PROBLEM SET
Chapter 4 -- Probability and Statistics

P. 75, #1. Assuming a 1:1 sex ratio, what is the probability that a family of five children will consist of a) three daughters and two sons?
In this case we have three daughters and two sons that are not born in any particular order. The probability of this occurrence is given by the binomial theorem, where
n = the number of trials
s = the number of daughters
t = the number of boys
p = the probability of a daughter on any given birth
q = the probability of a son on any given birth
                     5!
Prob(3D,2S) =   ----------- (.5^3)*(.5^2)
                 (3!)*(2!)

Prob(3D,2S) = 10/32
b) alternating sexes, starting with a son?
In this case we are specifying the sex of each child at the time of birth (the order is defined). Thus we want to know the probability of a family that consists of son, daughter, son, daughter, son. This is given by the PRODUCT RULE, in that it is the probability of a son on the first birth times the probability of a daughter on the second birth times the probability of a son on the third birth times the probability of a daughter on the fourth birth times the probability of a son on the fifth birth.
Prob(S,D,S,D,S) = 0.5 x 0.5 x 0.5 x 0.5 x 0.5 = 1/32

c) alternating sexes?
There are two ways that we could have alternating sexes, one way starts with a son and the second way starts with a daughter.
Prob(S,D,S,D,S) = 1/32 (see above)
Prob(D,S,D,S,D) = 1/32 (product rule again)
As there is two ways to have alternating sexes, we use the sum rule to get the total probability
Prob(alternating sexes) = Prob(S,D,S,D,S) + Prob(D,S,D,S,D)
Prob(alternating sexes) = 1/32 + 1/32 = 2/32

d) all daughters?
We are again specifying the birth of each child at each birth, in case this daughters on each birth. The probability of all daughters is again given by the PRODUCT RULE, such that
Prob(D,D,D,D,D) = 0.5 x 0.5 x 0.5 x 0.5 x 0.5 = 1/32

e) all the same sex?
There are two ways that we could have five children who are all the same sex, all sons or all daughters.
Prob(D,D,D,D,D) = 1/32 (see above)
Prob(S,S,S,S,S) = 1/32 (product rule again)
As there are two ways to get all the same sex, we use the sum rule to get the total probability.
Prob(all same sex) = Prob(D,D,D,D,D) + Prob(S,S,S,S,S)
Prob(all same sex) = 1/32 + 1/32 = 2/32

f) at least four daughters?
There are two ways to do this, one way with only one son and one way with all daughters. Thus, we need the probability of these two events. Prob(4 daughters, 1 son) is given by the binomial theorem (see above).
                    5!
Prob(4D,1S) =  -----------(.5^4)*(.5^1) = 5/32
                (4!)*(1!)

Prob(5D) = 1/32   (see d above)

Then the probability of at least 4 daughters is given by the sum rule, where
Prob(at least 4 daughters) = Prob(4D,1S) + Prob(5D)
                           = 5/32 + 1/32 = 6/32
P. 75, #4. On the average, about one child in every ten thousand live births in the United States has phenylketonuria (PKU). What is the probability that
a) the next child born in a Boston Hospital will have PKU?

The answer is 1 in 10,000 (0.0001). The fact that the birth occurs in a Boston Hospital has no effect on the probability.
b) after a PKU child is born, the next child will have PKU.
The answer is again 0.0001. The birth of the first child has no effect on the probability of a PKU child on the second birth. They are independent events.
c) two children born in a row will have PKU?
In this case we are asking for the probability of first one PKU birth and then a second PKU birth. This is given by the PRODUCT RULE.
Prob(two PKU births in a row) = Prob(PKU birth) X Prob(PKU birth)
                              = 0.0001 x 0.0001
                              = 0.00000001
P. 75, #6. The following data are from Mendel's original experiments. Suggest a hypothesis for each set and test this hypothesis with the chi-square test. Do you reach different conclusions with different levels of significance?

a. Self-fertilization of round-seeded hybrids produced 5,474 round seeds and 1,850 wrinkled ones.

In this case the cross is presumably a monohybrid cross (Rr x Rr). Thus, you would expect a 3 round to 1 wrinkled ratio in the offspring. The null hypothesis to be tested is that the 5,474 round and 1850 wrinkled seeds came from a distribution whose ratio is 3:1. We can test this with Chi-square analysis.
                      round             wrinkled       total
observed numbers        5474               1850         7324
expected ratio          0.75               0.25
expected numbers        5493               1831
((obs-exp)^2)/exp     0.0657             0.1972
chi-square(calculated) = 0.0657 + 0.1972 = 0.2629
chi-square(critical) (alpha = 0.05, d.f. = 1) = 3.841
As the calculated chi-square value is less that the tabled that we obtained from the table, we fail to reject our null hypothesis.

b. One particular plant from part (a) yielded 45 round seeds and 12 wrinkled ones.
In this case a segregating round seeded plant was allowed to self, which is really the same cross as above (Rr x Rr). Thus, you would expect a 3 round to 1 wrinkled ratio in the offspring. The null hypothesis to be tested is that the 45 round and 12 wrinkled seeds came from a distribution whose ratio is 3:1. We can test this with Chi-square analysis.
                       round             wrinkled       total
observed numbers          45                 12           57
expected ratio          0.75               0.25
expected numbers       42.75              14.25
((obs-exp)^2)/exp     0.1184             0.3553
chi-square(calculated) = 0.1184 + 0.3553 = 0.4737
chi-square(critical) (alpha = 0.05, d.f. = 1) = 3.841
As the calculated chi-square value is less than the value that we obtained from the table, we fail to reject our null hypothesis.

c. Of the 565 plants raised from F2 round-seeded plants, 372 gave both round and wrinkled seeds in a 3:1 proportion, whereas 193 yielded only round seeds when self-fertilized.
In this case, from what we know of a monohybrid cross, we would expect 2 segregating round seeds to 1 true-breeding round seeds among the round seeds in the F2 generation. Thus our null hypothesis is that 372 segregating and 193 true-breeding comes from a distribution of 2 segregating to 1 true-breeding. We can test this with Chi-square analysis.
                     segregating      true-breeding     total
observed numbers         372                193          565
expected ratio          0.66               0.33
expected numbers       376.7              188.3
((obs-exp)^2)/exp     0.0586             0.1173
chi-square(calculated) = 0.0586 + 0.1173 = 0.1759
chi-square(critical) (alpha = 0.05, d.f. = 1) = 3.841
As the calculated chi-square value is less than the value that we obtained from the table, we fail to reject our null hypothesis.

d. A violet-flowered, long-stemmed plant was crossed with a white-flowered, short-stemmed plant with the following offspring:
47 violet, long-stemmed plants
40 white, long-stemmed plants
38 violet, short-stemmed plants
41 white, short-stemmed plants
In this case we have two loci and that the phenotypes are in about a 1:1:1:1 ratio, we suspect that we have dihybrid testcross (VvLl, violet-flowered, long-stemmed x vvll, white-flowered, short-stemmed). Thus our null hypothesis is that 47 violet, long-stemmed; 40 white, long-stemmed; 38 violet, short-stemmed; and 41 white, short-stemmed came from a distribution that is 1 violet, long-stemmed : 1 white, long-stemmed : 1 violet, short-stemmed : 1 white, short-stemmed.
                      violet    white     violet     white
                      long       long      short      short        
observed numbers       47        40        38         41
expected ratio        0.25      0.25      0.25       0.25 
expected numbers      41.5      41.5      41.5       41.5
((obs-exp)^2)/exp    0.729     0.054     0.295      0.006
chi-square(calculated) = 0.729 + 0.054 + 0.295 + 0.006 = 1.084
chi-square(critical) (alpha = 0.05, d.f. = 3) = 7.815
As the calculated chi-square value is less than the value that we obtained from the table, we fail to reject our null hypothesis.

P. 76, #12. In human beings, the absence of molars is inherited as a dominant trait. If two heterozygotes have four children, what is the probability that
a. all will have no molars?

First you need to determine the probability that any given birth will result in a child with no molars. The cross between the two heterozygotes is simply a monohybrid cross with no molars dominant to molars. Thus, we should expect a 3:1 ratio of no molars to molars. For this we can determine that the probability of any given child having no molars is 0.75 and the probability of any given child having molars is 0.25.

To have four children all without molars is given by the product rule. The probability of the first child without molars times the probability of the second child without molars times the probability of the third child without molars times the probability of the fourth child without molars.
P(4 children without molars) = 0.75 x 0.75 x 0.75 x 0.75
P(4 children without molars) = 0.316

b. three will have no molars and one will have molars?
In this case we have four children (3 without molars, 1 with molars) who are not born in any particular order. The probability of this occurrence is given by the binomial theorem, where
n = the number of trials
s = the number of children without molars
t = the number of children with molars
p = the probability of no molars on any given birth
q = the probability of molars on any given birth
                     4!
Prob(3NM,1M) =   ----------- (.75^3)*(.25^1)
                 (3!)*(1!)

Prob(3NM,1M) = 108/256 =  0.4219

c. the first two will have molars and the second two will have no molars?
In this case the order of the births is specified and thus the answer is given by the product rule.The probability of the first child with molars times the probability of the second child with molars times the probability of the third child without molars times the probability of the fourth child without molars.
P(first 2 children with molars and then 2 children without molars) = 0.25 x 0.25 x 0.75 x 0.75
P(first 2 children with molars and then 2 children without molars) = 0.0352

P. 76, #17. A short-winged, dark-bodied fly is crossed with a long-winged, tan-bodied fly. All the F1 progeny are long-winged and tan-bodied. F1 flies are crossed among themselves to yield eight-four long-winged, tan-bodied; twenty-seven long-winged, dark-bodied; thirty-five short-winged, tan-bodied; and fourteen short-winged, dark-bodied flies.
a. What ratio do you expect in the progeny?

First, we have at least two loci because we have two different characters, wing length and body color. Second, all the F1's have long wings and tan bodies, which indicates that long wings is dominant to short wings and that tan body is dominant to dark body. Thus the short-winged, dark-bodied parent must be homozygous recessive. Also as all of the F1's were of the dominant phenotype, then we must conclude that the long-winged, tan-bodied fly in the P generation is homozygous dominant. The F1's are heterozygous for both loci and the F1 x F1 cross is simply a dihybrid cross. Thus we would expect to see the following ratio of offspring in the F2 generation
9 tan-bodied, long-winged :
3 tan-bodied, short-winged :
3 dark-bodied, long-winged :
1 dark-bodied, short-winged

b. Use the chi-square test to evaluate your hypothesis. Is the observed ratio within the expected range?
                      tan        dark     tan        dark
                      long       long     short      short        
observed numbers       85        27        35         14
expected ratio        9/16      3/16      3/16       1/16 
expected numbers       90        30        30         10
((obs-exp)^2)/exp    0.277     0.333     0.833      1.667
chi-square(calculated) = 0.222 + 0.333 + 0.833 + 1.667 = 3.043
chi-square(critical) (alpha = 0.05, d.f. = 3) = 7.815
As we failed to reject our null hypothesis, we can conclude that the observed ratio is within the expected range.

P. 76, #20. A plant that has the genotype AAbbccDDEE is mated with one that is aaBBCCddee. F1 individuals are selfed to produce an F2 generation. What is the chance of getting a plant whose genotype is identical to one of the parents?
In this case the F1's are heterozgyous at all the loci. When the F1's are mated, for any given locus there is a 1/4 chance that the locus will be homozygous in the offspring. The probability for an individuals to be homozygous at all five loci is given by the product rule. Thus for an individual to be AAbbccDDEE the chance is (1/4)^5 and the probability for an individual to be aaBBCCddee is (1/4)^5. The probability of being either of these is (1/4)^5 + (1/4)^5 from the sum rule. Thus the probability of an F2 having the same genotype as one of the original parents is 1/512 or 0.00195.


Last updated on 25 September 1996.
Provide comments to Dwight Moore at mooredwi@esumail.emporia.edu.
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