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castortr0y [4]
3 years ago
7

The electrical stimulation of the brain is conducted while the patient:

Biology
1 answer:
sattari [20]3 years ago
6 0
I believe the answer is "when hypnotized".
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During cellular respiration the largest amount of ATP is generated in
Degger [83]
The electron transport chain of the cellular respiration process produces maximum ATP.

There are three parts of cellular respiration:

Glycolysis, which produces 2 ATP.
Krebs cycle, which does not produce any ATP. It produces only the raw materials for ATP production.
Oxidative phosphorylation, which involves the electron transport chain that produces maximum ATP. It takes one NADH molecule via Channel 1 to produce 2.5 ATP and one FADH2 molecule via channel 2 to produce 1.5 ATP, for a total of 34 ATP.
3 0
3 years ago
A cross is made between homozygous wild-type female Drosophila (a^+ a^+ b^+ b^+ c^+ c^+) and triple-mutant males (aa bb cc) (the
marusya05 [52]

Answer:

a is the middle gene.

Distance [b-a]= 24.7 mu

Distance [a-c]= 15.8 mu

Distance [b-a} = 40.5 mu

Explanation:

A homozygous wild-type female drosophila (a⁺b⁺c⁺/a⁺b⁺c⁺) is crossed with a homozygous recessive male (abc/abc). <u>The order of the genes here is arbitrary. </u>

The F1 is heterozygous for the three genes (a⁺b⁺c⁺/abc). The F1 females were test crossed (crossed with abc/abc males).

The F2 shows the following phenotypic ratios:

  • 320 a⁺b⁺c⁺
  • 308 a b c
  • 102 a⁺ b c⁺
  • 112 a b⁺ c
  • 66  a⁺ b⁺ c
  • 59 a b c⁺
  • 18 a⁺ b c
  • 15 a b⁺ c⁺

Total = 1000

The male parent is homozygous recessive for the 3 genes, so the observed phenotypes of the offspring correspond to the gametes received from the mother.

Recombination during meiosis is a rare event, so the most abundant gametes are always the parentals:  a⁺b⁺c⁺ and abc.

The least abundant gametes, following the same logic, are the double crossovers (DCO): a⁺bc and ab⁺c⁺.

<h3><u>1st. Determine the gene order</u></h3>

Compare the parental and the DCO gametes. The allele that is switched corresponds to the middle gene. In this case, gene a is in the middle of the other two.

<h3><u>2nd Determine the single crossover gametes</u></h3>

The F1 mother that generated all 8 types of gametes had the genotype b⁺a⁺c⁺/bac (correct order of genes).

  • The single crossover (SCO) gametes resulting from recombination between genes b and a are b⁺ac and ba⁺c⁺.
  • The single crossover (SCO) gametes resulting from recombination between genes a and c are b⁺a⁺c and bac⁺.
<h3><u>3) Calculate the recombination frequencies between genes </u></h3>

Recombination frequency (RF) = #Recombinants/Total progeny

  • RF [b-a]= (102+112+18+15)/1000= 0.247
  • RF [f-br]= (66+59+18+15)/1000= 0.158
<h3><u /></h3><h3><u>4) Calculate the distance in map units </u></h3>

Distance (mu) = RF x 100

Distance [b-a]= 0.247 × 100 = 24.7 mu

Distance [a-c]= 0.158 × 100 = 15.8 mu

Distance [b-a} = 24.7 mu + 15.8 mu = 40.5 mu

<h3><u>The gene map therefore looks like: </u></h3>

b------------24.7 mu--------------------------a---------15.8 mu-----------c

4 0
3 years ago
The nephron is responsible for maintaining _______. select one:
stiv31 [10]
I say it is all of the above. 
4 0
3 years ago
Based on the survey data, what can be concluded about the market for coffee shops in the area
Mrac [35]

Answer:

saturation has been reached ; shoe stores

Explanation:

7 0
3 years ago
DNA polymerase breaks bonds between​
goldfiish [28.3K]

Answer:

The movement of the replication fork is accomplished by the enzyme helicase, which breaks hydrogen bonds between the paired bases and unwinds the double helix ahead of the advancing DNA polymerase.

4 0
3 years ago
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