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Tresset [83]
3 years ago
6

predict how the rate of cell division would differ between single-celled algae living in a sunny,nutrient-rich pond versus algae

living in a shady, nutrient-poor pond.How could you test your prediction?
Biology
1 answer:
Doss [256]3 years ago
4 0
Heat given off by the sun would give the cell energy aswell as the nutrients therefore the rate of cell division would be higher. the prediction could be tested by putting algae I'm both the places and recording the rate of cell division
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What could cause an igneous rock to become soil?
Troyanec [42]
The choices can be found elsewhere and as follows:

a. Increased temperature and pressure 
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<span>c. Exposure and weathering </span>
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PLZ HELP <br>Identify the process that orginsims use to grow and replace old or damaged cells
RUDIKE [14]
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4 0
3 years ago
22. A plant geneticist is investigating the inheritance of genes for bitter taste (Su) and explosive rind (e) in watermelon (Cit
mylen [45]

Answer:

See the answer below

Explanation:

Using the formula for calculating Chi square (X^2):

X^2  =  \frac{(O - E)^2}{E} where O = observed frequency and E = expected frequency.

The observed frequencies for the four phenotypes are 88, 62, 62, and 81 respectively.

For the expected frequency, the phenotypes are supposed to assort in 9:3:3:1 according to Mendel ratio.

Hence, expected frequencies are calculated as:

phenotype (1) = 9/16 x 293 = 164.81

phenotype (2) = 3/16 x 293 = 54.94

phenotype (3) = 3/16 x 293 = 54.94

phenotype (4) = 1/16 x 293 = 18.31

The X^2 is calculated thus:

Phenotype            O               E                             X^2

  1                          88            164.81            \frac{(88 - 164.81)^2}{164.81} = 35.80

  2                          62             54.94            \frac{(62 - 54.94)^2}{54.94} = 0.91

  3                           62             54.94              \frac{(62 - 54.94)^2}{54.94} = 0.91

  4                            81              18.31               \frac{(81 - 18.31)^2}{18.31} = 214.64

Total X^2 = 35.80 + 0.91 + 0.91 + 214.64 = 252.26

To the nearest tenth = 252

Degree of freedom = n - 1 = 4- 1 = 3

X^2 tabulated  = 7.815

<em>The calculated </em>X^2<em> exceeds the critical value, hence, the null hypothesis is rejected. The two genes did not assort independently.        </em>

4 0
3 years ago
Which one of the codons below would stop the translation of mRNA by ribosomal subunits?
Ostrovityanka [42]

Answer:

The correct option is D) UAG, UAA, UGA

Explanation:

The amino acid sequencing code or mRNA code contains specific codes which start and stop the process of translation at the right time. If the stop codon were not present then the ribosomal machinery would have made faulty proteins. If the stop codons are not at the right place, then it results in the production of faulty proteins. The stop codons which terminate the process of translation are UAG, UAA and UGA.

5 0
2 years ago
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