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1 year ago

Which part of the nephron receives fluid directly from the glomerular capsular space?

1 answer:

5 0

The proximal tubule is the segment of the nephron in kidneys which begins from the renal pole of the Bowman's capsule to the beginning of loop of Henle. The glomerulus is the site in the nephron where fluid and solutes are filtered out of the blood to form a glomerular filtrate. The proximal and distal tubules, the loop of Henle, and the collecting ducts are sites for the reabsorption of water and ions. Fluid and small solutes are forced under pressure to flow from the glomerulus into the capsular space of the glomerular capsule.

Reabsorption as this filtrate passes through the tubules, specific substances are reabsorbed back into the blood of the peritubular capillaries. Bowman's capsule surrounds the glomerular capillary loops and participates in the filtration of blood from the glomerular capillaries.

To learn more about glomerular filtrate ,here

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When pink sweet peas were self-pollinated and the seeds were collected and sown, the following flower colors were obtained: Red

Pavlova-9 [17]

Answer:

c. 1:2:1

The results are consistent with incomplete dominance for this trait, with pink flowers being heterozygous.

Explanation:

If flower color were determined by a gene showing incomplete dominance, the possible genotypes and phenotypes are as follows:

RR- red
ww - white
Rw - pink

If pink sweet peas are self-pollinated, then a cross between two heterozygous individuals is done (Rw x Rw).

<u>From this cross the expected ratios are:</u>

1/4 RR (red)
2/4 Rw (pink)
1/4 ww (white)

So the null hypothesis is that the observed results exhibit a 1:2:1 ratio.

<h3><u>Chi square test</u></h3>

$X^{2} = \sum \frac{(Observed - Expected)^2}{Expected}$

<u>The observed frequencies were:</u>

34 Red
76 Pink
40 White

Total 150

<u>The expected frequencies for our null hypothesis are:</u>

1/4 x 150 = 37.5 Red
2/4 x 150 = 75 Pink
1/4 x 150 = 37.5 white

$X^{2} = \frac{(34- 37.5)^2}{37.5} + \frac{(76- 75)^2}{75} + \frac{(40- 37.5)^2}{37.5}$

$X^2=0.5067$

The degrees of freedom (DF) are calculated as number of phenotypes - 1; in this case DF = 3-1 = 2.

If we look at the Chi square table, for 2 DF and a probability of p0.05, the critical value is 5.991

Our X^2 value of 0.5067 is less than the critical value, so we do not reject the null hypothesis. The results are consistent with incomplete dominance for this trait, with pink flowers being heterozygous.

8 0

3 years ago

The level occupied by the wolf at the top of this energy pyramid is a ____________ _____________.

Leya [2.2K]

Answer:

不，你做

Explanation:

There's the answer you know what to use to get the answer

7 0

3 years ago

Read 2 more answers

Science-what is the correct arrange of this E,M,A,I,H,T,P,C,M,E,R

Mariana [72]

Answer:

Champetre, imprecate, metameric, Hemiptera ??

Explanation: Choose I'm out of ideas xD

4 0

3 years ago

A situation in which the coefficient of coincidence is greater than 1.0 would indicate that: A. no double crossovers were found

damaskus [11]

Answer:

B. there were more double crossovers in the progeny than would be expected based on probability

Explanation:

Crossing over or recombination can be defined as the exchange of genetic material between homologous chromosomes during meiosis. Moreover, the coefficient of coincidence is the number of double recombinants found in the progeny. The coefficient of coincidence can be estimated by the following equation:

Coefficient of coincidence (COC) = ADRF / EDRF

where ADRF = Actual Double Recombinant Frequency

and EDRF = Expected Double Recombinant Frequency

In the case above described, ADFR is higher than EDRF, and therefore COC will be higher than 1.

3 0

3 years ago

Ellen is jogging. As her body's metabolism becomes more active, carbon dioxide levels increase, resulting in a decrease in pH. F

SVETLANKA909090 [29]

Buffers in our blood are able to neutralize changes in the bloods pH.

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3 years ago