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worty [1.4K]
3 years ago
9

A Sub-Saharan African widowed immigrant woman lives with her deceased husband's brother and his family, which includes the broth

er-in-law's children and the widow's adult children. Each family member speaks fluent English. Surgery was recommended for the client. What is the best plan to obtain consent for surgery for this client?A. Obtain an interpreter to explain the procedure to the clientB. encourage the client to make her own decision regarding surgeryC. ask the family members to provide a clarification of the surgeon's explanation to the clientD. tell the surgeon that the brother-in-law will decide after explanation of the proposed surgery is provided to him and the widow
Biology
1 answer:
Strike441 [17]3 years ago
3 0

Answer:

The correct option is<em> D. Tell the surgeon that the brother-in-law will decide after explanation of the proposed surgery is provided to him and the widow</em>

Explanation:

In the Sub- Saharan region, the customary law allows the inheritance of wife and the culture of having more than one wife is also practised in this region. According to the law, the widow will become the inherited wife of her deceased husband's brother.

In those regions, the woman are suppressed and the regions are male dominant regions. The women have to obey men's command. Hence, it will more likely be that the brother- in- law will make the decision.

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All of the following are macromolecules except:
Harman [31]

Answer:

The correct answer is c. Fatty acids

Explanation:

There are four major types of macromolecules present in nature and that are carbohydrates(polysaccharides), proteins, lipids, and nucleic acids. These macromolecules are polymers and are made up of monomer units.  

The monomeric unit of polysaccharides is mainly glucose, of protein is amino acids, of nucleic acid is nucleotides and the monomeric unit of lipid is fatty acids. Ribosomes are macromolecules because it is made up of RNA  and proteins.

So fatty acid is a monomer which joins together to form large macromolecules like lipids. Fatty acids are made up of a hydrocarbon chain which have one attached COOH group at the terminal position.

Therefore the correct answer is c. Fatty acids.

6 0
3 years ago
Animals have two major types: _____. aquatic and terrestrial, vertebrates and invertebrates, coelomates and acoelomates, sexual
Alexxandr [17]
The correct answer is the second option. Animals have two major types: vertebrates and invertebrates. Vertebrates are animals that have backbone while the invertebrates do not have one. Examples for vertebrates are birds and mammals while the latter are insects and squid.
6 0
4 years ago
Where do the light dependent reactions take place?
weqwewe [10]

Answer: C. Thylakoid membrane

Explanation: which is placed in chloroplast... And chloroplast is in plants and they can use light to produce glucose.

4 0
4 years ago
What is the AIDS Memorial Quilt is an example of? a. Surrealist art b. Community art making c. Art made in a workshop d. Shamani
Westkost [7]

Answer:

b. Community art making

Explanation:

The AIDS Memorial quilt ia a very large and enormous memorial which celebrates the lives of the people who became victim to AIDS and died due to causes related to AIDS. It is the largest portion of the community folk art.

As the <u>community art making</u> is defined as the method of art-making involving the professional artists and the communities which are self defined through the collaberative artistic expression.

Hence option B ) Community art making is the right option.

8 0
3 years ago
Explain how we know that DNA breaks and rejoins during recombination.
alisha [4.7K]

Answer:

It occurs through homologous recombination

Explanation:

GENERAL RECOMBINATION OR HOMOLOGIST

           Previously we defined its general characteristics. We will now describe a molecular model of this recombination, based on the classic Meselson and Radding, modified with the latest advances. Do not forget that we are facing a model, that is, a hypothetical proposal to explain a set of experimental data. Not all points of this model are fully clarified or demonstrated:

           Suppose we have an exogenote and an endogenote, both consisting of double helices. In recombination models, the exogenote is usually referred to as donor DNA, and the endogenote as recipient DNA.

1) Start of recombination: Homologous recombination begins with an endonucleotide incision in one of the donor double helix chains. Responsible for this process is the nuclease RecBCD (= nuclease V), which acts as follows: it is randomly attached to the donor's DNA, and moves along the double helix until it finds a characteristic sequence called c

Once the sequence is recognized, the RecBCD nuclease cuts to 4-6 bases to the right (3 'side) of the upper chain (as we have written above). Then, this same protein, acting now as a helicase, unrolls the cut chain, causing a zone of single-stranded DNA (c.s. DNA) to move with its 3 ’free end

2) The gap left by the displaced portion of the donor cut chain is filled by reparative DNA synthesis.

3) The displaced single chain zone of the donor DNA is coated by subunits of the RecA protein (at the rate of one RecA monomer per 5-10 bases). Thus, that simple chain adopts an extended helical configuration.

4) Assimilation or synapse: This is the key moment of action of RecA. Somehow, the DNA-bound RecA c.s. The donor facilitates the encounter of the latter with the complementary double helix part of the recipient, so that in principle a triple helix is formed. Then, with the hydrolysis of ATP, RecA facilitates that the donor chain moves to the homologous chain of the receptor, and therefore matches the complementary one of that receptor. In this process, the chain portion of the donor's homologous receptor is displaced, causing the so-called "D-structure".

It is important to highlight that this process promoted by RecA depends on the donor and the recipient having great sequence homology (from 100 to 95%), and that these homology segments are more than 100 bases in length.

Note that this synapse involves the formation of a portion of heteroduplex in the double receptor helix: there is an area where each chain comes from a DNA c.d. different parental (donor and recipient).

5) It is assumed that the newly displaced chain of the recipient DNA (D-structure) is digested by nucleases.

6) Covalent union of the ends originating in the two homologous chains. This results in a simple cross-linking whereby the two double helices are "tied." The resulting global structure is called the Holliday structure or joint.

7) Migration of the branches: a complex formed by the RuvA and RuvB proteins is attached to the crossing point of the Holliday structure, which with ATP hydrolysis achieve the displacement of the Hollyday crossing point: in this way the portion of heteroduplex in both double helices.

8) Isomerization: to easily visualize it, imagine that we rotate the two segments of one of the DNA c.d. 180o with respect to the cross-linking point, to generate a flat structure that is isomeric from the previous one ("X structure").

9) Resolution of this structure: this step is catalyzed by the RuvC protein, which cuts and splices two of the chains cross-linked at the Hollyday junction. The result of the resolution may vary depending on whether the chains that were not previously involved in the cross-linking are cut and spliced, or that they are again involved in this second cutting and sealing operation:

a) If the cuts and splices affect the DNA chains that were not previously involved in the cross-linking, the result will be two reciprocal recombinant molecules, where each of the 4 chains are recombinant (there has been an exchange of markers between donor and recipient)

b) If the cuts and splices affect the same chains that had already participated in the first cross-linking, the result will consist of two double helices that present only two portions of heteroduplex DNA.

8 0
4 years ago
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