Central Dogma

Central Dogma
1. Replicate the following DNA:

3’C A G T G T T T A C A C T G A G C G A T C C G A A T A T T T5’

2. Transcribe the following DNA:

3’ T A C T T A C C C A G A T A T C G A A C C G C A T C C A T C 5’

5’ 3’

3. Using the one letter code listed on the chart, convert the RNA you just made to a protein.

A table shows the codons and their matching amino acids. The first mRNA base is on the 5 prime end of the codon, the second base in the codon is in the middle, and the third mRNA base is the three prime end of the codon. U U U and U U C match the amino acid P H E. U U A and U U G match the amino acid L E U. U C U, U C C, U C A and U C G match the amino acid S E R. U A U and U A C match the amino acid T Y R. U A A and U A G are both a stop codon. U G U and U G C match the amino acid C Y S. U G A is a stop codon. U G G match the amino acid T R P. C U U, C U C, C U A and C U G match the amino acid L E U. C C U, C C C, C C A and C C G match the amino acid P R O. C A U, C A C match the amino acid H I S. C A A and C A G match the amino acid G L N. C G U, C G C, C G A and C G G match the amino acid A R G. A U U, A U C and A U A match the amino acid I L E. A U G is either the animo acid M E T or a start codon. A C U, A C C, A C A and A C G match the amino acid T H R. A A U and A A C match the amino acid A S N. A A A and A A G match the amino acid L Y S. A G U and A G C match the amino acid S E R. A G A and A G G match the amino acid A R G. G U U, G U C, G U A and G U G match the amino acid V A L. G C U, G C C, G C A, and G C G match the amino acid A L A. G A U and G A C match the amino acid A S P. G A A and G A G match the amino acid G L U. G G U, G G C, G G A and G G G match the amino acid G L Y.

4. Determine what type of mutations have occurred in each example. A codon chart has been provided for your convenience. Write the translation under the strand and the mutation next to the word mutation on each.

Original strand

5’-AUGCAGGGCUUUUAA-3’

Mutation 1:

5’-AUGUAGGGCUUUUAA-3’

Mutation 2:

5’-AUGCAGGGCUUCUAA-3’

Mutation 3:

5’-AUGCAGGGCUUUUUAA-3’

A table shows the codons and their matching amino acids. The first mRNA base is on the 5 prime end of the codon, the second base in the codon is in the middle, and the third mRNA base is the three prime end of the codon. U U U and U U C match the amino acid P H E. U U A and U U G match the amino acid L E U. U C U, U C C, U C A and U C G match the amino acid S E R. U A U and U A C match the amino acid T Y R. U A A and U A G are both a stop codon. U G U and U G C match the amino acid C Y S. U G A is a stop codon. U G G match the amino acid T R P. C U U, C U C, C U A and C U G match the amino acid L E U. C C U, C C C, C C A and C C G match the amino acid P R O. C A U, C A C match the amino acid H I S. C A A and C A G match the amino acid G L N. C G U, C G C, C G A and C G G match the amino acid A R G. A U U, A U C and A U A match the amino acid I L E. A U G is either the animo acid M E T or a start codon. A C U, A C C, A C A and A C G match the amino acid T H R. A A U and A A C match the amino acid A S N. A A A and A A G match the amino acid L Y S. A G U and A G C match the amino acid S E R. A G A and A G G match the amino acid A R G. G U U, G U C, G U A and G U G match the amino acid V A L. G C U, G C C, G C A, and G C G match the amino acid A L A. G A U and G A C match the amino acid A S P. G A A and G A G match the amino acid G L U. G G U, G G C, G G A and G G G match the amino acid G L Y.

Genetics
5. If a true breeding green frog is crossed with a less occurring true breeding yellow frog, then what will be the genotypic and phenotypic ratio of their grandchildren? (Hint: you’ll need to do 2 Punnett squares for this.)

Genotype: Phenotype:

6. In humans, the allele for albinism (a) is recessive to the allele for normal pigmentation (A). A normally pigmented woman whose father is an albino marries an albino man whose parents are normal. They have three children, two normal and one albino. Give the genotypes for each person. You may need to draw Punnett squares on separate paper.

PERSON

GENOTYPE

The woman

Her father

The albino man

His mother

His father

3 children (use commas between each)

7. A color-blind man and a normal vision carrier woman are expecting twin girls. What is the percent chance of the babies being color blind? Note: Color-blindness is X-linked recessive.

Percent chance:

8. Some babies were mixed up at the hospital, so the hospital blood typed everyone to try to figure out who the parents are. Move the babies around to match their parents.

Parents: IAi x IAi IBi x ii IBIB x ii IAIA x IBi

B

A

AB

O

Babies: