8 moles FeS

For 16.0 moles of Sulfur we need 8*16.0 = 128.0 moles

To react with 16.0 moles of sulfur we need 128.0 moles of iron (Fe).

A mixture contains N a H C O 3 together with unreactive components. A 1.68 g sample of the mixture reacts with H A to produce 0.561 g of C O 2 . What is the percent by mass of N a H C O 3 in the original mixture

Answers

There is 65% of NaHCO3 in the sample.

The equation of the reaction is;

HA + NaHCO3 -----> NaA + CO2 + H2O

Amount of CO2 formed = mass/molar mass

mass of CO2 = 0.561 g/44 g/mol = 0.013 moles

From the balanced reaction equation;

1 mole of NaHCO3 yields 1 mole of CO2

0.013 moles of Na2CO3 yields 0.013 moles of CO2

Hence, mass of NaHCO3 in the sample = 0.013 moles × 84 g/mol = 1.092 g of NaHCO3

Percentage by mass of NaHCO3 = 1.092 g/1.68 g ×100/1

= 65%

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Answer:

63.75%.

Explanation:

The first step here is to write out the reaction showing the chemical reaction between the two chemical species. Thus, we have;

HA(aq) + NaHCO3 --------------> CO2(g) + H20(l) + NaA(aq).

Therefore, the mole ratio is 1 : 1 : 1 : 1 that is go say one mole of HA reacted with one mole of NaHCO3 to give one mole of CO2 and one .ole of NaA.

Hence, the number of moles of CO2 = mass/molar mass = 0.561/44 = 0.01275 moles.

Thus, the number of moles of NaHCO3 = number of moles of CO2 = 0.01275 moles.

Therefore, we have ( 0.01275 moles × 84 g/mol) grams = 1.071 g NaHCO3 in the mixture.

Therefore, the percent by mass of N a H C O 3 in the original mixture = 1.071/1.68 × 100 = 63.75%.

In their elemental forms, the halogens are Select one: a. strong oxidizing agents. b. strong reducing agents. c. strong acids. d. strong bases. e. amphoteric.

Answers

Answer:

A

Explanation:

Halogens have 7 electrons in their valence shell and requires just one electron to achieve their stable structure. Because of this they have high electron affinity and hence high electronegativity and they are strong oxidizing agents. Oxidation is the loss of electrons . Halogens have a high readiness to take electrons which a metal loses thereby oxidizing it. The other options are wrong as they involve compounds and not the elemental form as stated in the question.

In the important industrial process for producing ammonia (the Haber Process), the overall reaction is: N2(g) + 3H2(g) → 2NH3(g) + 24,000 calories A yield of ammonia, NH3, of approximately 98% can be obtained at 200°C and 1,000 atmospheres of pressure. How many grams of N2 must react to form 1.7 grams of ammonia?

Answers

Answer:

Mass of N2 required = 1.429 g

Explanation:

The given reaction is:

N2(g) + 3H2(g) → 2NH3(g)

Mass of NH3 formed = 1.7 g

Molar mass of NH3 = 17 g/mol

Moles(NH3) = (Mass)/(Molar mass) = (1.7)/(17) = 0. 1

Based on the reaction stoichiometry:

1 mole of N2 forms 2 moles of NH3

Therefore, moles of N2 required to produce 0.1 moles of NH3 is:

= (1 mole(N2)*0.1moles(NH3))/(2 moles(NH3)) =0.05 moles(N2)

Molar mass of N2 = 28 g/mol

Mass of N2 required = moles*molar mass = 0.05*28 = 1.4 g

This is the theoretical mass corresponding to a 100% yield. Since the yield of NH3 is 98%, the corresponding mass of N2 required would be:

=(1.4)/(0.98) =1.429 g

28g N2/(17 x 2 x 0.98) g NH3


1.7 g NH3 x 28g N2/(17 x 2 x 0.98) g NH3 =

47.6 / 33.32 = 1.43 g N2