Problem Solving: Punnett Square
QUESTION:
It is known that having a widow’s peak is a dominant trait. If a man who is heterozygous dominant
marries a woman who is homozygous dominant, what is the probability that their children will not have
a widow’s peak?
GIVEN:
-
Assign the allele for widow’s peak (which is a dominant trait) the letter ‘W’
The man is heterozygous dominant Ww
The woman is homozygous dominant  WW
REQUIRED:
-
Find the probability that offspring of the above couple WILL NOT have a widow’s peak
ANALYSIS:
-
Draw a Punnet square
SOLUTION:
W
w
W
WW
Ww
W
WW
Ww
Offspring that are homozygous dominant (possess widow’s peak): 2 out of 4 = ½ = 50%
Offspring that are heterozygous dominant (possess widow’s peak): 2 out of 4 = ½ = 50%
Offspring that are homozygous recessive (Do not possess widow’s peak): 0 out of 4 = 0%
Offspring that are heterozygous recessive (Do not possess widow’s peak): 0 out of 4 = 0%
PARAPHRASE:
All of the offspring of this couple will have a widow’s peak.
Problem Solving: Half Life calculation
QUESTION:
When a certain rock formed, it contained 100.0 g of carbon-14. After 11460 years, the rock now
contains 25.0 g of carbon. What is the half life of carbon?
GIVEN:
-
Initial mass of rock = 100.0 grams
Final mass of rock = 25.0 grams
Total time of decay = 11 460 years
REQUIRED:
-
Number of half-lives = ? half-lives
Half-life = ? years
ANALYSIS:
Total time of decay = number of half-lives Xnumber of years / half life)
Number of years / half life = total time of decay / number of half lives
SOLUTION:
1. fraction of sample remaining = final mass of rock / initial mass of rock
= 25g/100g = ¼
2. after one half-life = ½
after two half-lives = ½ X ½ = ¼
So two half-lives have passed if only ¼ of the rocks carbon is still available
3. number of years / half life =11 460 y / 2 half – lives = 5730 y / 1 half – life
So 1 half life of carbon-14 = 5730 y
PARAPHASE:
The half life of carbon is 5730 years.