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:
Lactose is more likely to be utilised by E. Coli than Arabinose because Lactose will yield more energy (ATP) and lactose breakdown will give glucose and galactose and these will enter into the glycolytic pathways to pyruvate for ATP generation until Arabinose which will undergo Pentose phosphate pathway and this does not produce enough energy.
Answer:
Explanation:
human activities should no be more than the capacity of ecosystem
we should pay attention to carrying capacity of eco system
another one i don't no sorry for that
hope you like my answer and hope it helps you
Answer:
By preventing the synthesis of DNA halting cell growth.
Explanation:
Fluorouracil and methotraxate prevent the synthesis of the neucleoside Thymidine thus preventing DNA replication and elongation. Methotraxate has a structure analogous to Folic acid which is important for thymidine synthesis. Thus, it acts as a competitive inhibitor on dihdrofolate reductase an enzyme that is essential for tetrahydrofolate formation, a folic acid derivative.
Fluorouracil acts by inhibiting thymidylate synthase which catalyses an essential step in Thymidine synthesis.