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3 years ago

HEYYY GUYSSS HELPPPP MEEEE PLEEASSEEEE ILL MARKK YOUU BRAINLYYY :)

Biology

1 answer:

Len [333]3 years ago

4 0

Q1. broke down into carbon dioxide through a series of energy-extracting reactions collectively called the Krebs cycle.

Q2. Uses the high energy electrons from glycolysis and the Krebs cycle to convert ADP to ATP.

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Cheetahs have come close to extinction due to hunting, drought and disease. This has caused a decrease in the genetic variation

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It's B. The cheetah populations are less likely to be able to adapt to environmental changes.

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3 years ago

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In flies dumpy wings and ebony bodies are each mutations recessive to wild type, and you expect them to behave in a Mendelian fa

Dimas [21]

Answer:

See the answers below

Explanation:

Assuming that D (d) represents the allele for wing type and E (e) represents the allele for body type, crossing DdEe with DdEe will yield the following offspring according to the Punnet's square (see the attached image):

(a) 9/16 D_E_ wild type

3/16 D_ee wild wing, ebony body

3/16 ddE_ dumpy wing, wild body

1/16 ddee dumpy wing, ebony body

(b) Chi square $X^2$ = $\frac{(O - E)^2}{E}$ , where O = observed frequency and E = expected frequency.

Phenotype O E $X^2$

Wild type 473 9/16 x 830 = 466.875 $\frac{(473 - 466.875)^2}{466.875}$ = 0.08

wild w/ebony b 156 3/16 x 830 = 155.625 $\frac{(156 - 155.625)^2}{155.625}$ = 0.0009

Dumpy w/wild b 149 3/16 x 830 = 155.625 $\frac{(149 - 155.625)^2}{155.625}$ = 0.28

dumpy w/ebony b 52 1/16 x 830 = 51.875 $\frac{(52 - 51.875)^2}{51.875}$ = 0.0003

Total $X^2$ = 0.3612

Degree of freedom = 4 - 1 = 3

Tabulated value of $X^2$ (0.05)= 7.815

(c) Since the calculated Chi square value is less than the tabulated value, we conclude that the observed outcome agrees with the expected outcome and that the cross followed the standard Mendelian pattern.

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3 years ago

Describe the difference between the messenger RNA transcript produce by rna polymerase in the nucleus and the mRNA transcript th

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Answer:

The mRNA transcript that is produced in the nucleus is in immature form, it had to get matured, and then its capping and tailing is done. 5' cap is added and a poly A tail is added at 3' end of this mRNA. This helps it to get matured. moreover for its maturation the non-coding regions (introns) are removed and exons are packed up again to form a mature mRNA.

This mature mRNA then proceeds to its way to the ribosomes for translation of proteins.

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3 years ago

Hey! pls help i’ll give brainliest

Serhud [2]

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I think ans . living and please make me brainilist

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2 years ago

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