9.4.3 Laws of Probability: The Additive Rule

Question 1

note

Answer 1

• Review: According to the multiplicative rule, the probability of two independent events co-occurring is the product of their individual probabilities.
  • The additive rule states that the probability of one or another of two mutually exclusive (or separate) events occurring equals the sum of their individual probabilities.
  • Genetic questions involving dihybrid and trihybrid crosses can be answered by applying the multiplicative and additive rules.
• Review: The multiplicative rule can be used to determine the probability that two independent events will co-occur. For example, if two heterozygotes (Ss) are crossed, you can determine the probability that each will independently produce an S allele (1/4), giving rise to an SS individual. The probability is determined by taking the product of their individual probabilities:
• 1/2 X 1/2 = 1/4
• The multiplicative rule alone is not enough to determine the probability that two heterozygotes will give rise to an Ss individual. The reason is that two mutually exclusive events may give rise to an Ss individual. Parent one may contribute an “S” and parent two may contribute an “s”. Another possibility is that parent one may contribute an “s” and parent two may contribute an “S”. The additive rule states that the probability of one or another of two mutually exclusive (or separate) events occurring equals the sum of their individual probabilities.
• The multiplicative and additive rules must be used to solve the problem. Use the multiplicative rule to determine that the probability of producing an “Ss” by either of the two possibilities is 1/4. Use the additive rule to determine the probability the offspring will have an Ss genotype: 1/4 + 1/4 = 1/2.
• Because more than one genotype can give rise to the same phenotype, the additive rule is used to determine the probability that a given cross will produce a particular phenotype.
• For example, determine the probability that two individuals of the Yy genotype will produce yellow offspring. First, use the multiplicative rule to determine all possible combinations that will result in the yellow phenotype. Then use the rule of addition and sum their individual probabilities.
• Using the multiplicative rule shows that each event that results in a yellow pea has a 1/4 chance of occurring. Using the rule of addition, the three separate events are added together to yield the answer: 3/4. Of the 16 plants, 3/4 of them (12) are expected to be yellow.
• Determine the probability that two individuals of the YySs genotype will produce yellow wrinkled offspring.
• Use the multiplicative rule to determine all possible
  combinations that produce yellow wrinkled offspring. Three possible combinations produce the phenotype. The probability of each of 1/16. Using the rule of addition, sum the three events together. The answer is 3/16. Out of 16 individuals, three are expected to be yellow and wrinkled.

Question 2

If you have three fair dice, what is the probability of rolling 2 fours and a three on any dice?

Answer 2

• 1/72

Question 3

Two black dogs were mated and produced a litter of nine black and three white puppies.

Assuming that color is controlled by two alleles at one locus, what is the probability that, of two randomly chosen black puppies removed together from the litter, one will be homozygous (BB) and one will be heterozygous (Bb)?
(Find the probability that the 1st puppy is homozygous (BB) and the 2nd puppy is heterozygous (Bb) and the probability that the 1st puppy is heterozygous (Bb) and the 2nd puppy is homozygous (BB).)

Answer 3

• 4/9

Question 4

Tay–Sachs disease is a recessive genetic disorder that occurs in humans. If both parents are heterozygous for the disease, what is the probability of one of their offspring also being heterozygous for Tay–Sachs disease?

Answer 4

• 50%, or 1/2

Question 5

What is the probability of the offspring of the cross RrssTt × rrssTt having exactly two recessive phenotypes?

Answer 5

• 50%, or 4/8

Question 6

Two parents are heterozygous for cystic fibrosis, a recessive genetic disorder. What is the probability that their child will be heterozygous for cystic fibrosis?

Answer 6

• 50%, or 2/4

Question 7

A couple is hoping to have two girls and a boy. What is the probability of this occurring?

Answer 7

• 3/8

Question 8

Two consecutive children in a family have a recessive genetic condition. Which set of genotypes could not represent the parents?

Answer 8

• One parent with the condition; one parent homozygous normal

Question 9

Which of the following is an example of two mutually exclusive events?

Answer 9

• Pulling a queen or a king from a standard deck of cards.

Question 10

You have two bags and each one has a green, blue, and a red ball. If you picked one ball from each bag, what is the probability of obtaining one green and one blue ball?

Answer 10

• 2/9

Question 11

You have three jars of marbles as follows:
Jar 1: 400 blue and 200 white
Jar 2: 800 red and 100 white
Jar 3: 50 yellow and 600 white
You blindfold a friend and have your friend select one marble from each jar. What is the probability that your friend will obtain one blue and two white marbles?

Answer 11

• 24/351

Question 12

In a dihybrid cross, what is the probability of an offspring being heterozygous for both traits?

Answer 12

• 25%

Question 13

The offspring of a particular cross have a 25% chance of being homozygous for the dominant trait and a 25% chance of being homozygous for the recessive trait. What are the genotypes of the parents?

Answer 13

• Both heterozygous.

Question 14

A wife and her husband are both heterozygous for the recessive condition of blue eyes. If the couple has dizygotic twins, what is the probability that they will both have the same phenotype for eye color?

Answer 14

• 5/8

Question 15

The fictitious animal Astragalus domesticus has two pairs of chromosomes. The probability that two traits selected at random will be on the same chromosome is:

Answer 15

• 1/2

Question 16

A plant that is heterozygous for two characters is designated AaBb. After allowing the plant to self-fertilize, what is the probability of producing a plant that is homozygous recessive in at least one of the characters?

Answer 16

• 7/16

Question 17

If you cross two plants with the genotype CcXX, what is the probability of an offspring being heterozygous for the C gene and homozygous for the X gene?

Answer 17

• 50%, or 8/16

Question 18

You have three jars of marbles as follows:
Jar 1: 500 purple and 200 white
Jar 2: 600 black and 100 white
Jar 3: 50 green and 700 white
You blindfold a friend and have your friend randomly choose one marble from each jar. What is the probability that your friend will obtain at least one white?

Answer 18

• 705/735

Question 19

Which law or laws, if any, would be used to figure out the probability of a child receiving the genotype XxYyZz from parents with the following genotypes: XXYyzz and XxYyZZ.

Answer 19

• The multiplicative law and the additive law

Question 20

Let C be a gene for genetic condition and c be a normal gene. What is the probability of a child having the phenotype of the dominant genetic condition if both parents are heterozygous for the condition?

Answer 20

• 75%, or 3/4

Question 21

You have an F1 plant that is heterozygous for flower color. It is allowed to self-fertilize. What is the probability that the resulting plant will be heterozygous?

Answer 21

• 1/2