Bring on the tough stuff
1. A floriculturist is interested in generating roses with larger petals. He measures the following components for petal radius in a population of plants from his farm.
Additive genetic variance (VA): 5.1
Dominance genetic variance (VD): 1.1
Interaction genetic variance (VI): 0.5
Environmental variance (VE): 3.0
Genetic-environmental variance (VGE): 0.1
Based on your estimations of heritability, if this floriculturist starts a breeding program to generate flowers with larger petals, do you think he will succeed?
2. Why do you use model organisms (mouse, fruit flies, and so on) instead of humans to study genetics?
3. The CFTR gene codes for a specific transport protein found in cell membranes. Cystic fibrosis is an inherited illness, where a mutation in the CFTR gene causes chloride ions to build up inside the cells, which causes them to absorb more water by osmosis. In the lungs and pancreas, this occurrence leads to the production of thick, sticky mucus that can affect breathing and the production of digestive enzymes.
a. From the pedigree diagram, what can you say about how cystic fibrosis is inherited? Explain your answer.
b. Write out possible genotypes for individuals 1 and 2. Use CF to represent the normal allele and cf to represent the disease allele.
c. What is the probability that individual number 3 is heterozygous? What is the probability that they are homozygous for the normal allele? Use a Punnett square to help you. How will this affect their chances of having a child with cystic fibrosis?
4. Coat color in rabbits is controlled by four alleles: C is the agouti allele, which is dominant to all others; cch is the allele for the chinchilla color (grey); ch is the allele for the Himalayan coat color, which is white with brown ears, nose and paws; and c is the albino allele, which is recessive to all the other alleles.
A rabbit breeder wants to buy a female chinchilla rabbit, but he needs it to be homozygous for the cch allele. How could he find out if the rabbit he has his eye on is true-breeding for the chinchilla coat color? Explain your answer.
5. In a mythical dragon species, the pretend gene "glittery" determines if the dragon's scales are sparkly or drab. However, sparkly individuals vary greatly in how much they glisten: some are blinding in the Sun, while others barely reflect the light. What are some possible explanations for this observation?
6. You scratch the inside of your ear, and you discover you have dry earwax. You know from carefully reading the "To Pea or Not to Pea" section of this genetics unit that "dry" earwax is recessive, while "wet" earwax is dominant. You ask your mom, and she also has dry earwax, but your dad is not into checking his earwax's consistency. However, what can you deduce about your father's genotype for earwax?
7. A coat color gene in cats, located on the X chromosome, has two alleles: XB (black) and Xb (ginger).
a. Complete the diagram above by filling in the missing female genotypes.
b. Are the following statements true or false?
i. A cross between a male black cat and a female ginger cat would give female tortoiseshell kittens.
ii. The same cross would give male tortoiseshell kittens.
iii. A tortoiseshell female can have tortoiseshell kittens by whatever colored male she mates with.
c. A cat breeder wants to produce a pure-breeding strain of tortoiseshell cat. Is this possible?
8. You are writing a plot for a forensics show. The story involves the alleged kidnapping of a baby, named Clara, many years ago after the death of her mother. A couple, the Stuarts, raised the girl and claimed that she is their biological daughter. The parents of the deceased woman, the Sullivans, appeal to a court to try to get custody of the girl they believe is their granddaughter.
a. Could mitochondrial DNA testing be useful for this case? In what way?
b. Whose mitochondrial DNA would need to be tested? Why?
c Assume the relevant people were tested. If the Stuarts are telling the truth and Clara is their biological daughter, what are the results of the test? If, on the contrary, the Sullivans are Clara's grandparents what are the results of the test?
9. Often, the F2 generation resulting from crossing two pure-breeding strains for a quantitative trait present phenotypes that were not present in either the parentals or the F1 generation. Often, these phenotypes are extreme, for example, lighter color than the lightest or taller than the tallest. Can you explain why this might happen?
10. Two true breeding strains of maize plants, one with yellow round seeds and the other with white shrunken seeds, were crossed. The F1 was found to have yellow round seeds. A test cross was then carried out and the numbers of seeds with different phenotypes was counted. In this F2, there were 332 yellow round seeds, 340 white shrunken seeds, 14 yellow shrunken seeds, and 11 white round seeds.
a. From the information given, what can you say about the relationship between the alleles for yellow and white seed color, and the alleles for round and wrinkled seeds? Explain your answer.
b. The recombination frequency, or crossover value, between genes can be used as an indicator as to whether or not two genes are linked. The lower the crossover value, the closer the two genes are to each other. It can be calculated using the following equation:
Cross-over value (%) = Number of recombinant phenotypes x 100
Total number of offspring
Calculate the recombination frequency for the F2 cross using the above equation.
c. What does this tell you about the genes controlling seed color and shape in maize?
d. What phenotypic ratio would you have expected to see in the F2 if the genes had not been linked?
11. A plant breeder crosses 2 pink snapdragon plants and obtains 780 plants: 197 red, 182 white, and 401 pink. His daughter has been studying plant genetics and decides to use her Dad's results to revise the χ2-test.
a. What phenotypic ratio would the student be expecting from this cross and why?
b. What would the student's null hypothesis be?
c. Carry out a χ2 analysis of this cross. Present your calculations in a table.
d. How many degrees of freedom are there for this cross?
e. Compare the χ2 result to the χ2 probability table here at a significance level of 0.05. Can the student accept her null hypothesis? Explain your answer.
1. Not necessarily. Because the heritability ranges from 52% to 68%, this range suggests that there are large environmental factors also acting on petal size, not just what genes the plants have.
2. Model organisms tend to produce lots of offspring, which is necessary to clearly see any patterns in inheritance. They also breed more quickly than humans, who produce only small numbers of offspring and have a nine-month gestation period. Some model organisms, such as fruit flies, are also comparatively cheap to work with. Experimenting on humans could also be considered unethical, particularly in an enforced breeding program.
3a. Cystic fibrosis is a recessive disease because unaffected parents can have affected children. It is not sex-linked because it affects both males and females, and men can be carriers as well as women.
3b. Individuals 1 and 2 are both heterozygote carriers, so their genotypes will both be CFcf.
3c. There is a 50% probability that they will be heterozygous and therefore a carrier for cystic fibrosis and a 25% probability that they will be homozygous for the normal allele. If they are carriers and have a child with another carrier, then the child could have cystic fibrosis. If they are homozygous for the normal allele, then there is virtually no chance that they will have a child with cystic fibrosis.
4. He could carry out a test cross with an albino rabbit. Because the albino allele c is recessive to all the other coat color alleles, it will reveal what the exact genotype of the other parent is. If the chinchilla rabbit is true-breeding, then all of the offspring would be chinchilla; if there are any other alleles present, such as the Himalayan or albino allele, then there will be other-colored offspring in the phenotypic ratio of 1 chinchilla : 1 Himalayan/albino.
5. Some possibilities are a second, epistatic gene, variable expressivity, and possibly a quantitative trait.
6. He must be either homozygous for dry earwax, or be heterozygous. If he were homozygous for wet earwax, or the dominant condition, you'd have wet earwax, too.
7a. Row II: ginger female – XbXb, tortoiseshell female – XBXb; Row III: black female – XBXB, tortoiseshell female XBXb
7b. i. T, ii. F, and iii. T.
7c. No. A tortoiseshell female will produce tortoiseshell kittens whether she mates with a black or a ginger male, but there is no such thing as a male tortoiseshell cat, as you need both an XB and an Xb allele to get the tortoiseshell coloring.
8a. Yes. It would allow Clara's mother to be identified because Clara will have inherited all of her mtDNA from her mom.
8b. The DNA of Clara, Mrs. Stuart, and Mrs Sullivan would need to be tested. Mrs Sullivan is the dead woman's mother, so she would have passed her mtDNA onto her daughter, who would have passed it on to Clara—if Clara is her daughter!
8c. If the Stuarts are telling the truth, then Clara and Mrs. Stuart will have the same mtDNA, and Mrs. Sullivan's will be different. If the Stuart's aren't telling the truth, then Clara and Mrs. Stuart will have different DNA, and Clara's mtDNA will match that of Mrs. Sullivan.
9. There could be multiple loci involved, there could be an environmental effect on gene expression, there could be an epistatic effect, or there could be any combination of the above!
10a. Yellow is dominant to white, and round is dominant to wrinkled because when the two true-breeding parental strains were crossed, the F1 generation, or the heterozygotes, all had round yellow seeds.
10b. 25/697 = 0.036 x 100 = 3.6%.
10c. They are linked because the recombination frequency is very low.
10d. 1:1:1:1 because independent assortment of the alleles for both genes would give roughly equal numbers of all four possible phenotypes. Remember: a test cross is between an organism with the dominant phenotype and a homozygous recessive organism.
11a. 1 red : 2 pink : 1 white—incomplete dominance
11b. The null hypothesis would be that there is no difference between the observed numbers of flowers and the expected numbers of flowers from this cross
Phenotype Observed Expected (O-E) (O-E)2 (O-E)2/E
Red 197 195 2 4 0.02
Pink 401 390 11 121 0.31
White 182 195 -13 169 0.87
Total 780 780 1.20
Calculated χ2 value = 1.20.
11d. 2. There are three phenotypes, so df = 3 - 1 = 2.
11e. Yes, she can, because the calculated χ2 value is less than the value in the χ2 probability table for 2 degrees of freedom and a confidence level of 0.05. There is no statistically significant difference between the number of flowers her father bred and the number that were expected theoretically.
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