Note-taking can be an effective study strategy if you:

Biology

2 answers:

Ira Lisetskai [31]3 years ago

7 0

It is c. Take notes by hand. have a nice day :)

Sholpan [36]3 years ago

5 0

Answer:

c. take notes by hand

Hope this helps.

You might be interested in

Why is it important for DNA to replicate during interphase, before mitosis

77julia77 [94]

So that everything is not messed up during mitosis

4 0

3 years ago

For about reaction C4H10 plus 02 yields C02 plus H20 match the proper coefficient to the chemical substance it goes way for a ba

marusya05 [52]

The balanced reaction would be:

2 C4H10 + 13 O2 ----> 8 CO2 + 10 H2O

Like this both side of the reaction would have:
8 atoms of C
20 atoms of H
26 atoms of O

5 0

3 years ago

Each of the four pedigrees that follow represents a human family within which a genetic disease is segregating. Affected individ

Ne4ueva [31]

Answer:

The following four traits are -:

Pedigree 1 - A recessive trait (autosomal recessive) is expressed by pedigree 1.
Pedigree 2- Recessive inheritance is defined by Pedigree 2.
Pedigree 3 - The inheritance of the dominant trait (autosomal dominant) is illustrated by Pedigree 3.
Pedigree 4- An X-like dominant trait is expressed by Pedigree 4.

Explanation:

Explaination of each pedigree chart-

Pedigree 1 demonstrates the recessive trait 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. $X^A$ In this, both parents are autosomal recessive trait carriers, so the child will be affected by a 1/4 (aa)
Recessive inheritance is defined by Pedigree 2. This is X-related inheritance as autosomal recessive 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 ( $X^A$ ) x 1/2 (probability of Y from father II-6) = 1/8. An affected daughter's likelihood is 0 because a typical $X^A$ must be contributed by II-6.
The inheritance of the dominant trait is demonstrated by Pedigree 3 because affected children still have affected parents (remember that all four diseases are rare). The trait must be autosomal dominant 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.
Pedigree 4 is an X-linked dominant function 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.