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AveGali [126]
3 years ago
7

Dichotomous keys ask

Biology
1 answer:
astraxan [27]3 years ago
3 0
Wut I don’t get it : /
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Which of the following is a final product of aerobic respiration? A. pyruvate B. oxygen C. glucose D. carbon dioxide
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Answer:

carbon dioxide (and water)

Explanation:

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Which term best describes a community in which people work mostly in factories?
Ymorist [56]

Answer:

Industrial

Explanation:

Industrial is generally used to description under the term of <em>industry</em>, which would imply that people worked with the necessary things within that said industry, generally involving equipment, produced in factories.

Note however, that while factories fall under industrial, industrial includes more than just factories.

~

4 0
2 years ago
Please use your own words!
Bad White [126]
Latitude or distance from the equator – Temperatures drop the further an area is from the equator due to the curvature of the earth. In areas closer to the poles, sunlight has a larger area of atmosphere to pass through and the sun is at a lower angle in the sky.
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2 years ago
A drawing of a cat is an example of which type of model?<br><br> 1-D<br> 2-D<br> 3-D<br> 4-D
nekit [7.7K]

The drawing of a cat is an example of a 3-D drawing as applied in biology.

<h3>What is 3-D drawing?</h3>

A 3-D drawing is such the height, depth and the width of the object ios shown. Three dimemsional drawings are common in architechture as well as in biology.

Hence, the drawing of a cat is an example of a 3-D drawing as applied in biology.

Learn more about 3-D drawing: brainly.com/question/21333148

7 0
2 years ago
The trait for 'male-pattern baldness' is a recessive trait encoded for by "b". Non-balding is encoded for by a dominant allele e
wariber [46]

Answer:

<h2>Allele frequencies for B and b. </h2><h2>"b"(q) allele frequency = 0.60 </h2><h2> </h2><h2>"B"(p) allele frequency = 0.40 </h2><h2></h2><h2>Genotype frequencies; </h2><h2>BB = 0.16 </h2><h2>Bb = 0.48 </h2><h2>Bb = 0.36 </h2>

Explanation:

Given

Non-baldness (B) is dominant on baldness(b), so B is dominant over b.

Homozygous  pattern baldness male (bb) = 360,

Heterozygous  non- baldness (Bb)=  480,

Homozygous non-baldness (BB)= 160.

So, we can also denote then by genotypes  only,

BB= 160;

Bb= 480;

bb= 360;

Total= 1000

Allele frequency q² (bb) = 360/1000=0.36

allele fequency for q( b)= √o.36=0.60.

Allele frequency for p²(BB) = 160/1000=0.16

allele frequency p(B)= √0.16 = 0.4

 

Expected genotype frequencies;

BB = 160/1000  = 0.16

 

Bb = 480/1000= 0.48

 

Bb = 360/1000= 0.36

6 0
3 years ago
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