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Elis [28]
3 years ago
9

Do all eukaryotic cells contain a cell wall

Biology
1 answer:
romanna [79]3 years ago
3 0

Many kinds of prokaryotes and eukaryotes contain a structure outside the cell membrane called the cell wall. With only a few exceptions, all prokaryotes have thick, rigid cell walls that give them their shape. Among the eukaryotes, some protists, and all fungi and plants, have cell walls.


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What was the independent variable in Francesco Redi's experimental setup?
Furkat [3]
THe maggots is the correct answer
4 0
3 years ago
Read 2 more answers
What term is defined as the ability of a test to detect small concentration of an antigen or antibody?
valina [46]
The answer is 5

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7 0
2 years ago
An atom that has 19 protons, 21 neutrons, and 18 electrons has a mass of ____? *
Paul [167]

Answer:

19 - Potassium - K.

Explanation:

3 0
2 years ago
2. List three sources of error that could account for the differences between your values for the enthalpy of fusion of water an
Dvinal [7]

1 trial :  nothing is given for result comparision - so we have no idea if it's a mistake.

2nd trial : The results can be compared - if varies, one may go wrong, but which one?

3rd trial : If 3rd result is different from 1st and 2nd, it is unreliable.

calculating enthalpy of fusion. M, C and m,c = mass and specific heat of calorimeter and water, n, L = mass and heat of fusion of ice; T = temperature fall.

L = (mc+MC)T/n.

c=4.18 J/gK. assuming copper calorimeter , so C=0.385 J/gK.

1. M = 409g, m = 45g. T = 22c, n = 14g

L = (45*4.18+409*0.385)*22/14 = 543.0 J/g.

2. M = 409g, m = 49g, T = 20c, n = 13g

L = (49*4.18+409*0.385)*20/13 = 557.4 J/g.

3. M = 409g, m = 54g, T = 20c, n = 14g

L = (54*4.18+409*0.385)*20/14 = 547.4 J/g.

(i) Estimate error in L from spread of 3 results.

Average L = 549.3 J/g.

squared differences average (variance) = (6.236^2+8.095^2+1.859^2)/3 = 35.96

standard deviation = 5.9964

standard error = SD/(N-1) = 5.9964/2 = 3 J/g approx.

% error = 3/547 x 100% = 0.5%.

(ii) Estimate error in L from accuracy of measurements:

error in masses = +/-0.5g

error in T = +/-0.5c

For Trial 3

M = 409g, error = 0.5g

m = 463-409, error = sqrt(0.5^2+0.5^2) = 0.5*sqrt(2)

n =(516-463)-(448-409)=14, error = 0.5*sqrt(4) = 1.0g

K = (mc+MC)=383, error = sqrt[2*(0.5*4.18)^2+(0.5*0.385)^2] = 2.962

L = K*T/n

% errors are

K: 3/383 x 100% = 0.77

T: 0.5/20 x 100% = 2.5

n: 1.0/14 x 100% = 7.14

% errors in K and T are << error in n, so ignore them.

% error in L = same as in n = 7% x 547.4 = 40

The result is (i) L= 549 +/- 3 J/g or (ii) L = 550 +/- 40 J/g.

Both are very far above  334 J/g, so there is at least one systematic error  

e.g: calorimeter may not be copper, so C is not 0.385 J/gK. (If it was polystyrene, which absorbs/ transmits little heat, the effective value of C would be very low, reducing L.)

Using +/- 40 is best.

However, the spread in the actual results is much smaller

* measurements were "fiddled" to get better results; other Trials were made but only best 3 were chosen.

<h3>Other sources of error: </h3>

L=(mc+MC)T/n is too high, so n (ice melted) may be too small, or T (temp fall) too high - why?

* we have assumed initial and final temperature of ice was 0c, it may actually have been colder, so less ice would melt -which explain small values of n

* some water might have been left in container when unmelted ice was weighed (eg clinging to ice) - again this could explain small n;

* poor insulation - heat gained from surroundings, melting more ice, increasing n - but this would reduce measured L below 334 J/g not increase it.

* calorimeter still cold from last trial when next one started, not given time to reach same temperature as water - this would reduce n.

3 0
3 years ago
A geneticist crossed fruit flies to determine the phenotypic ratio. The geneticist crossed a fly with blistery wings and spinele
kondor19780726 [428]

Complete question:

A geneticist crossed fruit flies to determine the phenotypic ratio. The geneticist crossed a fly with blistery wings and spineless bristles (bbss) with a heterozygous fly that had normal wings and normal bristles (BbSs). Which proportion of offspring that are dominant for both traits in would you not expect based on Mendel's law of independent assortment? 1/2 , 4/16, 25% , or 1/4

Answer:

1/2 is the proportion of the offspring that is NOT expected among individuals that are dominant for both traits.

4/16 = 1/4 = 25% of the progeny and the correct expected proportion of individuals that are dominant for both traits.

Explanation:

<u>Available data</u>:

  • Cross:  a fly with blistery wings and spineless bristles with a heterozygous fly that had normal wings and normal bristles
  • Recessive trait: blistery wings and spineless bristles
  • Dominant trait: normal wings and normal bristles

Let us say that:

  • B is the dominant allele for normal wings
  • b is the recessive allele for blistery wings
  • S is the dominant allele for normal bristles
  • s is the recessive allele for spineless bristles

Parentals)        bbss       x        BbSs

Gametes)  bs, bs, bs, bs     BS, Bs, bS, bs

Punnett square)    BS        Bs         bS        bs

                     bs    BbSs    Bbss     bbSs    bbss

                     bs    BbSs    Bbss     bbSs    bbss

                     bs    BbSs    Bbss     bbSs    bbss

                     bs    BbSs    Bbss     bbSs    bbss

F1)  4/16 = 1/4 = 25%  of the progeny is expected to be BbSs, dyhibrid individuals, expressing normal wings and normal bristles

     4/16 = 1/4 = 25% of the progeny is expected to be Bbss, expressing normal wings and spineless bristles

     4/16 = 1/4 = 25% of the progeny is expected to be bbSs, expressing  blistery wings and normal bristles

     4/16 = 1/4 = 25% of the progeny is expected to be bbss, expressing  blistery wings and spineless bristles    

5 0
3 years ago
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