Concept explainers
In 1927, the ophthalmologist George Waaler tested 9049 schoolboys in Oslo, Norway, for red-green color blindness and found 8324 of them to be normal and 725 to be color blind. He also tested 9072 schoolgirls and found 9032 that had normal color vision while 40 were color blind.
a. | Assuming that the same sex-linked recessive allele c causes all forms of red-green color blindness, calculate the allele frequencies of c and C (the allele for normal vision) from the data for the schoolboys. (Hint: Refer to your answer to Problem 12a.) |
b. | Does Waaler’s sample demonstrate Hardy-Weinberg equilibrium for alleles of this gene? Explain your answer by describing observations that are either consistent or inconsistent with this hypothesis. On closer analysis of these schoolchildren, Waaler found that there was actually more than one c allele causing color blindness in his sample: one kind for the prot type (cp ) and one for the deuter type (cd ). (Protanopia and deuteranopia are slightly different forms of red-green color blindness.) Importantly, some of the apparently normal females in Waaler’s studies were probably of genotype cp /cd . Through further analysis of the 40 color-blind females, he found that 3 were prot (cp /cp ), and 37 were deuter (cd /cd ). |
c. | Based on this new information, what are the frequencies of the cp, cd, and C alleles in the population examined by Waaler? Calculate these values as if the frequencies obey the Hardy-Weinberg equilibrium. (Note: Again, refer to your answer to Problem 12a.) |
d. | Calculate the frequencies of all genotypes expected among men and women if the population is at equilibrium. |
e. | Do these results make it more likely or less likely that the population in Oslo is indeed at equilibrium for red-green color blindness? Explain your reasoning. |
a.
To determine:
The allele frequencies of c and C.
Introduction:
George Waaler conducted a survey on color blindness. This survey was conducted in the year 1927. Around 9049 school boys and 9072 school girls were tested during this survey. The aim of this survey was to detect the average number of boys and girls that suffered from color blindness.
Explanation of Solution
Color blindness is a recessive trait. It is an X-linked disorder. This reflects that males are hemizygous for this trait. As a result, boys are the common sufferers of color-blindness.
The given information is as follows;
C is the allele for normal vision while c is the allele for color-blindness.
The formula to be used is as follows:
Substituting the given information in the above formula:
The allele frequencies of c and C are 0.92 and 0.08.
b.
To determine:
Whether Waaler’s sample demonstrated Hardy-Weinberg equilibrium for alleles.
Introduction:
Geoffrey H. Hardy was a scientist who proposed the concept of Hardy-Weinberg equilibrium. This concept is used to associate the allele frequency with the genotype frequency. The populations that have allele frequency and the genotypic frequency at equilibrium follow the concept of Hardy-Weinberg equilibrium.
Explanation of Solution
In case the population is at Hardy-Weinberg equilibrium, then the allele frequency of girls should be equal to the allele frequency of boys.
The given information is as follows:
Thus,
In case, the allele frequency of girls is at Hardy-Weinberg equilibrium, then
However, the allele frequency of c in boys is 0.08. This reflects that Waaler’s sample does not demonstrate Hardy-Weinberg equilibrium for alleles.
c.
To determine:
The frequencies of the cP, cd, and C alleles when the values of frequencies obey Hardy-Weinberg equilibrium:
Introduction:
Waaler discovered that there are two types of c alleles that are responsible for color blindness. These are prot type c allele (cp) and deuter type c allele (cd). The prot allele codes for protanopia color blindness while deuter allele codes for deuteranopia color blindness.
Explanation of Solution
The given information is as follows:
The people suffering from protanopia have cpcp while people deuteranopia has cdcd .
The formula to be used is as follows:
According to Hardy-Weinberg equilibrium:
Where:
p is the allele frequency of C
q is the allele frequency of c
The allele frequency of c (q) has been calculated as 0.082.
The frequency of C can be calculated by using the above formula:
Thus, frequencies of the cP, cd, and C alleles are 0.018, 0.064 and 0.918 respectively.
d.
To determine:
The frequencies of all genotypes if the population is at equilibrium.
Introduction
The set of the alleles in DNA that carries the information for the expression of a trait in an individual is known as its genotype. For example, genotype ‘TT’ expresses the tallness in plants.
Explanation of Solution
In case the population is at equilibrium, then the allele frequency and genotype frequencies of boys must be equal to the allele and genotype frequencies of girls.
Thus, frequencies of the cP, cd, and C alleles in boys are as follows:
The genotype frequencies in girls are as follows:
e.
To determine:
Whether the population in Oslo is more likely or less likely at equilibrium for color blindness.
Introduction:
The survey that was conducted by George Waaler was done on the school boys and school girls of Oslo. This survey helped in understanding the importance of Hardy-Weinberg equilibrium in studying red-green color blindness.
Explanation of Solution
The allele frequency of C is same in both boys and girls. The allele frequency of c in boys is also same as the allele frequency of c in girls. The frequencies of genotypes with normal and color blind vision are same in both boys and girls. This reflects that the population in Oslo is more likely at equilibrium for color blindness.
Want to see more full solutions like this?
Chapter 21 Solutions
Genetics: From Genes to Genomes
- Tay–Sachs disease is caused by recessive alleles on anautosome. In which case(s) could two parents with anormal phenotype have a child with Tay–Sachs?a. Both parents are homozygous for a Tay–Sachs allele.b. Both parents are heterozygous for a Tay–Sachsallele.c. One parent is homozygous for a Tay–Sachs allele,and the other is heterozygous.arrow_forwardA heterozygous individual is crossed with a homozygous recessive individual. a. Draw a Punnett square to represent this cross. b. What is the probability that an offspring will have a homozygous genotype? c. What is the probability that an offspring will have a dominant phenotype? d. What is the probability that three offspring will be produced that all carry the recessive allele but do not express the recessive phenotype?arrow_forwardThe condition phenylketonuria is caused by a recessive allele. There are two carriers who have progeny.a. Give the gene notation. b. Give the expected genotypic and phenotypic ratios. c. What is the probability that their child will be heterozygous if they have a normal child?d. What is the probability of having two affected children and one normal child if they have three children?arrow_forward
- Deafness is often inherited in humans as an autosomal recessive trait. Assume that this is the case here. Two severely deaf people meet and marry. They have four children and all of them have normal hearing. What is the MOST reasonable explanation for this outcome? Select one: a. The mutant allele for deafness shows variable expressivity and the normal children may have some hearing loss that is difficult to detect. b. Complementation has occurred in the children, indicating that the deafness mutations in the parents involved different genes. c. Deafness in this family is most likely caused by an epigenetic change such as the addition of methyl groups to the DNA. d. Deafness in this family shows a genetic maternal effect with the condition being determined by the genotype of the mother.arrow_forward15. The following pedigree shows inheritance of Huntington's disease, a fatal genetic disorder that causes neurodegeneration. Since signs and symptoms usually do not appear until adulthood, many who are carriers may not realize their risk of passing on the disease-causing allele. The following pedigree represents a family in which some people are affected by Huntington's disease. Reeessive Trit er btmnt be Mec yplicalty Hinheteearrow_forwardA mutant allele in persons with familial hypercholesterolemia (FH) causes death due to a lack of liver receptors for LDL. Susceptible persons have half the normal number of receptors, while other individuals have the normal number of receptors. In a phenotypically normal couple, the man had a female 1 cousin (on his father’s side) who died from FH; the woman had a maternal uncle with FH. a. What is the probability that neither of the couple might be susceptible. b. What is the probability that one of them might be susceptible, but the other is not. c. What is the probability that they will have an FH child if a test discloses that both of them are susceptible.arrow_forward
- Bronze coloured turkey is controlled by a dominant allele, B. Red coloured turkey is homozygous for a recessive allele bb. Dominant gene N produce normal feather and recessive genotype produce ‘hairy’ feather nn. In a cross between homozygous bronzed coloured turkey with ‘hairy’ feather and homozygous red coloured turkey with normal feather. What are the ratios of F2 progeny with Bbnn genotypes? (I want to check if what i'm doing is correct)arrow_forwardQuestions a to e are answerable by yes or no. Indicate the possible parental genotypes if your answer is yes.a. Can a man with hairy ears have a hairy-eared daughter?b. Can two normal parents produce a colorblind son?c. Can two normal parents produce a colorblind daughter?d. Can a colorblind woman have a normal son?e. Can a bald man have a nonbald daughter?arrow_forwardIn cats, the gene for calico (multicolored) cats is both sex-linked and codominant. Due to a phenomenon known as dosage compensation, females that receive a B and an R gene have black and orange splotches on white Males can only be black or orange, but never calico. a. What would a calico cat’s genotype be? b. Show the cross of a female calico cat with a black male. What percentage of the kittens will be black and male? c. What percentage of the kittens will be calico and male? d. What percentage of the kittens will be calico and female? e. Show the cross of a female black cat with a male orange cat. f. What percentage of the kittens will be calico and female? What color will all the male cats be?arrow_forward
- In classical Mendelian genetics, how can one check the genotype of a parent (A) expressing the characters of a dominant allele? Select one: a. By performing a back cross with a recessive homozygote parent (B). If the A parent is homozygote for the dominant allele, then all the individuals from the F1 will display the dominant character. If the parent A was, instead, a heterozygote, then 50% of the F1 progeny will express the recessive character (homozygote recessive) and 50% the dominant one (heterozygotes). b. It is impossible to check such genotype without using specific molecular assays. c. By performing a back cross with a dominant homozygote parent (B). If the A parent is homozygote for the dominant allele, then all the individuals from the F1 will display the dominant character.arrow_forwardTay-Sachs disease is a rare human disease in which toxic substances accumulate in nerve cells. The recessive allele responsible for the disease is inherited in a simple Mendelian manner. For unknown reasons, the allele is more common in populations of Ashkenazi Jews of eastern Europe. A woman is planning to marry her first cousin, but the couple discovers that their shared grandfather’s sister died in infancy of Tay-Sachs disease.a. Draw the relevant parts of the pedigree, and show all the genotypes as completely as possible. b. What is the probability that the cousins’ first child will have Tay-Sachs disease, assuming that all people who marry into the family are homozygous normal?arrow_forwardJ. W. McKay crossed a stock melon plant that produced tan seeds with a plant that produced red seeds and obtained the following results (J. W. McKay. 1936. Journal of Heredity 27:110–112). Cross F1 F2 tan ♀ × red ♂ 13 tan seeds 93 tan, 24 red seeds a. Explain the inheritance of tan and red seeds in this plant. b. Assign symbols for the alleles in this cross and give genotypes for all the individual plants.arrow_forward
- Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage Learning