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Last updated: 10/27/01
GST 2020 -
Changing Life on Earth
Supplement #2 for Agenda 7
Contents:
Full Directions for Punnett Square Problems: Steps in completing a Punnett Square problem for the case of a single gene:
Review:
Step |
What to do | Comment |
1 |
Draw the square (two-by-two) | |
2 |
Write the genotypes for the first set of gametes to the left side of the square, aligned with the rows in the square (either genotype can be on the top row). | Results in single letters; one gene for each trait. |
3 |
Write the genotypes for the second set of gametes above the top row of the square, aligned with the columns in the square (either genotype can be on the left column). | Results in single letters, one gene for each trait. |
NOTE: The first set of gamete genotypes can be on the top, instead. |
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4 |
Start in any one of the squares. Write down, side by side, the gamete genotype on the left edge of the same row, and the gamete genotype on the top of the same column. | Results in two letters. This is the result of the fertilization of the two gamete in the same column and the gamete in the same row. |
5 |
Do this for the three remaining cells in the Punnett Square. | The four cells now have all possibilities (neglecting mutations) for the genotypes of the offspring of those two parents. |
6 |
Write down the list of all distinct genotypes (for example, if Aa appears in more than one cell, only list it once). For each genotype, the odds for it appearing are 25% for each square in which it appears (e.g. 25% for one square, 50% for two squares, etc.). Label this list "Genotypes." | If you do the odds for each genotype correctly, addign up all of the odds will be 100%. |
7 |
Write down the list of all distinct phenotypes (i.e. AA and Aa result in A (dominant) and aa results in a (recessive)). The odds for it appearing are the sum of all of the odds for the genotypes that lead to it. Label this list "Phenotypes." | If you have done the odds correctly, these will also sum to 100%. |
NOTE: A Quiz or Exam question does does not have to ask for all of these steps.
Punnett Square Animation: To replay the animation, click on the "Reload" or "Refresh" button up near the top of your browser screen. There should be five seconds between each step in the animation.

Notice: Both parents have the dominant phenotype, and yet 25% of their offspring will have the recessive phenotype - will look like neither parent! A high rate of variation - that is what this illustrates, and that is the evolutionary advantage of sexual reproduction.
DNA, RNA and protein questions
Molecular structures:
| What happens? | When does it happen? | How does it happen? |
| A. Chromosome (DNA double helix) is duplicated inside the cell nucleus | Both types of cell division (mitosis and meiosis) have chromosome duplication as their first step | The two strands of the DNA double helix separate and a new DNA double helix is grown onto each separated strand, with A - T pairing and C - G pairing |
| B. Making proteins | During normal cell growth and repair | See steps 1 and 3 below |
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The two strands of the DNA double helix separate, and a single strand of RNA is grown on one of them, using pairing like DNA's, except that U in the RNA substitutes for T in DNA | |
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Each three bases of RNA are a "codon" that results in one amino acid in the protein. |
Examples:
I. Chromosome (DNA) duplication
Problem: Given the A-T-C-G nucleotide sequence of one of the two strands of chromosomal DNA after the double helix separates, write the nucleotide sequence for the new DNA strand that forms on the separated strand.
Solution: Use A-T and C-G pairing.
| A.Example | Given | G | A | T | T | G | A | A | T | C |
| You write | ? | ? | ? | ? | ? | ? | ? | ? | ? | |
| Partial solution below: T opposite A, A opposite T, C opposite G, G opposite C | ||||||||||
| B. Repeat | Given | G | A | T | T | G | A | A | T | C |
| You write | C | T | A | A | C | ? | ? | ? | ? | |
| Full solution below | ||||||||||
| C. Repeat | Given | G | A | T | T | G | A | A | T | C |
| You write | C | T | A | A | C | T | T | A | G | |
II. As the first step in making a protein, given the nucleotide sequence of one of the two strands of chromosomal DNA after the double helix separates, write the nucleotide sequence of the new RNA that forms on the separated DNA strand.
Solution: just like DNA duplication except that in RNA, U substitutes for T in DNA
| Example | Given (DNA) | G | A | T | T | G | A | A | T | C |
| You write (RNA) | C | U | A | A | C | U | U | A | G |
III. As the second step in making a protein, given the nucleotide sequence of the RNA strand, write the amino acid sequence for the protein that is formed.
| Example | Given (RNA) | C | U | A | A | C | U | U | A | G |
| You write (amino acids - 1 for every 3 RNA nucleotides) | ? | ? | ? | |||||||
For the yellow cell, the first RNA nucleotide is C, the second is G and the third is A, or in other words, CGA is the first codon. For easy tracing, we color-code this first codon: CUA so that we can locate the amino acid in the genetic code table below. Use the four rows selected by the C in the light red cell in the left-hand column. Then the second and third bases point to the leu in the yellow cell.
The second amino acid is thr, shown with an aqua background and the third is the STOP code, shown with a lime background.
| Example | Given (RNA) | C | U | A | A | C | U | U | A | G |
| You write (amino acids - 1 for every 3 RNA nucleotides) | leu | thr | STOP | |||||||
| FIRST BASE | SECOND BASE OF A CODON | THIRD BASE | |||
| U | C | A | G | ||
| U | phe | ser | tyr | cys | U |
| phe | ser | tyr | cys | C | |
| leu | ser | STOP | STOP | A | |
| leu | ser | STOP (3rd) | try | G | |
| C | leu | pro | his | arg | U |
| leu | pro | his | arg | C | |
| leu (1st) | pro | glu | arg | A | |
| leu | pro | glu | arg | G | |
| A | iso | thr (2nd) | asp | ser | U |
| iso | thr | asp | ser | C | |
| iso | thr | lys | arg | A | |
| met / START | thr | lys | arg | G | |
| G | val | ala | asp | gly | U |
| val | ala | asp | gly | C | |
| val | ala | glu | gly | A | |
| val | ala | glu | gly | G | |
The resulting amino acid chain is two long: leu & thr