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professor190 [17]
3 years ago
15

You cross two fruit trees. One tree produces lemons with spiky leaves and bumpy fruit. The other produces lemons with smooth lea

ves and smooth fruit. Your F1 generation produces lemons with smooth leaves and spiky leaves and bumpy fruit. What are the genotypes of the parents
Biology
1 answer:
Talja [164]3 years ago
3 0

Answer:

<h2>LLss and llSs </h2><h2 />

Explanation:

Given;

L =for lemons,

l = for limes,

S= with smooth leaf,

s =with spiky leaf.

So parents are LLss and llSs

Cross between LLss and llSs

F1 generation are   LlSs, Llss, LlSs. Llss;

L is dominant on l;

S is dominant on s;

LlSs = Lemon with smooth leaves;

Llss= Lemon with spicy leaves.

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The sequence of nucleotides in the template strand of DNA could code for the polypeptide sequence Phe-Ser-Gln is AAG, AGG, and GUU.

<h3>What are Nucleotides?</h3>

Nucleotides may be defined as a molecule that consists of a nitrogen-containing base, a phosphate group, and a pentose sugar.

The codons that codes for the given amino acids are as follows:

  • Phe = UUC
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  • Gln = CAA.

Therefore, The sequence of nucleotides in the template strand of DNA could code for the polypeptide sequence Phe-Ser-Gln is AAG, AGG, and GUU.

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brainly.com/question/26929548

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assume that life on Mars requires cell potential to be 100mV, and the extracellular concentrations of the three major species ar
svetlana [45]

Answer:

Explanation:

From the information given:

The cell potential on mars E = + 100 mV

By using Goldman's equation:

E_m = \dfrac{RT}{zF}In \Big (\dfrac{P_K[K^+]_{out}+P_{Na}[Na^+]_{out}+P_{Cl}[Cl^-]_{out} }{P_K[K^+]_{in}+P_{Na}[Na^+]_{in}+ P_{Cl}[Cl^-]_{in}}      \Big )

Let's take a look at the impermeable cell with respect to two species;

and the two species be Na⁺ and Cl⁻

E_m = \dfrac{RT}{zF} In \dfrac{[K^+]_{out}}{[K^+]_{in}}

where;

z = ionic charge on the species = + 1

F = faraday constant

∴

100 \times 10^{-3} = \Big (\dfrac{8.314 \times 298}{1\times 96485} \Big) \mathtt{In}  \Big ( \dfrac{4}{[K^+]_{in}}   \Big)

100 \times 10^{-3} = 0.0257 \Big ( \dfrac{4}{[K^+]_{in}}   \Big)

3.981= \mathtt{In} \Big ( \dfrac{4}{[K^+]_{in}}   \Big)

exp ( 3.981) = \dfrac{4}{[K^+]_{in}} \\ \\  53.57 = \dfrac{4}{[K^+]_{in}}

[K^+]_{in} = \dfrac{4}{53.57}

[K^+]_{in}  = 0.0476

For [Cl⁻]:

100 \times 10^{-3} = -0.0257 \  \mathtt{In} \Big ( \dfrac{120}{[Cl^-]_{in}}   \Big)

-3.981 =  \  \mathtt{In} \Big ( \dfrac{120}{[Cl^-]_{in}}   \Big)

0.01867 =  \dfrac{120}{[Cl^-]_{in}}

[Cl^-]_{in} = \dfrac{120}{0.01867}

[Cl^-]_{in} =6427.4

For [Na⁺]:

100 \times 10^{-3} = 0.0257 \Big ( \dfrac{145}{[Na^+]_{in}}   \Big)

53.57= \Big ( \dfrac{145}{[Na^+]_{in}}   \Big)

[Na^+]_{in}= 2.70

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Answer:

Explanation

Given that 36% are recessive in traits

100-36 = 64% for dominant traits considering a whole population to be 100%

P=dominant allele

q= recessive allele

P2= dominant genotype

q2= recessive genotype

according to hardyweinberg principle, p+q=1

64/100= 0.64 frequency for dominant traits or genotype, therefore

p2=0.64

then

P=√0.64

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Therefore, dominant allele frequency (p) for the population is 0.8

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Use the information to answer the following question.
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Answer:

c

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