What are the genotypic and phenotypic ratios

2 answers:

8 0

- Genotype is the genetic make-up of an organism that describes the nature of each trait. (the letters we often write for a certain trait)

- Phenotype is the morphology of a living organism .(the trait you usually observe while looking/tracking the organism)

Example:

- In the flower of a pea plant there are a pink color(RR) or (Rr) and a white color(rr).

- When crossing two hybrid ( with two different letters) pink colors,

P : Rr × Rr

G : R r R r

F1 : RR Rr Rr rr

- the phenotype ratio will be (3 pink : 1 white)

- the genotype ratio will be [1 (RR) : 2 (Rr) : 1 (rr)]..

hichkok12 [17]3 years ago

4 0

RR Rr rr

1:2:1

Hope this helps

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Which of the following observations indicate an Atom has neutrons

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Rutherford postulated the existence of some neutral particle having mass similar to proton but there was no direct experimental evidence. Several theories and experimental observations eventually led the discovery of neutron.

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4 years ago

Using the following reaction (depicted using molecular models), large quantities of ammonia are burned in the presence of a plat

Mila [183]

Answer:

17.65 grams of O2 are needed for a complete reaction.

Explanation:

You know the reaction:

4 NH₃ + 5 O₂ --------> 4 NO + 6 H₂O

First you must know the mass that reacts by stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction). For that you must first know the reacting mass of each compound. You know the values of the atomic mass of each element that form the compounds:

N: 14 g/mol
H: 1 g/mol
O: 16 g/mol

So, the molar mass of the compounds in the reaction is:

NH₃: 14 g/mol + 3*1 g/mol= 17 g/mol
O₂: 2*16 g/mol= 32 g/mol
NO: 14 g/mol + 16 g/mol= 30 g/mol
H₂O: 2*1 g/mol + 16 g/mol= 18 g/mol

By stoichiometry, they react and occur in moles:

NH₃: 4 moles
O₂: 5 moles
NO: 4 moles
H₂O: 6 moles

Then in mass, by stoichiomatry they react and occur:

NH₃: 4 moles*17 g/mol= 68 g
O₂: 5 moles*32 g/mol= 160 g
NO: 4 moles*30 g/mol= 120 g
H₂O: 6 moles*18 g/mol= 108 g

Now to calculate the necessary mass of O₂ for a complete reaction, the rule of three is applied as follows: if by stoichiometry 68 g of NH₃ react with 160 g of O₂, 7.5 g of NH₃ with how many grams of O₂ will it react?

$mass of O_{2} =\frac{7.5 g of NH_{3} * 160 g of O_{2} }{68 g of NH_{3} }$