Question

We are going to test the ratio of offspring given the parental genotypes above for Huntington’s Disease by using a coin toss. The procedure is:

1. Obtain two coins. One represents the possible genotype(s) of the egg. The other represents the possible genotype(s) of the sperm.

2. Get one coin that represents the gametes of the mother. Each side of the coin represents one allele in a single gamete. Repeat for the father.

3. One student represents the mother and will flip the mother’s coin. A second student represents the father and will flip the father’s coin at the SAME time.

4. Note the two gametes that are face up. This represents the genotype of the offspring. Tally this genotype in the table below.

5. Repeat steps 1-4 24 times.

Table 2. Data for Huntington’s Disease

genotype	Total # genotype after 24 tosses	(Total # genotype / 24) x 100 in %	Probability from Punnett Square in %	phenotype	Total # phenotype after 24 tosses	(total # phenotype / 24) x 100 in %	Probability from Pun nett Square in %
HH
Hh
hh

Answer 1

Answers:

The test ratio of offspring,which is given the parental genotype for Huntington'sngton's disease by using coin toss , with the help of following procedure:

Obtained two coins:

• phenotype of parents: mother: expressed and father : not express
• Genotype of parents : mother : Hh and Father : hh
• Gametes genotype (s) : Egg : Hh and sperm : hh

Punnett square:

• ​​​​​​​In the given diagram represent the pridicted frequency of all of the potential genotype among offspring due to each reproduction:

​​​​​​​

• Punnet square is used to determine or pridict frequency of genotype cross or breeding experiment
• If the 100% of the offspring is heterozygous (Hh) , so the H (mother) allele is dominant over the h ( father allele) of offspring,
• From the parents genotype both have heterozygous Hh genotype, there will be 0% HH ,50% Hh and 50% hh offspring on average.
• Probability of phenotype from punnet square is useful for genetic of the production rule of offspring.
• When probability of 1 for events mean that is guaranteed to happen. While probability of 0 for events mean that is guaranteed not to happen.
• The probability of offspring with dominant phenotype H from mother and H from father) + ( probability of H from mother and h from father) = 1/2 + 1/2 = 2/4

  Genotype	genotype probability	Probability in %	phenotype	phenotypic probability	phetype probability%
  HH	0/4	0%	express	0	0
  Hh	2/4	50%	express	2/4	50%
  hh	2/4	50%	not express	2/4	50%

• On the basis of that we can find phenotype probability is 1/4.
• Offspring genotype ratio : 1:2:1
• Offspring phenotype ratio: 3:1
• This table shows genotypic and phenotypic data for hungiton's disease:

Genotype	Total genotype after 24 tosses	Total genotype/24 )× 100 in %	probability from punnet square in%	phenotype	Total phenotype after 24 tosses	total phenotype/24)×100 in %	probability from punnet square in%
HH	0	0	0	express	8	33.3%	50%
Hh	10	41.7%	50%	express	0	0%
hh	14	58.3%	50%	not express	16	66.6%	50%