BIOLOGY 1406 GENETICS

HOW TO SOLVE GENETICS PROBLEMS

Write the genotypes and phenotypes from the information given in the problem. Designate a dominant allele with a capital letter and a recessive allele with a small (lower case) letter. When more than one pair of alleles is involved, write the alleles together, instead of separately. Example, SSVV instead of SVSV.

If a gene is sex-linked (X-linked), the female human being or fruit fly (XX) possesses two alleles, where as the male (XY) has only one X-chromosome, and therefore only one allele. Because of the singular X-chromosome, the male is said to be homozygous. This X-chromosome is inherited by all the daughters. Sons inherit their father’s Y-chromosome. When solving problems involving X-linked genes, this information must be kept in mind. Write the X-chromosome with a superscript symbol (XS) located on the X-chromosome.

If information on the genotype is incomplete, set down as much as possible from the direct information given. Enter the symbols that are known and leave an underlined blank for those which are unknown. Then proceed to enter the symbols for the appropriate alleles from the indirect information supplied. EXAMPLE: Jane is a normally pigmented (A) girl with normal vision (N). INCOMPLETE GENOTYPE: A_ N_. Her father was born albino, and color blind, aann. Jane must have received "a" and "n" from her father, thus her genotype is AaNn

If raw data, the number of individuals having each trait or combination of traits are given for the offspring, reduce these to the simplest ratio. Examine the ratio since it may furnish information on the genotype of the parents. If the offspring fall into, groups of approximately equal size (ratio 1:1) regardless of the sex, then one parent is homozygous and the other is heterozygous. If the offspring fall into a 3:1 ratio, the parents are both heterozygous. If the offspring are all identical the parents are homozygous.

If a solution to the problem does not immediately come to mind, think of problems that have been worked previously. The solution to a previous problem may furnish ideas needed to work the present problem.

If the genotypes of the parents are known, the genotypes of the offspring can be ascertained by means of a Punnett Square. Place the gametes of the male across the top of the square and the female gametes vertically along the left side of the square. Write the genotypes of the progeny in the square where they intersect. After completing the square the genotypic and phenotypic ratios can be ascertained.

In problems on human inheritance, the pedigree is a useful instrument for tracing descent of alleles through several generations. Females are designated with circles and males with squares. A marriage is indicated by a horizontal line connecting husband and wife, while progeny are attached to a second horizontal line, attached to the marriage line by a vertical line. The progeny are arranged in serial order, left to right, with the oldest sibling first. Known genotypes may be written below the symbols.

Because genetic analysis requires large samples, the small size of most human families could handicap geneticists in their study of man. However, pedigrees on a particular trait may be pooled and analysis of this larger group can furnish information of the genetic basis of a trait. A feature of a good pedigree is that it reveals the probable variety of gene(s) behind a trait, such as dominant, codominant, recessive, X-linked and so forth. Since expression of a recessive trait requires two recessive alleles (one if X-linked in the male), a rare recessive trait customarily appears erratically in a pedigree. The trait may not appear in every generation, and aa affected person may have normal parents. With a completely dominant allele, an affected person must have an affected parent, and the trait does not skip a generation.

BIOLOGY 1406

GENETICS PROBLEMS: FIRST SET
  1. In peas, axial flower position (A) is dominant to terminal position (a). Give the expected phenotypic ratios for the following crosses. (A) Homozygous axial X Terminal    (B) Heterozygous X Terminal   (C) Heterozygous X Heterozygous
  2. In cattle, hornless (H) is dominant to the horned condition (h). A certain hornless bull is bred to three cows. With cow A, which is horned, a hornless calf is produced. With cow B, also horned, a horned calf is produced, and with cow C, which is hornless, a hornless calf is produced. What is the genotype of the three cows and the bull?
  3. An albino man marries a woman with normal skin pigmentation, but her mother is albino. Show the type of children the couple may have and the proportions of each type. Albinism is a recessive condition.
  4. Red coat color (R) is codominant with white coat (R’) in short horn cattle; with the heterozygous condition producing the roan condition. A farmer has a roan bull and a red herd of cows. How can he produce a pure breeding white herd, using only his present herd. Explain your answer.
  5. Red-green color blindness is due to a recessive X-linked allele (n), and normal color vision results from the dominant allele (N). If a woman of normal color vision, whose father is color blind marries a normal vision male whose father was also color blind, what is the expected phenotypic ratio of their possible children?
  6. A man with type B blood, whose mother was type O, marries a woman with type AB. Give the expected phenotypic and genotypic ratios of their possible children.
  7. A man and his wife have normal color vision, but have a color blind daughter. Red-green color blindness is caused by a X-linked recessive allele. The man sues his wife for divorce on grounds of infidelity. Can genetics provide evidence supporting his case? Explain your answer.
  8. In four O’clock flowers, red flowers (R) show incomplete dominance to white flower (R’). The heterozygous condition produces pink flowers. Give the expected genotypic and phenotypic ratios for the offspring of a cross between two pink flowering plants.
  9. In Guinea pigs black hair (B) is dominant to brown hair (b), while smooth coat (S) is dominant to rough coat (s). Give the phenotypic and genotypic ratios for the F-1 and F-2 of a cross between an individual homozygous for black rough hair and a brown individual homozygous for smooth hair.
  10. In garden peas, yellow peas (Y) are dominant to green peas (y) and red flowers (R) are dominate to white (r). Give the expected phenotypic ratios of a cross between an individual homozygous for both yellow peas and white flowers and an individual heterozygous for both pair of alleles.

 

GENETICS PROBLEMS: SECOND SET
  1. A man with cystic fibrosis of the pancreas wishes to marry a woman in whose family there has been no evidence of the disease for at least four generations. Cystic fibrosis is an inborn error of metabolism, a recessive condition, in which an enzyme for fat digestion is lacking. Could this couple have a child with cystic fibrosis? Write the possible genotypes for the man, the woman and their possible children.
  2. In cats, short hair (S) is dominant over long hair, Angora (s). An Angora male is mated with a short-haired female. She produces a litter of eight kittens, six which are short-haired and two which are Angora. Give the genotypes of the parents. What is the expected ratio of the offspring of this mating. Give the expected numbers of each phenotype, if these same two cats produce fifty kittens in successive litters.
  3. In man, aniridia, a type of blindness is due to a dominate gene, and migraine, a type of sick headache, is the result of a different dominant gene on another chromosome. A man with aniridia, whose mother was not blind, marries a woman with normal vision, but who suffers from migraine. Her father did not have migraine. Give the phenotypic ratio of their expected children.
  4. In Drosophila, white eyes "r" is an X-linked recessive, with red eyes "R" being the dominant allele. In one cross the following number of offspring is produced. Give the genotypes and phenotypes of the parents.  273 red-eyed males 258 white-eyed males   304 red-eyed females 243 white eyed female
  5. Hemophilia, bleeder’s disease, is associated with an X-linked recessive allele. The normal daughter of a hemophilic man marries a normal man. What is the probability of having an affected child?
  6. The weight of a certain variety of squash is determined by three pair of genes: AABBCC produces fruit weighting 6 pounds, while aabbcc produces 3 pound fruit. Each gene represented by a capital letter adds 1/2 pound to the weight of the fruit. When a plant producing 6 pound fruit is crossed with a plant producing 3 pound fruit, all of the offspring produce 4 1/2 pound fruit. Give the weights of the fruit and proportions of each for the offspring resulting from a cross between two of the 4 1/2 pounds producing plants derived from the previous cross.
  7. The X-linked barred locus in chickens controls the pattern of feathers, with the allele (B) for barred pattern and (b) for no bars. If a barred male is mated to a non-barred female, what will be the appearance of the male and female offspring.

 

Return To: NLC Biology or Biology 1406 Lecture

Last Updated: 11/13/98