Answer:
No, when the concentration of carbon dioxide is high, such as in peripheral tissues, CO2 binds to hemoglobin and the affinity for O2 decreases, causing it to release.
Explanation:
The O2 molecule is reversibly combined with the heme portion of the hemoglobin. When the partial pressure of O2 is high, as in the case of pulmonary capillaries, for example, the binding of O2 to hemoglobin and the release of carbon dioxide are favored, this is known as the Haldane effect. If, on the contrary, when the concentration of carbon dioxide is high, such as in peripheral tissues, CO2 is bound to hemoglobin and the affinity for O2 decreases, causing it to release, this is known as the effect Bohr.
Hey there!
The answer is B. the policy holder When you sign up for an insurance policy, your insurer will charge you a premium. This is the amount you pay for the policy or the total cost of your insurance. Policyholders may choose from a number of options for paying their insurance premiums.
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The answer is 0.47 or 47%.
Let's first distinguish some terms and their frequencies.
R - the dominant allele that causes red eyes
r - the recessive allele that causes green eyes
Since R is dominant over r, lizards with at least one dominant R allele will have red eyes. Therefore, the genotypes and the phenotypes are as following:
genotypes - phenotypes
RR - the lizards with red eyes
Rr - the lizards with red eyes
rr - the lizards with green eyes
Now, we will use some Hardy-Weinberg equations:
p + q = 1
p² + 2pq + q² = 1
where:
p - the frequency of dominant allele R
q - the frequency of recessive allele r
p² - the frequency of lizards with genotype RR
2pq - the frequency of lizards with genotype Rr
q² - the frequency of lizards with genotype <span>rr
</span>
We are interested in <span>the frequency p of the dominant R allele.
</span>So, we have the frequency (72%) of the lizards that have red eyes. However, lizards with genotypes RR and Rr will have red eyes. Since their frequencies are p² and 2pq, respectively, we have:
p² and 2pq = 72% = 72/100 = 0.72
Now, use this in the equation p² + 2pq + q² = 1:
0.72 + q² = 1
q² = 1 - 0.72 = 0.28
From here, we will calculate q and later p using the formula p + q = 1:
q² = 0.28
q = √0.28 = 0.53
p + q = 1
p + 0.53 = 1
p = 1 - 0.53
p = 0.47
They are related because they share the same body. The parasite<span> lives off the </span>host's<span> body.</span>
Answer:
See Explanation
Explanation:
The question is incomplete as the diameters of the Willapa Bay urchins were not given.
However, I'll give a general way to solve this problem.
Average is calculated using:
i.e. Sum of all diameters divided by the number of bay urchins.
The diameter of urchins is usually between 6cm to 12cm; So, I will assume the following diameters for 10 urchins.
10cm, 8cm, 10cm, 7cm, 9cm, 10cm, 6cm, 10cm,8cm, 8cm
Using the above data, the average is:
