1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
otez555 [7]
3 years ago
7

Which of these cannot be placed in a respirometer? Insects Humans Germinating seeds

Biology
1 answer:
yarga [219]3 years ago
8 0

Answer:

the awnser is germinating seeds.

i got this awnser correct on my test :)

You might be interested in
PLS tell me the biochemistry of Thyme
Inessa05 [86]

Answer:

Thymus vulgaris (Thyme)

This plant is generally known as thyme. The characterization of thyme oil by GC and GC-MS methods indicated that thymol (40.5%), p-cymene (23.6%), carvacrol (3.2%), linalool (5.4%), β-caryphyllene (2.6%), and terpinen-4-ol (0.7%) are present in thyme EO.

Explanation:

mark me branliest

5 0
3 years ago
Which of the following is not true of a dissecting microscope
Aleksandr-060686 [28]

I think c

Im not really sure sorry

Good luck


8 0
3 years ago
Read 2 more answers
Tetrads form during___<br> A. Prophase I B. Metaphase I C.prophase. II D. Metaphase II
yan [13]
The answer to your questions should be letter B! Hope this helps :)
6 0
3 years ago
7. Pyrethrums are chemicals used to kill insects. In bed bugs, the mutant-type allele, r, confers resistance, but only in homozy
natita [175]

Answer:

Allele frequency

Normal allele  = 0.9

Mutant r allele = 0.1

Genotype frequency

Homozygous normal bugs = 0.81

Homozygous mutant bug = 0.01

Heterozygous normal bug with one mutant r allele and one normal allele = 0.18

Explanation:

It is given that 99% of the bugs were killed after the spray of  pyrethrum. This suggests that 1% of the bugs that were not killed must be homozygous for the mutant type allele "r"

Thus, the frequency of homozygous "rr" species i.e q^2 = 0.01

From this we can evaluate the frequency of mutant "r" allele.

Thus, q = \sqrt{0.01} \\

q = 0.1

As per Hardy-Weinberg first equilibrium equation, p + q = 1

Substituting the value of q in above equation, we get

p = 1 - q\\p = 1 - 0.1\\p = 0.9

Thus, the frequency of homozygous normal bug is equal to

p^2 = 0.9^2 = 0.81

As per Hardy-Weinberg second equilibrium equation-

p^2 + q^2 + 2pq = 1\\

Substituting all the available values we get -

0.81 + 0.01+ 2pq = 1\\2pq = 0.18

Allele frequency

Normal allele  = 0.9

Mutant r allele = 0.1

Genotype frequency

Homozygous normal bugs = 0.81

Homozygous mutant bug = 0.01

Heterozygous normal bug with one mutant r allele and one normal allele = 0.18

3 0
3 years ago
Which two molecule do green plants use to make glucose
anygoal [31]

Answer:

Carbon Dioxide and Water

4 0
3 years ago
Other questions:
  • What are the 6 steps of photosynthesis
    9·1 answer
  • What are segments of DNA that code for a particular trait
    12·1 answer
  • Vascular tissue in plants consists of
    15·2 answers
  • The energy required for muscle contraction comes from
    9·1 answer
  • What conditions cause cells to break down fat molecules?
    11·1 answer
  • 3 facts about skin disease and disorders
    7·2 answers
  • A city experiences a heavy rainfall. Days later, when no rainwater is present, rock, debris, and soil fall down a steep hill. Wh
    7·2 answers
  • a 27yr old uses opioid, was found unresponsive with no breathing, but he has a strong pulse, you suspect a opioid - associated l
    12·1 answer
  • Mutations:
    9·1 answer
  • What type of weak bond is formed when the oxygen of one water
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!