Pedigree 2 from Part A is shown below. Recall that this pedigree shows the inheritance of a rare, autosomal dominant condition.Fill in the genotypes for the indicated individuals in the pedigree by dragging the best label to the appropriate location. Labels can be used once, more than once, or not at all.
  • 1. 2/32. 1/23. 1/12
  • I- DdII- Dd, dd, DdIII- dd, Dd, dd
  • autosomal dominant reaction, autosomal dominant reaction, autosomal recessive.
  • Red shows incomplete dominance over white.
Based on this incomplete information, calculate the probability that if this couple has a child, the child will have sickle-cell disease. Express your answer as a fraction using the slash symbol and no spaces (for example, 1/16).
  • 1/9
  • 1/16
  • 1/6
  • 1/2
Consider pea plants with the genotypes GgTt and ggtt . These plants can each produce how many type(s) of gametes?
  • BB x bb
  • iiI A iI B iI A I B
  • all are 1/4
  • four ... one
Jane and John are considering having another child. Given the pedigree you constructed and the mode of inheritance for galactosemia, what is the risk that their next child will have the disorder?
  • 1/4 (because they are both heterozygotes)
  • all are 1/4
  • Red shows incomplete dominance over white.
  • Albino (b) is a recessive trait; black (B) is a dominant trait.
Part A: If two phenotypically normal tigers that are heterozygous at this locus are mated, what percentage of their offspring will be cross-eyed? Express your answer using the percent symbol and no spaces (for example, 10%).
  • all are 1/4
  • 100%
  • 25%
  • 1/2
What is an allele?
  • a combination of polygenic inheritance and environmental factors
  • It is homozygous at two loci.
  • iiI A iI B iI A I B
  • an alternative version of a gene
Part A: A black guinea pig crossed with an albino guinea pig produces 12 black offspring. When the albino is crossed with a second black one, 7 blacks and 5 albinos are obtained.What is the best explanation for this genetic outcome?
  • gametes: 1/2 B and 1/2 b (heterozygous parent) and b; offspring: 1/2 Bb and 1/2 bb
  • Albino (b) is a recessive trait; black (B) is a dominant trait.
  • all are 1/4
  • ultrasound imaging ... chorionic villus sampling ... amniocentesis
Part C: What is the genotype of the child?
  • ii
  • BB x bb
  • I A i
  • I B i
Individuals III-3 and III-4 are expecting their first child when they become aware that they both have a family history of this recessive condition. As their genetic counselor, you can calculate the probability that they are carriers and that their child will be affected with the condition.
  • 12:3:1
  • iiI A iI B iI A I B
  • 1. 2/32. 1/23. 1/12
  • I- DdII- Dd, dd, DdIII- dd, Dd, dd
Part A: What is the genotype of the man?
  • I B i
  • Bb x bb
  • ii
  • I A i
You cross a true-breeding red-flowered snapdragon with a true-breeding white-flowered one. All of the F1 are pink. What does this say about the parental traits?
  • These homologous chromosomes represent a maternal and a paternal chromosome.
  • 1/4 (because they are both heterozygotes)
  • They do not interact at all.
  • Red shows incomplete dominance over white.
Select the correct explanation for the fact that a carrier of a recessive genetic disorder does not have the disorder.
  • Affected individuals have recessive mutations, but a carrier has a dominant mutation.
  • These homologous chromosomes represent a maternal and a paternal chromosome.
  • They do not interact at all.
  • gametes: 1/2 B and 1/2 b (heterozygous parent) and b; offspring: 1/2 Bb and 1/2 bb
Quantitative characters vary in a population along a continuum. How do such characters differ from the characters investigated by Mendel in his experiments on peas?
  • Affected individuals have recessive mutations, but a carrier has a dominant mutation.
  • Quantitative characters are due to polygenic inheritance, the additive effects of two or more genes on a single phenotypic character. A single gene affected all but one of the pea characters studied by Mendel.
  • The chance that an individual taken at random from the F2 generation produces wrinkled seeds is 25% and the chance that the same individual produces yellow seeds is 75%.
  • 1/4 (because they are both heterozygotes)
In his breeding experiments, Mendel first crossed true-breeding plants to produce a second generation, which were then allowed to self-pollinate to generate the offspring. How do we name these three generations?
  • P ... F1 ... F2
  • He or she will show some symptoms of the disease.
  • I- rrII- Rr, RR, RrIII- R_, R_, R_
  • It is very likely that at least one of Woody Guthrie's parents also have had the allele for Huntington's disease.
In maize (corn plants) a dominant allele I inhibits kernel color, while the recessive allele i permits color when homozygous. At a different locus, the dominant allele P causes purple kernel color, while the homozygous recessive genotype pp causes red kernels.If plants heterozygous at both loci are crossed, what will be the phenotypic ratio of the offspring?_____ colorless: _____ purple: _____ red
  • 12:3:1
  • 1/9
  • 25%
  • 1/16
Note that individual II-3 has no family history of this rare condition.Fill in the genotypes for the indicated individuals in the pedigree by dragging the best label to the appropriate location. Labels can be used once, more than once, or not at all.
  • 1/4 (because they are both heterozygotes)
  • I- rrII- Rr, RR, RrIII- R_, R_, R_
  • 1. 2/32. 1/23. 1/12
  • I- DdII- Dd, dd, DdIII- dd, Dd, dd
Various procedures can be used to detect genetic disorders before birth. Among the tests discussed in this chapter, which is the least invasive (list first), and which two allow the chromosomes of the fetus to be examined?
  • a combination of polygenic inheritance and environmental factors
  • Albino (b) is a recessive trait; black (B) is a dominant trait.
  • These homologous chromosomes represent a maternal and a paternal chromosome.
  • ultrasound imaging ... chorionic villus sampling ... amniocentesis
Part E: What are the genotypes of the gametes and offspring in the second cross?
  • BB x bb
  • gametes: B and b; offspring: all Bb
  • gametes: 1/2 B and 1/2 b (heterozygous parent) and b; offspring: 1/2 Bb and 1/2 bb
  • iiI A iI B iI A I B
Part C: What are the genotypes of the gametes and offspring in the first cross?
  • gametes: B and b; offspring: all Bb
  • It is homozygous at two loci.
  • iiI A iI B iI A I B
  • gametes: 1/2 B and 1/2 b (heterozygous parent) and b; offspring: 1/2 Bb and 1/2 bb
If Jane and John want to have another child, they plan to see a genetic counselor to find out when it would be best to test for galactosemia. A newborn with galactosemia must be put on a lactose- and galactose-free diet as soon as possible after birth. Even on this diet, affected individuals may still suffer from learning disabilities, ovarian failure (in young women), late-onset cataracts, and early death.Which of the following tests would be most useful for Jane and John to have?
  • Albino (b) is a recessive trait; black (B) is a dominant trait.
  • autosomal dominant reaction, autosomal dominant reaction, autosomal recessive.
  • newborn screening (either assaying for the GALT enzyme or measuring excess galactose in the newborn's blood)
  • Affected individuals have recessive mutations, but a carrier has a dominant mutation.
Part B: What is the genotype of the woman?
  • I B i
  • ii
  • I A i
  • BB x bb
Two organisms with genotype AaBbCcDdEE mate. These loci are all independent. What fraction of the offspring will have the same genotype as the parents?
  • 25%
  • autosomal recessive
  • 1/6
  • 1/16
Part D: What are the genotypes of the parents in the second cross (the cross that produced 7 black and 5 albino offspring)?
  • I A i
  • iiI A iI B iI A I B
  • I B i
  • Bb x bb
Part B: What are the genotypes of the parents in the first cross (the cross that produced 12 black offspring)?
  • I B i
  • BB x bb
  • gametes: B and b; offspring: all Bb
  • I A i
Part E: In what frequencies would you expect the offspring genotypes? Indicate the frequency of each genotype by dragging the labels to the table. Labels may be used once, more than once, or not at all.
  • all are 1/4
  • BB x bb
  • Albino (b) is a recessive trait; black (B) is a dominant trait.
  • iiI A iI B iI A I B
Look at the Punnett square, which shows the predicted offspring of the F2 generation from a cross between a plant with yellow-round seeds (YYRR) and a plant with green-wrinkled seeds (yyrr). Select the correct statement about wrinkled yellow seeds in the F2 generation.
  • Albino (b) is a recessive trait; black (B) is a dominant trait.
  • Top row (left to right): Hanna's father, Harry's father2nd row (left to right): Hilda, Hope, Holly, Hanna, Harry3rd row (left to right): Jen, Joe, Jane, John4th row (left to right): Les, Lee, Leah(THIS IS THE ONLY ONE THAT IS RED)
  • The chance that an individual taken at random from the F2 generation produces wrinkled seeds is 25% and the chance that the same individual produces yellow seeds is 75%.
  • 1/4 (because they are both heterozygotes)
Assuming that both parents are heterozygous for the gene that causes the disease, what is the probability that the second child will also have the disease? Express your answer as a fraction using the slash symbol and no spaces (for example, 1/2).
  • 1/16
  • 1/6
  • 1/2
  • 1/9
Folk singer Woody Guthrie died of Huntington's disease, an autosomal dominant disorder. Which statement below must be true?
  • He or she will show some symptoms of the disease.
  • These homologous chromosomes represent a maternal and a paternal chromosome.
  • Affected individuals have recessive mutations, but a carrier has a dominant mutation.
  • It is very likely that at least one of Woody Guthrie's parents also have had the allele for Huntington's disease.
What percentage of cross-eyed tigers will be white? Express your answer using the percent symbol and no spaces (for example, 10%).
  • 1/16
  • 1/2
  • 100%
  • 1/6
0:0:1



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