Plant Chapter 14 Concept Check

11 Questions | Attempts: 288

Plant Quizzes & Trivia

Take this quiz to know more about plants. Score a perfect and tell everyone that you know everything about nature.

Questions and Answers
  • 1. 
    A  pea plant heterozygous for inflated pods (Ii) is crossed with a plant homozygous for constricted pods (ii).  Draw a Punnett square for this cross.  Assume pollen comes from the ii plant. A cross of Ii x ii would yield offspring with a genotypic ratio of  :  and a phenotypic ratio of  inflated :  constricted
  • 2. 
    Pea plants heterozygous for flower position and stem length (AaTt) are allowed to self-pollinate, and 400 of the resulting seeds are planted.  Draw a Punnett square for this cross.  How many offspring would be predicted to have terminal flowers and be dwarf? According to the law of  ,  plants are predicted to be aatt, or recessive for both characters. The actual result is likely to differ slightly from this value.
  • 3. 
    List the different gametes that could be made by a pea plant heterozygous for seed color, seed shape, and pod shape (YyRrIi).  How large a Punnett square would you need to predict the offspring of a self-pollination of this “trihybrid”? The plant could make eight different gametes (  ,  ,  ,  ,  ,  ,  ,  ) To fit all the possible gametes in a self-pollination, a Punnet square would need  rows and  columns.  It would have spaces for the  possible unions of gametes in the offspring.
  • 4. 
    For any gene with a dominant allele C and recessive allele c, what proportions of the offspring from a CC x Cc cross are expected to be homozygous dominant, homozygous recessive, and heterozygous? homozygous dominant (CC),  homozygous recessive (cc), and  heterozygous (Cc)
  • 5. 
    An organism with the genotype BbDD is mated to one with the genotype BBDd.  Assuming independent assortment of these two genes, write the genotypes of all possible offspring from this cross and use the rules of probability to calculate the chance of each genotype occurring. BBDD; BbDD;  BBDd;  BbDd
  • 6. 
    Three characters (flower color, seed color, and pod shape) are considered in a cross between two pea plants (PpYyIi x ppYyii).  What fraction of the offspring would be predicted to be homozygous recessive for at least two of the three characters? The genotypes  and frequencies to fulfill this condition are 
  • 7. 
    Incomplete dominance and epistasis are both terms that define genetic relationships.  What is the most basic distinction between these terms? describes the relationship between two alleles of a , whereas  relates o the genetic relationship between two 
  • 8. 
    If a man with type AB blood marries a woman with type O blood, what blood types would you expect in their children? of the children would be expected to have to have type A blood and  type B blood.
  • 9. 
    A rooster with gray feathers is mated with a hen of the same phenotype.  Among their offspring, 15 chicks are gray, 6 are black, and 8 are white.  What is the simplest explanation for the inheritance of these colors in chickens?  What phenotypes would you expect in the offspring of a cross between a gray rooster and a black hen? The black and white alleles ar expressing , with gray in color. A cross between a gray rooster and a black hen should yield gray and blackoffspring.
  • 10. 
    Beth and Tom each have a sibling with cystic fibrosis, but neither Beth nor Tom nor any of their parents have the disease.  Calculate the probability that if this couple has a child, the child will have cystic fibrosis.  What would be the probability if a test revealed that Tom is a carrier but Beth is not? (Since cystic fibrosis si caused by a  , Beth and Tom's siblings who have CF must be  )  Therefore, each parent must be a  .  Since neither Beth nor Tom has CF, this means they each have a  chance of being a carrier. If they are both carriers, there is a chance that they will have a child with CF. If a test proves Tom is a carrier and Beth is not, there isa % chance that they produce a child with CF.
  • 11. 
     Joan was born with six toes on each foot, a dominant trait called polydactyly.  Two of her five siblings and her mother, but not her father, also have extra digits.  What is Joan’s genotype for the number-of-digits character? Joan's genotype is  because the allele for polydactyly is  to the allele for five digits per appendage.  Because Joan's father does not have polydactyly, his genotype must be  , therefore Joan, who does have the trait, must be  .

Related Topics

Back to Top Back to top

Here's an interesting quiz for you.

We have other quizzes matching your interest.