## "S" PROBLEM SOLUTIONS: S-1 to S-4

Problem S-1: "Gene Size".

(a) "Typical" protein: (~60,000 daltons) / (~120 daltons/aa) = ~ 500 amino acids

(b) Coding region of gene: (500 aa) x (1 codon/aa) x (3 bp/codon) = ~1500 base pairs

(c) Entire gene: 1500 base pairs (coding region) is 10% of total, so the gene is about 15,000 base pairs.

(d) ( 25,000 genes ) x ( 15,000 bp/gene ) / ( 3,000,000,000 bp in genome ) = ~ 12.5 % of genome is "regions within genes". So, ~ 87.5 % of genome is "regions not within genes".

(e) 10% of 12.5% = ~ 1.25 % of genome is "coding regions within genes".

Problem S-2: "Drawing Chromosomes in Anaphase".

We are considering a hypothetical organism that has two pairs of chromosomes, #1 and #2, with gene alpha on #1 and gene beta on #2, with this particular individual being genotype AaBb.

a) Your drawing of anaphase of mitosis should look like that on page 139 (Fig. 4.3).
b) Your drawing of anaphase I of meiosis should look like that on page 145 (Fig. 4.7).
c) Your drawing of anaphase II of meiosis should look like that on page 144 (Fig 4.7).

Color and size of the chromosomes in your drawings should be as in Figure 4.11.

Your drawings should NOT show any "crossing over" (denoted as change of color within chromosomes). We will get to that added complexity in the next chapter.

The appropriate alleles of genes alpha and beta should be shown as the logical extension of the meiotic metaphase I situation shown on page 149 (Fig. 4.11).

Problem S-3: "Siblings Genotype".

a) The probability that two siblings have inherited the same chromosome #1 from their mother = 1/2.

b) The probability that two siblings have inherited the same pair of chromosomes #1 (one from the mother and one from the father) = 1/2 x 1/2 = 1/4.

c) The probability that two siblings have inherited the same pairs of both chromosomes #1 and #2 = 1/4 x 1/4 = 1/16.

d) The probability that two siblings have inherited the same pairs of all 22 autosomes
= (1/4) to 22'nd power
= approximately ten to the minus 13'th power.
= approximately 1 chance in a "thousand billion".

e) There are about 6 billion people alive now, and perhaps about a hundred billion humans have lived. So, it is unlikely that there has EVER been a case of two regular siblings having the same pairs of all 22 autosomes from their parents.

Problem S-4: "Heterozygous parents, have three children".

P(3D,0r) = 3!/3! x (3/4)cubed = 27/64 = .42 ( 42% )

P(2D,1r) = 3!/2! x (3/4)squared x (1/4) = 3 x (9/16) x (1/4) = 27/64 = .42 ( 42% )

P(1D,2r) = 3!/2! x (3/4) x (1/4)squared = 3 x (3/4) x (1/16) = 9/64 = .14 ( 14% )

P(0D,3r) = 3!/3! x (1/4)cubed = 1/64 = .02 ( 2% )