Classical Genetics (30 points) 

Name:________John Gibson__________________________ 

Section # (801-819):______807_________________ 

Instructor/TA:________George Gikas_____________________ 

 

1. (2 pts) Starting with the P1 generation for the monohybrid cross, fill in a Punnett Square to represent the F1 offspring.  Then using two flies from F1, fill in a Punnett Square to represent the F2 offspring. 

              F1 Generation                                      F2 Generation 

 

F1	R	R
r	Rr	Rr
r	Rr	Rr

F2	R	r
R	RR	Rr
r	Rr	rr

 

 

 

 

 

 

 

 

2. (1 pt) What is the ratio of phenotypes for your F2 monohybrid cross in lab?  Divide the most frequent by the least frequent phenotype.  What was the expected ratio (from the Punnett Square in question 1)? 

Counted 93:28 red eyes to purple eyes. 93/28=3.32 , 3.32:1 red eyes to purple eyes ratio. 

Expected 3:1 red eyes to purple eyes ratio 

 

 

 

 

3. (1 pt) What combination of alleles (genotypes) can produce each phenotype?  Note: there may be more than one.  In your answer, please specify what phenotype the genotype produces. 

Red eyes phenotype can be produced by RR and Rr genotypes. 

Purple eyes phenotype can only be produced by rr genotype. 

 

 

 

 

 

 

 

 

 

4. You can test the validity of your data using a chi-square test (see additional information at the end of this lab).  This allows you to determine how well the experimental results (that you counted) correspond to the theoretical prediction (the expected results from the Punnett Square).  The formula for the chi-square calculation is: 

c2 = (o1 - e1)2 + (o2 - e2)2 = 

   e1                   e2 

c2 = the symbol for the term, chi-square 

o = observed count (number) (o1 =red eyes, o2 = purple eyes) 

e = the expected number according to the predicted ratio. Both e1 and e2 must be 

calculated for the cross. 

 

1. (2 pt) Calculate the expected numbers (e1 and e2) for the class monohybrid cross.  From your Punnett Square, you expect ¾ of the total population to be red-eyed and ¼ of the total population to be purple-eyed. 

e1 = (total # flies counted) x (3/4 or 0.75) = ___940*¾ = 705___________ 

e2 = (total # flies counted) x (1/4 or 0.25) = ___940*¾ = 235___________ 

 

 

5. (2 pts) You can now enter these values into the chi-square formula above and determine c2 for the class data and fill in the Table below.  Show your work! 

c2  

= (707-705)^2 / 705 + (233 – 235)^2 / 235 = 0.00567 + 0.0170 = 0.0227 

 

 

 

 

6. (2 pts) For the dihybrid cross, perform a Punnett Square analysis on two flies that are heterozygous for each trait to produce the F2 generation that you analyzed. 

	RT	Rt	rT	rt
RT	RRTT	RRTt	RrTT	RrTt
Rt	RRTt	RRtt	RrTt	Rrtt
rT	RrTT	RrTt	rrTT	rrTt
rt	RrTt	Rrtt	rrTt	rrtt

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

7. (2 pts) What is the ratio of phenotypes for your F2 flies?  To determine this, divide the number of the number of flies expressing the most frequent phenotype by the number of flies expressing the least frequent phenotype. 

Redeye Tanbody : Purpleeye Tanbody : Redeye Ebony : Purpleeye Ebony : Purpleeye Ebony =  

72 : 26 : 21 : 7 

72 / 7 = 10.3 ; 26 / 7 = 3.7 ; 21 / 7 = 3.0 ; 7 / 7 = 1 , so ratio is 10.3 : 3.7 : 3.0 : 1 . 

 

 

 

 

 

 

8. (2 pts) What combination of alleles (genotypes) can produce each phenotype?  Note: there are 9 genotypes that could produce each of the 4 phenotypes represented.  In your answer, please specify what phenotype the genotype produces. 

Red eye Tan body phenotype can have 4 genotypes: RRTT, RRTt , RrTT , RrTt  

Red eye Ebony body phenotype can have 2 genotypes: RRtt, Rrtt 

Purple eye Tan body phenotype can have 2 genotypes: rrTT, rrTt 

Purple eye Ebony body phenotype can only have 1 genotype: rrtt 

 

 

 

 

 

 

 

 

 

9. You can test the validity of the data using a chi-square test.  This allows you to determine how well the experimental results (that you counted) correspond to the theoretical prediction (the expected results from the Punnett Square).  The formula for the chi-square calculation is: 

c2 = (o1 - e1)2 + (o2 - e2)2  +(o3 – e3)2 + (o4 – e4)2 = 

                                                         e1                     e2                    e3                    e4 

c2 = the symbol for the term, chi-square 

o = observed count (number)  

e = the expected number according to the predicted ratio. Both e1 and e2 must be 

calculated for your particular sample. 

1. (2 pts) To calculate the expected numbers (e1, e2, e3, and e4) for the class F2 dihybrid cross: 

From your Punnett Square, you expect 9/16 of your total sample to be red-eyed/tan body. 

e1 = (total # seeds counted) x (9/16 or 0.5625) = ___976 * 9/16 = 549__________________ 

 

From your Punnett Square, you expect 3/16 of your total sample to be red-eyed/ebony body. 

e2= (total # seeds counted) x (3/16 or 0.1875) = ___976 * 3/16 = 183__________________ 

 

From your Punnett Square, you expect 3/16 of your total sample to be purple-eyed/tan body. 

e3 = (total # seeds counted) x (3/16 or 0.1875) = ___976 * 3/16 = 183__________________ 

 

From your Punnett Square, you expect 1/16 of your total sample to be purple-eyed/ebony body. 

e4= (total # seeds counted) x (1/16 or 0.0625) = ___976 * 1/16 = 61__________________ 

 

 

 

 

10. (2 pts) You can now enter these values into the chi-square formula above and determine c2 for the class data and fill in the Table below.  Show your work! 

 

 

c2 

= (549-549)^2 / 549 + (189-183)^2 / 183 + (181-183)^2 / 183 + (57-61)^2 / 183 = 0+0.0219+0.1967+0.2623 = 0.4809 

 

 

 

 

 

 

 

11. (2 pts) Fill in the following table with your results from above: 

Cross	Chi-Squared	Degrees of Freedom	Accept or Reject?
Class Monohybrid	0.0227	1	Accept
Class Dihybrid	0.4809	3	Accept

 

 

 

12. (4 pts) The following Punnett Squares have been partially filled out for you to assist in determining the gametes of each parent as well as the other two offspring.  Fill in the rest of the Punnett Squares and predict the likelihood of offspring being females with disease (affected), female carriers (unaffected), female non-carrier, males with disease (affected)  or male non-carrier.  Note the affected/mutant allele is recessive. 

 

 

1. A homozygous, non-carrier female x an affected male: 

	XA	XA
Xa	XAXa	XAXa
Y	XA Y	XAY

 

I predict that female offspring affected (disease) likelyhood is 0%. 

Female carriers likelyhood is 100% . 

Female non-carrier likelyhood is zero.  

Male with disease likelyhood is zero. 

Male non-carrier likelyhood is 100%.  

 

2. An unaffected carrier female x an unaffected male 

	XA	Xa
XA	XA XA	XAXa
Y	XA Y	XaY

I predict that female offspring affected (disease) likelyhood is 0%. 

Female carriers likelyhood is 50% . 

Female non-carrier likelyhood is 50%.  

Male with disease likelyhood is 50%. 

Male non-carrier likelyhood is 50%. 

 

 

 

 

 

 

13. (4 pts) The Punnett Squares below are similar to above, but they have NOT been partially filled in.  Fill in the following Punnett Squares and predict the likelihood of offspring being females with disease (affected), female carriers (unaffected), female non-carrier, males with disease (affected) or male non-carrier.  Note the affected/mutant allele is recessive. 

   

1. A homozygous affected female x an unaffected male 

	Xa	Xa
XA	XA Xa	XA Xa
Y	Xa Y	Xa Y

I predict that female offspring affected (disease) likelyhood is 0%. 

Female carriers likelyhood is 100% . 

Female non-carrier likelyhood is 0%.  

Male with disease likelyhood is 100%. 

Male non-carrier likelyhood is 0%. 

 

2. A homozygous affected female x an affected male 

 

	Xa	Xa
Xa	Xa Xa	Xa Xa
Y	Xa Y	Xa Y

 

I predict that female offspring affected (disease) likelyhood is 100%. 

Female carriers likelyhood is 100% . 

Female non-carrier likelyhood is 0%.  

Male with disease likelyhood is 100%. 

Male non-carrier likelyhood is 0%. 

 

 

 

 

 

 

 

 

 

 

 

14. (1 pt) X-linked diseases are commonly recessive and are displayed more often in males than in females.  Why is this so? 

Male only has 1 X chromosome. Inheriting from mother’s 1 disease chromosome causes disease. Father can not give X to the male son. 

Female has 2 X chromosomes. Inheriting from mother’s 1 disease chromosome can be rescued by father’s non-disease X chromosome. 

 

 

 

 

15. (1 pt) What were the genotypes of the parents in the X-linked cross performed in lab? 

My team’s (group 1) parents genotypes are: Father XA Y , Mother XaXa  

	Xa	Xa
XA	XA Xa	XA Xa
Y	Xa Y	Xa Y

Female offspring affected (disease) is 0%.  This was the observed phenotype. 

Female carriers likelyhood is 100% . 

Female non-carrier likelyhood is 0%.  

Male with disease is 100%.  This was the observed phenotype. 

Male non-carrier is 0%.  This was the observed phenotype implied genotype.