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sng the fodlowing Seenplate, epdicate DXINA Op) Template ATC.GTC TAA TCO COA-TOs

ID: 50202 • Letter: S

Question

sng the fodlowing Seenplate, epdicate DXINA Op) Template ATC.GTC TAA TCO COA-TOs use the following genetic to answer the below questions: 39,) List the translate the associated amino acid sequence for each. sequences of the mRNA molecules transcribed from the following DNA template sequences. Also m RNA Lis AA 40) Consider a population of 1000 lions: 750 are brown (BB), 50 are tan (T), and 200 are yellow (t). ir Determine each genotype, phenotype, and allele frequency. Be sure to show your work! (8pt)

Explanation / Answer

38) To replicate a DNA strand complementary bases are: A to T and C to G and vice versa.

Therefore, the complementary sequence to the template strand would be:

Template- 3'- ATC-GTC-TAA-TCG-CGA-TGC- 5'

Complementary- 5'- TAG-CAG-ATT-AGC-GCT-ACG- 3'

39) To form complementary mRNA sequence for a template replace A with T, C with G, G with C and T with U.

Use the mRNA codon from the codon table to decipher the amino acid name. Thus the mRNA sequence and amino acid would be:

Template- TGT-TAG-TCG-CTT-GAG-AGT

mRNA- ACA-AUC-AGC-GAA-CUC-UCA

Amino A- Thr- Ile- Ser- Glu- Leu- Ser

40) Out of 1000 lions- 750 are brown (BB), 50 are tan (Tt) and 200 are yellow (tt)

To find genotype, phenotype and allele frequency we use the following:

Allele frequency:

To calculate allele frequency for t we use:

f(t) = (Tt) + 2 (tt) / 2 * (TT) + 2 * (Tt) + 2 * (tt)

= 50 + 400/ 2000

= 450/ 2000 = .225

Thus, the allele frequency for t is 0.225

Now the allele frequency for T would be:

1- 0.225= .775

Thus, the allele frequency for T is 0.775

Genotype:

Thus, calculating genotype frequencies we get:

f (TT) = p2 = (.775)2 = .60

f (Tt) = 2pq= (2* .225 * .775) = .348

f (tt) = q2 = (.225)2 = .052

The phenotypes can be calculated as:

.60 * 1000 = 600 brown, .348 * 1000 = 348 tan and .052 * 1000 = 52 yellow lions.