A cladogram is a branching diagram that shows the cladistic relationship between a number species. It comes from the greek clados meaning branch and gramma meaning character. It is not to be confused with an evolutionary tree since it does not show the relationship between ancestors and descendants plus it also lacks the ability to show how they have changed over time. The major components of a cladogram are the tip (the start of the lineage), root (the end of the lineage) and node (where two or more lineages combine).
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.
Answer:Phosphorylation cascades are similar to a row of dominoes falling down, instead of one domino knocking down the next, a phosphate being added activates the message. In this way, a series of different _(phosphate group) are each _(added ) after another. Inactive protein kinase 1 gets a __(phosphate group_)___ added and now it is active protein kinase 1. Active protein kinase 1 transfers a (phosphate group) and now inactive protein kinase 2 is now ___(active)_____. This continues until the desired (protein ) is activated to cause a cellular response.
Explanation: