Question

2 Questions: 1st one I did the chart, but unsure of the a and b questions:

2nd question, just a double check to see if I have done it correctly or not. If not, can someone please help me.

Question 1)

Phenotypes	Expected Probabilities	Observed Probabilities
				Data Set 1	Data Set 2
Reg Red	9/16 =0.56	26/59=0.44	846/1500=0.56
Reg Yellow	3/16=0.19	15/59=0.25	273/1500=0.18
Potatoe Red	3/16= 0.19	6/59=0.10	287/1500=0.19
Potatoe Yellow	1/16=0.06	12/59=0.20	94/1500=0.06

1. Data set 1 is a from a small sample size and data set 2 is from a large sample size. Compare each data set to the expected probabilities. Explain any differences that you see in how different the observed results are from the expected results between the two data sets.
2. Compare the expected probabilities of each phenotypic class to the observed probabilities for data set 2 (large sample size) to determine if the genes for leaf shape and fruit colour assort independently.

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Question 2: Finger mid-digital hair (M) is dominant over the recessive trait of no mid-digital hair (m). Freckles (F) is dominant over the recessive trait of no freckles (f).

a) woman that is heterozygous for freckles and has no mid-digital hair is married to a man with no freckles that is heterozygous for mid-digital hair. The couple is expecting a child.

b) what is the probability that their offspring will have mid-digital hair?

c)hat is the probability that two offspring in a row will have no freckles?

d)what is the probability of two offspring in a row having mid-digital hair and freckles?

my answer:

a) Ffmmx ffMm

b) FfMm, and ffMm will be mid digital hair 1/4+1/4 = 2/8 = 50%

c)Genotypes with no freckles: ffmm and ffMM

d)Genotype: FfMm would be 25% =0.25x0.25= 0.0625 (6.25%)

Answer 1

Question 1

a) Okay, so here we are just going to compare by the naked eye the values. There's is one thing that can be observed in the data sets compared to the expected values: the larger the population, the closer the values get to the expected values. Both have very close values compared to the expected frequencies, but data set 2 tends to have closer values. This might be due to reaching a more stable state.

b) Here we are going to discuss the same likeliness but with the use of a statistical test, let's use a Chi square to prove it.

$X^{2}= \sum \frac{(O-E)^{2}}{E}$

Data set 1

Let's do a chart to summarize it:

Phenotype	O	E	(O-E)2/E
Reg red	0.44	0.56	0.0257
Reg yellow	0.25	0.19	0.0189
Potatoe red	0.10	0.19	0.0426
Potatoe yellow	0.20	0.06	0.3266
			0.4138

Now let's check the critical value in the Chi square distribution table. We have 3 degrees of fredom, the critial value is 7.815. Since our value 0.4138 is smaller, then the frequencies for data set 1 are regular, as expected.

Data set 2

Phenotype	O	E	(O-E)2/E
Reg red	0.56	0.56	0
Reg yellow	0.18	0.19	0.0005
Potatoe red	0.19	0.19	0
Potatoe yellow	0.06	0.06	0
			0.0005

Our value is smaller than 7.815, the the frequencies are regular, described by the expected distribution.

The traits assort independently, there is no other force affecting the frequencies.

Question 2

Let's confirm or correct these questions:

a) Female: Ffmm and Male: ffMm (Confirmed)

b) For this we'll correct some things, let's make a punnet square:

	Fm	fm
fM	Ff Mm	ff Mm
fm	Ff mm	ff mm

The probability of child to have mid-digital hair is 2/4, because two options out of four are heterozygous for such trait (there are no dominant homozygous posible). So the probability is 50% or 0.5

c) Let's correct here

For this we have to know the probability for a single kid with no freckles, which is 2/4 or 0.5, and then apply the product rule. The probability for such event to occur two times in a row is the product of their independent probabilities. So (0.5)(0.5) = 0.25 or 25%

d) Let's correct here

We are going to do the same as the previous question but this time the kid in question has two trait specifications. The probability for a kid having mid-digital hair and freckles is 1/4 or 0.25. Now the probability for two kids in a row with such traits is (0.25)(0.25) = 0.0625 or 6.25%