The R and S loci are 35 m.u. apart. If a plant of genotype is selfed, what progeny phenotypes will be seen and in what proportions

Answers

Answer 1

Complete question:

The R and S loci are 35 m.u. apart. If a plant of genotype RS/rs is selfed, what progeny phenotypes will be seen and in what proportions

Answer:

R-/S- = 0.6052 R-/ss = 0.1443rr/S- = 0.1443rr/ss = 0.1056

Explanation:

Available data:

R-S are 35 mu apartCross: RS/rs   x   RS/rs

First, we need to recognize the parental gametes and the recombinant ones.

Cross:

Parentals)     RS/rs    x      RS/rs

Gametes) RS  Parental   ⇒  Equal to the parental genotype

                 rs   Parental   ⇒  Equal to the parental genotype

                 Rs  Recombinant   ⇒  Product of recombination

                 rS   Recombinant   ⇒  Product of recombination

We know that genes are 35 MU apart from each other. The map unit is the distance between the pair of genes for which every 100 meiotic products, one of them results in a recombinant one.  

To calculate the recombination frequency, we have to know that 1% of recombinations = 1 map unit = 1cm. And that the maximum recombination frequency is always 50%.  

According to this information, if 1 MU = 1% recombination frequency, then

35MU --------- 35% recombination frequency = 0.35

Now, let us calculate the frequency of each gamete type.

35 map units = 35 % of recombination in total  

                      = % Rs + % rS  

                      = 17.5% Rs + 17.5% rS

Now, if the recombination frequency equals 35% then the parental frequency is 100% - 35% = 65%

65 % parental frequency = % of RS + % rs  

                                          = 32.5% RS + 32.5% rs

The frequency of each gamete is

RS  ⇒ Parental  ⇒  32.5% (65% / 2) = 0.325rs  ⇒ Parental ⇒ 32.5% (65% / 2) = 0.325Rs  ⇒ Recombinant  ⇒ 17.5% (35% / 2) = 0.175rS  ⇒  Recombinant ⇒  17.5% (35% / 2) = 0.175

F1 Genotypes and Proportions

RS/RS = 0.325 x 0.325 = 0.1056     RS/Rs = 2x 0.325 x 0.175 = 0.1137RS/rS = 2x 0.325 x 0.175 =0.1137   rs/rs = 0.325 x 0.325 = 0.10562rs/RS = 2x 0.325 x 0.325 = 0.2112rs/Rs = 2x 0.325 x 0.175 = 0.1137rs/rS = 2x 0.325 x 0.175 = 0.1137Rs/rS = 2x 0.175 x 0.175 = 0.061Rs/Rs = 0.175 x 0.175 = 0.0306rS/rS = 0.175 x 0.175 = 0.0306

F1 Phenotypes and Proportions

R-/S- = 0.1056 + 0.1137 + 0.1137 + 0.2112 + 0.061 = 0.6052 R-/ss = 0.1137 + 0.0306 = 0.1443rr/S- = 0.1137 + 0.0306 = 0.1443rr/ss = 0.1056

Answer 2

If R and S loci are 35 m.u. apart and a plant of genotype RS/rs is selfed, then the progeny phenotypes will be 0.606 RS; 0.106 rs; 0.144 Rs and 0.144 rS.

In this case, it is imperative to estimate gamete frequencies

RS = 0.65% / 2 = 0.325rs  = 0.65% / 2 = 0.325Rs  (recombinant) = 0.35 / 2 = 0.175rS (recombinant) = 0.35 / 2 = 0.175

In consequence, the F1 genotypes will be RS/RS = 0.1056  (i.e., 0.325 x 0.325); RS/Rs = 0.1137; RS/rS = 0.1137; rs/rs = 0.10562; rs/RS = 0.2112; rs/Rs = 0.1137; rs/rS = 0.1137; Rs/rS = 0.06; Rs/Rs = 0.0306; and rS/rS = 0.0306.

Finally, it is possible to obtain phenotypic frequencies by summing genotypic frequencies for each case in particular.

In conclusion, if R and S loci are 35 m.u. apart and a plant of genotype RS/rs is selfed, then the progeny phenotypes will be 0.606 RS; 0.106 rs; 0.144 Rs and 0.144 rS.

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Answer:

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Answer:

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Explanation:

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Answer:

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Answer:

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Answers

Answer:

The correct answer is -

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Answers

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Answers

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Hope this helps:)

The given observation best indicates that the phytoplankton provides nutrients to consumers in the ecosystem.

What do you mean by Ecosystem?

An ecosystem may be defined as a place or an area where organisms of different species are present together and interact with one another for their need.

Phytoplanktons belong to lower trophic levels in the aquatic ecosystem. The organisms of higher trophic levels depend on it for the need of energy.

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Answer:

The correct answer is - Swimmerets.

Explanation:

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Ocean pollution is caused by the introduction of toxic materials and harmful pollutants such as agricultural and industrial waste, chemicals, oil spills, and plastic litter into the ocean waters.

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Answers

Answer:

Due to conjugation.

Explanation:

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X 74% 94% 62%

Y 32% 12% 18%

Z 45% 67% 31%


Based on the results of the experiment, the company concluded that insecticide X would be the best one to use. Upon which of the following lines of evidence did the company base its conclusion?

A.

Insecticide X was more effective against sugar ants than against carpenter ants or fire ants.


B.

Insecticide X killed a greater percentage of two out of the three species than did insecticides Y and Z.


C.

Insecticide X killed a greater percentage of all three species than did insecticides Y and Z.


D.

Insecticide X was more effective against carpenter ants than against sugar ants or fire ants.

Answers

Answer:

C.

Explanation:

i just did this question 5 minutes ago.

A company recently tested the effectiveness of three new insecticides, and based on the results of the experiment, the company concluded that insecticide X would be the best one to use because it killed a greater percentage of all three species than did insecticides Y and Z, which are in option C.

What is the insecticide?

The insecticide is given for the destruction or killing of the insects, and the intensity of the insecticides varies depending on the insects and their poison. There are three insecticides: X, Y, and Z, and X kills a higher percentage of all ants than the other two, so it is preferred.

Hence, A company recently tested the effectiveness of three new insecticides, and based on the results of the experiment, the company concluded that insecticide X would be the best one to use because it killed a greater percentage of all three species than did insecticides Y and Z, which are in option C.

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Answers

Answer:

Skin contains tightly packed cells.Skin is tough, flexible, and waterproof.You cough when a pathogen enters your mouth.You sneeze when a pathogen enters your nose.

Explanation:

The physical barriers of the immune system prevent infections since they stop or expel any pathogen that enters the body. The skin with its cells tightly packed and characteristics, the endothelia, and mucous membrane, like the one in our nose, are physical barriers. They stop antigens from infecting our body since they can remove them by sneezing, coughing, or shedding.

Answer:

skin contains tightly packed cells

skin is tough, flexible, and waterproof

you cough when a pathogen enters your mouth

you sneeze when a pathogen enters your nose

Explanation:

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Answers

Because we would die without oxygen

Explanation:

Oxygen is also called a life supporting gas because oxygen is required in the process of cellular respiration.

A high concentration of which substance is indicated by a green coloration of estuarine water?
A.
oxygen
B.
salt
C.
silt
D.
chlorophyll

Answers

Answer:

The most important light-absorbing substance in the oceans is chlorophyll, which phytoplankton use to produce carbon by photosynthesis. Due to this green pigment - chlorophyll - phytoplankton preferentially absorb the red and blue portions of the light spectrum (for photosynthesis) and reflect green light.
the answer is d, chlorophyll

When O2 binds to heme in hemoglobin, the ion is drawn into the plane of the porphyrin causing a conformational change that is transmitted to adjacent subunits enhancing the for additional O2 binding. Group of answer choices Fe2 ; affinity Mg2 ; folding Mg2 ; planarity Fe3 ; affinity

Answers

Answer: Fe2 ; affinity

Explanation:

Hemoglobin is the specialised protein which are present in red blood cells which gives it the ability to carry oxygen to the tissues and return carbondioxide from the tissues to the lungs.

The hemoglobin is made up of

--> a heme group: this consists of a ferrous iron ( Fe2) and a surrounding porphyrin ring.

--> globin.

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Also, when the partial pressure of oxygen (PO2) in the blood is high, the beta chains move close together, O2 binds to heme in hemoglobin, the ion Fe2 is drawn into the plane of the porphyrin causing a conformational change that is transmitted to adjacent subunits enhancing the AFFINITY for additional O2 binding.

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