Question

1- A dihybrid cross was made between a true breeding maize cultivar with purple and smooth kernel and a true breeding cultivar with white and wrinkled kernel. The F1 kernels are all purple and smooth. The F₁ plants were then crossed with a true breeding cultivar with white and winkled kernels. The following data were obtained regarding phenotypes of kernels

F₂ generation:    110 Purple and smooth

                        90 Purple and wrinkled

                          95 White and smooth

                        105 White and wrinkled

Based on the comparison of your calculated Chi-square value and the Chi-square value in the table, does the inheritance of those genes in maize follow the law of independent assortment?

Group of answer choices

No

Yes

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2.

A dihybrid cross was made between a true breeding maize cultivar with purple and smooth kernel and a true breeding cultivar with white and wrinkled kernel. The F1 kernels are all purple and smooth. The F₁ plants were then crossed with a true breeding cultivar with white and winkled kernels. The following data were obtained regarding phenotypes of kernels

F₂ generation:    110 Purple and smooth

                        90 Purple and wrinkled

                          95 White and smooth

                        105 White and wrinkled

Perform a chi-square analysis to demonstrate if the inheritance pattern obeys the law of independent assortment. What will be the Chi-square value based on your calculation?

Group of answer choices

2.521

114.667

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3-

A dihybrid cross was made between a true breeding maize cultivar with purple and smooth kernel and a true breeding cultivar with white and wrinkled kernel. The F1 kernels are all purple and smooth. The F₁ plants were then crossed with a true breeding cultivar with white and winkled kernels. The following data were obtained regarding phenotypes of kernels

F₂ generation:    110 Purple and smooth

                        90 Purple and wrinkled

                          95 White and smooth

                        105 White and wrinkled

Suppose the inheritance of these genes follow the law of independent assortment.

What will be expected numbers of those phenotypes in the offspring?

Group of answer choices

225 Purple and smooth, 75 Purple and wrinkled, 75 White and smooth, 25 White and wrinkled

100 Purple and smooth, 100 Purple and wrinkled, 100 White and smooth,100 White and wrinkled

110 Purple and smooth, 90 Purple and wrinkled, 95 White and smooth,105 White and wrinkled

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4-

A dihybrid cross was made between a true breeding maize cultivar with purple and smooth kernel and a true breeding cultivar with white and wrinkled kernel. The F1 kernels are all purple and smooth. The F₁ plants were then crossed with a true breeding cultivar with white and winkled kernels. The following data were obtained regarding phenotypes of kernels

F₂ generation:    110 Purple and smooth

                        90 Purple and wrinkled

                          95 White and smooth

                        105 White and wrinkled

Suppose the inheritance of these genes follow the law of independent assortment.

What will be ratio of those phenotypes: Purple and smooth: Purple and wrinkled: White and smooth: White and wrinkled?

Group of answer choices

9:3:3: 1

2: 1: 0.9: 1

1:1:1:1

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5-

Chloroplasts are transmitted only through the cytoplasm of the egg in plants. In four o’clock flower plants, green phenotype is due to the synthesis of chlorophyll by normal chloroplasts, whereas, white phenotype is due to chloroplast’s that carry mutant allele that diminishes green pigmentation. What would be the phenotypes of the offspring if you cross a variegated plant (as female parent) with a green plant (as male parent)?

Group of answer choices

Green, white, and variegated

Green

White

Variegated

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6-

How may Barr bodies would you expect to find in humans with the compositions of sex chromosomes XXX?

Group of answer choices

0

2

1

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7-

On rare occasions, an abnormal human male is born who is somewhat feminized compared with normal males. Microscopic examination of the cell of one such individual revealed that he has a single Barr body in each cell. What is the chromosomal composition of this individual?

Group of answer choices

XYY

XX

XXY

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8-

If a cross produced 100 offspring and the rate of mitochondrial paternal leakage was 3%, how many offspring would be expected to contain paternal mitochondria?

Group of answer choices

10

5

3

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9-

For non-Mendelian inheritance patterns, it is often not enough to know the combination of alleles on chromosomes in the nucleus to predict the offspring's phenotype. What do you need to know to predict the outcome of Dwarfism in mice due to a mutant IGF2 allele?

Group of answer choices

The genotypes of mother

The genotype of father

The genotypes of the offspring

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10-

41. Myoclonic epilepsy and ragged-red fiber (MERRF) is an extremely rare disorder that affects neuromuscular systems. MERRF results from a mutation in mitochondrial DNA (mtDNA) that impairs protein synthesis, oxygen consumption, and energy production. When a normal male and an affected female reproduce, which of the following best predicts the expected phenotypic ratios of the offspring?

Group of answer choices

All offspring will be affected

One-fourth of the offspring will be affected

None of the offspring will be affected

Answer 1

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Answer 1:

Given parents are

P1: PPSS (True breeding Purple Smooth kernels)

P2: ppss (True breeding White Wrinkled kernels)

In the F1 generation, P1xP2 gives Purple Smooth progeny.

This means, Purple (P) is dominant over White (p), and Smooth (S) is dominant over Wrinkled (s).

Therefore, the F1 generation has the genotype PpSs.

Now, crossing F1 progeny with true breeding true breeding cultivar with white and winkled kernels, we get

PpSs x ppss

Constructing a Punnett square, we get

	PS	Ps	pS	ps
ps	PpSs (purple smooth)	Ppss (purple wrinkled)	ppSs (white smooth)	ppss (white wrinkled)

Here, the genotype ratio is 1:1:1:1.

It means all there is an equal expectation of all the genotypes.

Now, according to the experiment, we have the following observations.

Phenotype	Observed Value	Expected Value
Purple Smooth	110	100
Purple Wrinkled	90	100
White Smooth	95	100
White Wrinkled	105	100
Total	400	400

Now, we have to check the validity using Chi-square test.

The formula for Chi-square test is

χ²

where,

O_i is the observed frequency

E_i is the expected frequency

In the above experiment, we can have the following results:

Phenotype	Observed Value (O)	Expected Value (E)	Observed – Expected (O-E)	(Observed – Expected)2 (O-E)2
Purple Smooth	110	100	10	100
Purple Wrinkled	90	100	-10	100
White Smooth	95	100	-05	25
White Wrinkled	105	100	05	25

We assume the hypothesis,

H1: The observed results are a matter of chance.

Calculating the value of χ²

χ² =

χ² =

χ² = 2.5

Now, calculating the degrees of freedom,

Degrees of freedom = n- 1, where n is the number of different expected phenotypes.

Degrees of freedom = (4-1) = 3

From the table of critical values of χ² distribution, we get

Probability lies between 0.90 and 0.1

This means that there is 90% probability that the deviation between observed and expected values is due to chance.

Since, probability is greater than 0.5, we accept that the chance may be responsible for the deviation between the observed and the expected values.

Therefore,

The hypothesis H1: The observed results are a matter of chance, is accepted.

The Law of Independent Assortment says that when there are two pairs of contrasting characters, the distribution of the members of one pair into the gametes is independent of the distribution of the other pair.

Here, we can see that this condition is satisfied.

Therefore,

The correct answer is YES.