Answer:
The correct answer is D. involves proteins in plasma membranes.
Explanation:
Animal cells have several types of proteins embedded in their cell membrane. These proteins play an important role in the transport of molecules, cell recognition, and cell communication.
Recognition proteins are a type of glycoproteins present in the plasma membrane that allow one cell of the body to recognize the other body cells by making contact with recognition proteins of other cells. Receptors proteins allow cell-cell communication by receiving extracellular proteins.
These proteins are important for proper growth and development of the cell. Therefore the correct answer is D. involves proteins in plasma membranes.
Answer:
<u> The following four traits are -: </u>
- <u>Pedigree 1 -</u> A recessive trait (autosomal recessive) is expressed by pedigree 1.
- <u>Pedigree 2- Recessive inheritance is defined by Pedigree 2. </u>
- <u>Pedigree 3</u> - The inheritance of the dominant trait (autosomal dominant) is illustrated by Pedigree 3.
- <u>Pedigree 4-</u> An X-like dominant trait is expressed by Pedigree 4.
Explanation:
<u>Explaination of each pedigree chart</u>-
- Pedigree 1 demonstrates the <u>recessive trait </u>since their children have been affected by two unaffected individuals. If the characteristics were X-linked, in order to have an affected daughter, I-1 would have to be affected.
In this, both parents are autosomal recessive trait carriers, so the child will be affected by a 1/4 (aa) - <u> Recessive inheritance</u> is defined by <u>Pedigree 2</u>. This is<u> X-related inheritance as autosomal recessive</u> inheritance has already been accounted for in part 1. This inference is confirmed by evidence showing that the father (I-1) is unaffected and that only the sons exhibit the characteristic in generation II, suggesting that the mother must be the carrier. The individual I-2 is a carrier for this X-linked trait. A typical Xa chromosome is attached to the unaffected father (I-1), so the chance of carrier II-5 is 1/2. Probability of an affected son = 1/2 (probability II-5 is a carrier) x 1/2 (probability II -5 contributes (
) x 1/2 (probability of Y from father II-6) = 1/8. An affected daughter's likelihood is 0 because a typical
must be contributed by II-6. - The inheritance of the<u> dominant trait</u> is demonstrated by <u>Pedigree 3 </u>because affected children still have affected parents (remember that all four diseases are rare). The trait must be <u>autosomal dominant</u> because it is passed down to the son by the affected father. There is a 1/2 risk that the heterozygous mother (II-5) would pass on mutant alleles to a child of either sex for an autosomal dominant feature.
- <u>Pedigree 4</u> is an <u>X-linked dominant function</u> characterized by the transmission to all of his daughters from the affected father but none of his son. On the mutant X chromosome, the father (I-1) passes on to all his daughters and none of his sons. As seen by his normal phenotype, II-6 therefore does not bear the mutation. An affected child's likelihood is 0.
In the question the pedigree chart was missing ,hence it is given below.
That person will exhibit <span>monosomy,</span>
Answer:
It destroys endangered animal habitats
Explanation:
D. because if you think about the dna is the same as thier genetic material/genes whatever you want to call it.