Question

1. You cross a true-breeding parental with yellow flowers and oval petals with a true-breeding parental with purple flowers and round petals. The F1 offspring have purple flowers and oval petals.

a. Give the genotypes for the parentals and F1s (you can choose your own letters for the alleles):

yellow flower oval petal parental: _____

purple flower round petal parental: _____

purple flower oval petal F1: _______

b. You perform a cross of the F1s, to produce the F2 generation. You get the following counts of plants of four different phenotypes. Calculate the expected counts for each phenotype.

Phenotype Observed Expected

Purple flower, oval petals 270
Purple flower, round 109
petals

Yellow flower, oval petals 97
Yellow flower, round 24
petals
Total count 500 500

c. Determine the X2 value.

d. How many degrees of freedom? What is the critical value?                  

e. Do you accept or reject your hypothesis that this is a basic dihybrid cross?

Answer 1

Answer-

According to the given question-

Here we have a true-breeding parental plant having yellow flowers oval petals and and other plant is also   true-breeding parental plant having purple flowers and round petals and we make a cross between them and we get  F1 offspring that all having  purple flowers and oval petals.

Parental = yellow flowers oval petals   purple flowers and round petals

F1 offspring = purple flowers and oval petals

Here the color of plant   Yellow flower is recessive phenotype and purple flowers is dominant phenotype . In the same way the oval petals is dominant phenotype and round petal is recessive phenotype.

So based on above information let’s suppose the genotype of yellow flower is = yy

Genotype of Purple flower is = YY

Genotype of oval petals = OO

and Genotype of round petals petals= oo

Parental genotype = yellow flowers oval petals   purple flowers and round petals

yyOO YYoo

Genotype of yellow flower oval petal parental =  yyOO

Genotype of purple flower round petal parental: = YYoo

and we get F1 offspring = purple flowers and oval petals = YyOo (purple flowers and oval petals)

Genotype of purple flower oval petal F1= YyOo

On selfing YyOo (purple flowers and oval petals) we get following offspring-

YyOo YyOo

	YO	Yo	yO	yo
YO	YYOO (Purple flower with Oval petals)	YYOo (Purple flower with Oval petals)	YyOO(Purple flower with Oval petals)	YyOo (Purple flower with Oval petals)
Yo	YYOo (Purple flower with Oval petals)	YYoo (Purple flower with round petals)	YyOo (Purple flower with Oval petals)	Yyoo  (Purple flower with round petals)
yO	YyOO (Purple flower with Oval petals)	YyOo (Purple flower with Oval petals)	yyOO (Yellow flower with Oval petals)	yyOo (Yellow flower with oval petals)
yo	YyOo (Purple flower with Oval petals)	Yyoo (Purple flower with round petals)	yyOo (Yellow flower with Oval petals)	yyoo (Yellow flower with round petals)

The ratio of progeny having character such as -

Purple flower with Oval petals= 9

Purple flower with round petals= 3

Yellow flower with Oval petals= 3

Yellow flower with round petals= 1

so here we get the phylogeny in the ratio of 9:3:3:1which is following mendal dihybrid corss

According to the given question we have-

Purple flower with  oval petals = 270
Purple flower, with round petals= 109

Yellow flower with oval petals = 97
Yellow flower round petals= 24

Total offspring = 270 + 109 + 97 + 24 = 500

So expected phenotype of offspring are-

Purple flower with  oval petals= 9 / 16 * 500= 281.25

Purple flower, with round petals= 3 / 16 * 500 = 93.75

Yellow flower with oval petals = 3 / 16 * 500 =93.75

Yellow flower round petals= 1 / 16 * 500= 31.25

Total offspring= 281.25+ 93.75 + 93.75 + 31.25 = 500

	Observed frequency (Oi)	Expected frequency (Ei)	Oi - Ei	(Oi -Ei )2	(Oi -Ei )2 / Ei
Purple flower with  oval petals	270	281.25	270- 281.25= -11.25	126.5625	126.5625 / 281.25= 0.45
Purple flower, with round petals	109	93.75	109- 93.75= 15.25	232.5625	232.5625 / 93.75=  2.48
Yellow flower with oval petals	97	93.75	97- 93.75= 3.25	10.5625	10.5625 / 93.75=  0.11
Yellow flower round petals	24	31.25	24- 31.25= -7.25	52.5625	52.5625 / 31.25= 1.68
	Total = 500	Total = 500			X2 = 0.45 + 2.48 + 0.11 + 1.68 = 4.72

Chi square valve = X2 = 4.72

Degree of freedom = 4 - 1 = 3

For three degree of freedom the critical valve at the probability of 0.05 is = 7.815

So calculated X2 < Critical valve of X2

i.e. 4.72 < 7.815

So we accept the hypothesis for three degree of freedom for this type of dihybrid cross.