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ahrayia [7]
2 years ago
5

Fungi have the following characteristic(s)

Biology
1 answer:
nadya68 [22]2 years ago
4 0

Answer:

Photosynthesis is the answer hope it helps

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When oxygen is available, a cell will begin the Krebs cycle after
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The correct answer is B. Glycolysis

It is a process that deals with dissolving sugars which leads directly into the Krebs Cycles.
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I need help...thank you in advance!
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Read 2 more answers
Importance of environment
larisa86 [58]

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The enviornment is important because its where animals interact with nature, materials, and how they live life. Messing with the environment can kill a whole species or even a whole forest. The environment is where animals make their habitat and improve their own homes. Without environments we would not be able to live because the food we need would end up dying or just moving to find another location to live. Without prey or predators, the environment in incomplete and there's a high chance we will overpopulate or just die.

5 0
2 years ago
In watermelons, bitter fruit (B) is dominant over sweet fruit (b), and yellow spots (S) are dominant over no spots (s). The gene
NikAS [45]

Answer:

A) 9:3:3:1

B) All bitter fruit, yellow spotted offsprings

C) Phenotypes are bitter yellow spotted (4), bitter no spot (4), sweet yellow spot (4), and sweet no spot (4). 1:1:1:1

Explanation:

This is a typical dihybrid cross involving two genes, one coding for fruit taste and the other for spot color. The allele for bitter taste (B) and yellow spot (S) is dominant over the allele for sweet taste (b) and no spot (s) respectively.

Hence, a heterozygous F1 resulting from a cross between an homozygous dominant (bitter fruit, yellow spot) and homozygous recessive (sweet fruit, no spot) will have a BbSs genotype. The heterozygous F1 offsprings are self-crossed and produce gametes BS, Bs, bS, bs. (See punnet square). The F2 offsprings will have the following phenotypes: Bitter fruit, yellow spot (9)

Bitter fruit, no spot (3)

Sweet fruit, yellow spot (3)

Sweet fruit, no spot (1)

Back cross between a F1 offspring (BbSs) and homozygous dominant parent (BBSS) will produce all bitter fruit, yellow spot offsprings (see attached image). BBSS (4), BBSs (4), BbSS (4), and BbSs (4) are the offsprings' genotypes.

For the back cross between a F1 offspring (BbSs) and a homozygous recessive (bbss) parent, the Phenotypes with their proportions are as follows:

Bitter fruit, yellow spot (BbSs, 4)

Bitter fruit, no spot (Bbss, 4)

Sweet fruit, yellow spot (bbSs, 4)

Sweet fruit, no spot (bbss, 4).

4 0
3 years ago
You are testing for nitrate reduction. The first test, which involves adding Trommsdorff reagent and dilute H2SO4, produces a bl
Kamila [148]

Answer:

The correct answer is - option a. positive.

Explanation:

Nittarte reduction is the test for the determinantion of the production of nitrate reductase enzyme through various test. One of the test is trommsdroff test.

This test involves trommsdroff reagent and dilute sulfuric acid that produce blue color if the reduction occurs and reductase present. This color is formed due to the conversion of ammonia to nitrate.

The correct answer is - option a. positive.

8 0
3 years ago
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