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zhannawk [14.2K]
3 years ago
6

The specific valve that prevents backflow of blood into the right ventricle when the ventricles relax is the ______________ valv

e.
Biology
1 answer:
kaheart [24]3 years ago
8 0
Pulmonary valve - connects right ventricle and pulmonary trunk (which divides into right and left pulmonary arteries). It is a semilunar valve
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3 years ago
Offspring of certain fruit flies may have yellow or ebony bodies and normal wings or short wings. Genetic theory predicts that t
Rashid [163]

Answer:

Yes, the observed results are consistent with the theoretical distribution predicted by the genetic model. The population is in equilibrium.

Explanation:

<u>Available data:</u>

  • Offspring of fruit flies may have yellow or ebony bodies and normal wings or short wings.
  • Expected phenotypic ratio 9:3:3:1
  • 9 yellow, normal: 3 yellow, short: 3 ebony, normal: 1 ebony, short
  • Sample size , N = 100
  • Phenotypic distribution 59:20:11:10

To know if these results are consistent with the expected ones, we need to develop a chi-square analysis. To do it we need the following information

  • Chi square= ∑ ((Obs-Exp)²/Exp)

- ∑ is the sum of the terms

- Obs are the Observed individuals

- Exp are the Expected individuals

  • Freedom degrees = K – 1

- K =genotypes number = 4  

  • Significance level, 5% = 0.05
  • Table value = Critical value  

First, define the hypothesis:

Hypothesis: The allele of this population will assort independently. The population is in equilibrium

H₀= Individuals will be equally distributed.  

H₁ = Individuals will not be equally distributed.  

Second, we need to get the expected number of individuals with each phenotype. To do it, we will use the expected phenotypic ratio and the total number of individuals in the sample. We can just perform a three simple rule, as follows:

  16 ------------------------------------ 100 individuals in the sample ------- 100%

  9 yellow, normal ---------------- X = 56.25 individuals -------------------56.25%

  3 yellow, short ------------------- X = 18.75 individuals --------------------18.75%

  3 ebony, normal -----------------X = 18.75 individuals ---------------------18.75%

  1 ebony, short ---------------------X = 6.25 individuals ----------------------6.25%

Now that we know the expected numbers of individuals with each genotype, we can compare them with the observed ones.

                  <u>yellow, normal      yellow, short     ebony, normal    ebony, short</u>

Expected            <em>56.25                    18.75                   18.75                    6.25</em>

Observed  <em>          59                           20                      11                          10</em>

The chi-square value = Σ(Obs-Exp)²/Exp.

So now we need to calculate (Obs-Exp)²/Exp

  • <u>yellow, normal</u>

(Obs-Exp)²/Exp = (59-56.25)²/56.25 = 0.134

  • <u> yellow, short </u>

(Obs-Exp)²/Exp = (20 - 18.75)²/18.75 = 0.083

  • <u>ebony, normal</u>

(Obs-Exp)²/Exp = (11 - 18.75 )²/18.75 = 3.203

  • <u> ebony, short</u>

(Obs-Exp)²/Exp = (10 - 6.25)²/6.25 = 2.25

X² = Σ(Obs-Exp)²/Exp =  0.134 + 0.083 + 3.203 + 2.25 = 5.67

  • X² = 5.67
  • Significance level = 0.05
  • Degrees of freedom = genotypes number - one = 4 - 1 = 3
  • Critical value or table value = 9.348

P₀.₀₅ > X2

9.348 > 5.67

There is not enough evidence to reject the null hypothesis. The genotypes might be in equilibrium, and there might be an independent assortment.

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3 years ago
When the double stranded DNA helix is unwound, two single strands of DNA are formed. These strands have to be stabilized because
Pie

Answer:

single-strand DNA-binding protein

Explanation:

Based on the scenario being described within the question it can be said that these proteins are known as single-strand DNA-binding protein. They are protein mainly in the E. Coli bacteria, and bind to single-straded regions of DNA. They are a main functional part of everything related to DNA metabolism such as replication, recombination, and repair.

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