Question

You are studying tail length in squirrels. Short tail (S) is dominant to long tail (1). If you cross a homozygous dominant (SS) with a homozygous recessive (II) and you obtain the following offspring: 115 squirrels: 45 long tail, 70 short tail and you apply a chi-square test to determine if the trait is Mendelian or not What is the expected value for each phenotype? (value with 2 decimals) expected short 115 expected longo what would be your CALCULATED chi-square value? (value with 2 decimals) 17.61 With a critical chi square value of 3.84 (with one degree of freedom and p=0.05), would you conclude your traitis Mendelian? choose from the following options and paste the answer in the blank Yes, because the observed is significantly different from the expected. No, because the observed is significantly different from the expected. Yes, because the observed is not significantly different from the expected. No, because the observed is not significantly different from the expected. No, because the obs
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Answer 1

Answer-

According to the given question-

Here we have characteristics of tail length, where the short tail is dominant over the tall tail.

The allele for Short tail = S

The allele for long-tail = l

When we make a cross between the one parent with a homozygous short tail and another parent with homozygous tall, then in the F1 generation we get following offspring-

The genotype of homozygous short tail = SS

The genotype of homozygous long tail = ll

Cross between SS $\times$ ll

	l	l
S	Sl (Short tail )	Sl (Short tail )
S	Sl (Short tail )	Sl(Short tail )

Then in the F1 generation, all the offspring is having short tail.

But the observed frequency of offspring n the F1 generation has given -

Short tail = 70

Long tail = 45

Total offspring = 70 + 45 = 115

Expested number of offspring in the F1 generation will be-

Short tail = 115

Long tail = 00

Chi square characteristics-

Phenotype	Observed frequency (O)	Expected frequency (E)	O -E	(O -E)2	(O -E)2 $\div$ E
Short tail	70	115	70 - 115 = -45	2025	2025 $\div$ 115 = 17.61
Long tail	45	00	45-0= 45	2025	2025 $\div$ 0 = 0
	Total = 115	Total = 115			X2 = 17.61

Here the calculated value of Chi square = X² = 17.61

Degree of freedom = 2- 1= 1

The critical value for 1 degree of freedom at the ptobability of 0.05 = 3.841

Here the calculated value of Chi-square is more than the critical value of chi-square for 1 degree of freedom at the probability of 0.05 i.e. X² Calculated > X² Critical or 17. 61 > 3.841

Thus we reject the null hypothesis and it means that the observed trait is not according to the Mendelian law, and the observed value is different from the expected value.

Answer is (B)