A species that lacks adaptation to a changing environment is more likely to

Bone marrow transplant taken from a donor and infused through the central vein is coded to what icd-10-pcs code

slava [35]

The ICD-10-PCS code is 30240G4.

Firstly, select "Administration" (section 3) because the procedure is the administration of bone marrow to the patient. Secondly, select "Circulatory" (section 30) because it is being administered in a vein (circulatory system). Third, select "Transfusion" (section 302) because it is being done a transfusion of bone marrow. Then select "Central Vein" (section 3024) because that's the place of administration in the circulatory system. Lastly, go to "Bone Marrow" (section 30240G) as that is what's being transfused and then choose "Transfusion of Allogeneic Unspecified Bone Marrow into Central Vein, Open Approach" (30240G4) because it is not specified what type of transfusion it is.

5 0

3 years ago

pThe probability of having an offspring with both the dominant phenotypes for both genes is equal to the product of the individu

Elena-2011 [213]

The product probability rule states P(A∩B) = P(A) x P(B). In the exposed example, the probability of having a dominant phenotype is 9/16.

<h3>What is the product probability rule?</h3>

The product probability rule allows us to calculate the probability of occurrence of event A and event B at the same time.

It is about a joint probability of two or more events that might happen simultaneously, not excluding each other.

This rule is based on the dependence or independence of the events. If two events A and B are independent of each other, the occurrence of one of them does not affect the occurrence of the other one.

For independent events, the rule establishes that, given two different events A and B, the probability of occurrence of both together is

P(A∩B) = P(A) x P(B)

The probability of inheriting one of the alleles from each parent is 0.5 (beacause each parent has two alleles).

According to this, the probability of having a dominant phenotype for both traits depends on the paretal genotypes.

For instance, let us analyze the cross between two dihybrid individuals AaBb x AaBb.

Assuming the involved genes code for different traits and both are independent genes, for easier calculations, we can perform crosses that involve each of the genes separately -events-, and then multiply de results according to the genotype and phenotype we are interested in (A-B-)

Gene A

Parentals) Aa x Aa

Gametes) A a A a

Punnett square) A(1/2) a(1/2)

A(1/2) AA(1/4) Aa(1/4)

a(1/2) Aa(1/4) aa(1/4)

F1) Genotype

1/4 AA ⇒ 1/2A x 1/2A

2/4 = 1/2 Aa ⇒ (1/2A x 1/2a) + (1/2A x 1/2a)

1/4 aa ⇒ 1/2a x 1/2a

Phenotype

3/4 A- ⇒ 1/4AA + 2/4Aa

1/2 aa

Gene B

Parentals) Bb x Bb

Gametes) B b B b

Punnett square) B b

B BB Bb

b Bb bb

F1 Genotype

1/4 BB ⇒ 1/2B x 1/2B

2/4 = 1/2 Bb ⇒ (1/2B x 1/2b) + (1/2B x 1/2b)

1/4 bb ⇒ 1/2b x 1/2b

Phenotype

3/4 B- ⇒ 1/4 BB + 2/4 Bb

1/2 bb

The probability of having a dominant genotype and phenotype for both traits is,

Genotpe AABB ⇒ 1/4 AA x 1/4 BB = 1/16 AABB

Phenotype A-B- ⇒ 3/4 A- x 3/4 B- = 9/16 A-B-

You can learn more about the product probability rule in crosses at

brainly.com/question/13049972

brainly.com/question/14933381

4 0

3 years ago

NO G00GLE PLEASE <3 Define cell

oksian1 [2.3K]

Answer:

lol um

Explanation:

8 0

3 years ago

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Griding quartz crystal down to produce sand is an example of a

anyanavicka [17]

I do believe its a change of state, hope it helped

7 0

3 years ago

How might insects affect the interpretation of blood spatter pattern analysis?

Colt1911 [192]

The presence of insects at an incident scene involving blood is common. Flies that feed on blood at the scene produce small circular stains known as flyspeck which can be mistaken for alternative bloodstain such as expirated blood, also insects walking through a stain may spread the blood causing small additional stains.

4 0

3 years ago

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