Answer:
The surface of the rough endoplasmic reticulum (ER) is covered in ribosomes, while the smooth endoplasmic reticulum is not covered in ribosomes.
Explanation:
Hence their terms, rough and smooth endoplasmic reticulum, the smooth is like a fine tube whereas the rough consists of ribosomes on its surface. Both the smooth and rough endoplasmic reticulum work together, delegate tasks and are the site for production and storage of proteins.
Smooth ER is responsible for storage and lipids production, while the rough ER is responsible for production of proteins and some hormones.
have a darling weekend ^w^
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.
Both of the above, is the correct answer
Independent Variable is what you change in an experiment. A Dependent Variable is what you’re trying to find out. Constants are what you keep that same. Control Group is the standards to which comparisons are made.
Answer: My favorite song is see you again, That is my favorite I am not sure that I have another favorite song
Explanation:
