Question

1. Use the picture provided to count the number of yellow and purple corn kernels. You should count ALL of the kernels visible in the picture. Each kernel is a seed-seeds are baby plants. Therefore, each kernel represents an offspring of the plant cross. Enter these Observed Values into Table 1. 048 2. Tally the total number of offspring that are yellow and purple by adding them together. This is the total Observed (O) value. Fill into Table 2. Table 2 Offspring Type Observed Number of individuals (0) Purple 80 Yellow 33 113 Total number of offspring counted (Yellow and Purple added together) 3. Calculate the expected number of people with each phenotype based on the total number of offspring counted. For example, if you counted 979 offspring total, and you expected that 1/8 of those individuals would have the disease (your predicted fraction in Table 1), then your Expected (E) for those with disease = (1/8) X 979. 4. Fill in your phenotypes, fraction expected, and your calculated expected values in Table 3 Calculate the x 2 value using Table 3 and the following instructions 5. For each possible outcome, calculate the difference between observed (0) and expected (E) values. The expected values were calculated in step 2. 6. For each outcome, divide the square of the difference (0 - E) by the expected (E) value.

Answer 1

Answer-

According to the given question-

Here we have a condition, where a color characteristic is observed in a corn kernel seed, where the purple color kernel seed is dominant over the yellow color corn kernel seed.

The allele for purple color seed = P,

The allele for yellow color seed = p

When we make a cross between the two true-breeding parents, both are homozygous, i.e. homozygous purple color plant having genotype PP, and yellow color kernel having genotype pp,

then in the F1 generation, we get all the offspring having purple color kernel seed.

The Genotype of Purple color kernel = PP

The genotype of yellow color kernel = pp

Cross between Purple PP $1596183631466_blob.png$ pp Yellow

	p	p
P	Pp (Purple color seed)	Pp (Purple color seed)
P	Pp (Purple color seed)	Pp (Purple color seed)

When the offspring of the F1 generation was self crossed, then in the F2 generation we get following kernel color offspring-

A Cross between (Purple offspring ) Pp $1596183631565_blob.png$ Pp

	P	p
P	PP (Purple color seed)	Pp (Purple color seed)
p	Pp (Purple color seed)	pp (Yellow color seed)

The offspring observed in the F2 generation,-

Phenotype ratio-   Purple color seed:   Yellow color seed = 3: 1

Genotype ratio = PP : Pp :pp = 1 : 2 :1 .

- The observed frequency of the F2 generation offspring in the given image -

Offspring type	The observed number of individuals
purple kernel	80
yellow kernel	33
Total number of kernel = 80 + 33 = 113

We know that according to mendals law the ratio of the expected number of the kernel must be - 3: 1,

Thus the -

Number of purple kernel = 3 $1596183631561_blob.png$ 4 $1596183631605_blob.png$ 113 = 0.75 $1596183631614_blob.png$ 113 = 84.75

The number of yellow kernel = 1 $1596183631601_blob.png$ 4 $1596183631771_blob.png$ 113 = 0.25 $1596183631794_blob.png$ 113 = 28.25

Offspring type	Expected number of individuals
purple kernel	84.75
yellow kernel	28.25
Total number of kernel = 84.75 + 28.25 = 113

Chi-square characteristics-

Phenotype	Observed frequency (O)	Expected frequency (E)	O -E	(O - E)2	(O - E)2$1596183631826_blob.png$ E
purple kernel	80	84.75	80 - 84.75 = - 4.75	22.5625	22.5625 $1596183631832_blob.png$ 84.75 = 0.2662
yellow kernel	33	28.25	33 - 28.25 = 4.75	22.5625	22.5625 $1596183631834_blob.png$ 28.25 = 0.7986
	Total = 113	Total = 113			X2 = 0.2662 + 0.7986 = 1.0648

The calculated value of Chi square characteristics X² = 1.0648

Degree of freedom = 2-1= 1

The critical value of X² for the probability of 0.05, at 1 degree of freedom = 3.841

Thus the calculated value of chi-square < Critical value of chi-square, i.e. 1.0648 < 3.841

hence we accept the null hypothesis and observed that the cross between the purple color and yellow color kernel follows the same pattern of inheritance as we predicted.