Question

Probability and Genetics Lab

In heredity, we are concerned with the occurrence, every time an egg is fertilized, of the probability that a particular gene or chromosome will be passed on through the egg, or through the sperm, to the offspring. As you know, genes and chromosomes are present in pairs in each individual, and segregate as they go into the gametes (egg and sperm). There are two possible genes (alleles) that the egg or sperm might obtain from each pair, but it actually receives only one of them. If the probability of getting either one is equal, this probability can be expressed as 1/2, like the probability of getting heads or tails when you flip a penny. But one cannot examine the genes in a sperm or egg. One must wait until fertilization has occurred and a new individual has been produced, and some characteristic controlled by the genes has had time to develop. Thus, we are faced with the probability that it will go into the sperm, together with the probability that these will combine at fertilization.

Recall that the GENOTYPE is the combination of alleles that determine a particular trait. For example, Bb is a genotype for eye color.

The PHENOTYPE is the expression of the genotype, or what is actually visible to the human eye. For example, Bb is the genotype for BROWN eye color, WHILE BROWN is the phenotype.

PART A.

In the first section of the lab, you are going to use 2 pennies to determine the alleles inherited from each parent. The first penny will represent the female (mother) and the second penny will represent the male (father). Mark one of the pennies so you know which penny represents each parent. We are going to assume that both parents are heterozygous for all traits discussed (Bb). Heads represents the dominant allele (B) and tails represents the recessive allele (b). In this section we are going to look at the presence or lack of widow's peak in humans. Widow's peak is a V-shaped point in the hairline in the center of the forehead, and is a dominant trait. Both parents have the genotype: Bb.

PART B.

In the second section of this lab, you are going to use 2 pennies and 2 nickels to determine the alleles inherited from each parent. Therefore, we are looking at the inheritance of TWO traits rather than just one as seen in Part A. The pennies will represent one trait (widow's peak) and the nickels will represent an entirely different trait (cleft chin). The first penny and nickel will represent the female (mother) and the second penny and nickel will represent the male (father). Mark one of the pennies and one of the nickels so you know which two represents each parent. We are going to assume that both parents are heterozygous for all traits discussed (BbTt). Heads represents the dominant alleles (B & T) and tails represents the recessive alleles (b & t). In this section we are going to look at the presence or lack of widow's peak and cleft chin in humans. Widow's peak is a V-shaped point in the hairline in the center of the forehead, and is a dominant trait. Cleft chin is a dimple in the chin, and is a recessive trait. Both parents have the genotype: BbTt and the phenotype: widow's peak and smooth chin.

QUESTIONS: PLEASE SHOW WORK

1. Based on the parents' genotypes, what is the genotypic ratio expected from a monohybrid cross between the two parents for the trait?

2.

Based on the genotypic ratio from question 1, what is the expected probability of each genotype?

BB	Drag answer here
Bb	Drag answer here
bb	Drag answer here

0%

50%

75%

25%

100%

3. Toss the coins together 50 times and keep a tally sheet while doing so. They can only turn up as BB, Bb, or bb. Remember: each coin represents each parent and each toss can only turn up one way, therefore, a parent can give only one gene of a pair. How many tally's for each genotype did you get?

BB

Bb

bb

4. Determine the actual probability in percent, that each genotype occurred.

BB

Bb

bb

5. How close did your actual results come to your expected results? What caused the difference?

Answer 1

1. A monohybrid cross is one which involves only one trait.

Suppose parental traits are 'AA' and 'aa' , then the genotype will be 'Aa'.

2. If parents are 'Bb' and 'Bb' i.e. heterozygous, then BB:Bb:bb will be in 1:2:1. Probability of 'BB' and 'bb' is 25%. Probability of 'Bb' is 50%.

3. This is a practical question . I'm assuming a solution. Your answer will be different if you do the practical yourself.

BB=30; Bb= 5; bb=15

4. Probability of BB= 60%; Bb= 10%; bb=30%

5. It is different due to random mating. Results are quite different in comparison to expected.