The steps to determining the probabilities of genotypic and phenotypic frequency are

· Identify dominant and recessive alleles · Determine parental genotypes · Set up a Punnett square that shows the gametic contributions of the mother and father · Examine the probable genotypic and phenotypic frequencies in the offspring

- Cystic fibrosis is a disease that results from a defective plasma membrane protein and causes complications of the lungs, pancreas, and sweat glands. - A couple has a child with cystic fibrosis. Neither of the parents has the disease. What are the odds that their next child will not have cystic fibrosis, but will be a carrier (heterozygous)? - To solve the problem, follow the steps for determining the probabilities of genotypic and phenotypic frequencies in the offspring. - Identify dominant and recessive alleles. - The child has the disease, but the parents don’t; therefore, the cystic-fibrosis trait must be recessive. Let N = normal n = cystic fibrosis nn = cystic fibrosis - Determine parental genotypes. Because both parents are phenotypically normal, each parent has at least one dominant allele. P = normal X normal N_ N_ - Because the child has cystic fibrosis, his genotype must be nn. Therefore the parents each must have one n allele. - P = normal X normal Nn Nn - Set up a Punnett square that shows the gametic contributions of the mother and father. - Each side of the Punnett square shows the possible gametes of the mother or father. Each parent can contribute either an N or n allele. The squares inside the Punnett square portray the genotypic frequencies for a single offspring. Each offspring is an independent event with respect to the other offspring. - At birth the brother is phenotypically normal (does not have cystic fibrosis.) The nn genotype can be eliminated from the Punnett square. The probability that the child is heterozygous is 2/3.

A plant biologist has recently discovered a population of wild orchids in the Ecuadoran Amazon. The majority of the population has red flowers but a few have pink flowers. To test the inheritance of the trait, she crosses two pink flowers and observes that 1/3 of the offspring are red and 2/3 are pink. What could explain the set of observed frequencies?

- The recessive allele is lethal when homozygous

The fictitious variety of hamsters known as “tuxedo” is prized for their black and white coat pattern. You purchase a stud hamster named Henry for $300. All of the progeny sired by Henry in matings with a pure breeding tuxedo female display the black and white phenotype. Matings between certain pairs of offspring, however, produce an undesirable yellow and white coat color 25% of the time. Regarding the situation, which of the following statements is false?

- The pairs of offspring that produce the undesirable coat pattern are both homozygous for a recessive allele

Cystic Fibrosis (CF) is a recessive inherited disease. If two parents without CF have a child with CF, what must be true of the parents?

- They are both heterozygous for the recessive trait.

The Punnett square provides a simple method of calculating the probability of a particular genotypic or phenotypic outcome. In these squares, the dominant allele is often indicated by an uppercase letter, and the recessive by a lowercase letter (in this case the single trait is symbolized by the letters CF (or cf)). What are the genotypes of the offspring represented by cells B and D?

- heterozygous, homozygous recessive

9.2.2 Mendelian Inheritance Flashcards by Irina Soloshenko

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9.2.2 Mendelian Inheritance Flashcards

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