In: Biology
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-
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.