Practice: Use the product law to calculate the probability that mating two organisms with the genotype of AaBbCcDd will produce offspring with the genotype of AA bb Cc Dd?

Subjects

Sections | |||
---|---|---|---|

Mendel's Experiments and Laws | 18 mins | 0 completed | Learn |

Inheritance in Diploids and Haploids | 34 mins | 0 completed | Learn |

Monohybrid Cross | 33 mins | 0 completed | Learn |

Dihybrid Cross | 59 mins | 0 completed | Learn |

Sex-Linked Genes | 22 mins | 0 completed | Learn |

Probability and Genetics | 27 mins | 0 completed | Learn |

Pedigrees | 31 mins | 0 completed | Learn |

Practice: Use the product law to calculate the probability that mating two organisms with the genotype of AaBbCcDd will produce offspring with the genotype of AA bb Cc Dd?

Practice: In a family of five children what is the probability that… I. Three are males and two are females

Practice: In a family of five children what is the probability that… All are females

Practice: In a family of five children what is the probability that… Two are males and three are females

Practice: In a family of six children, where both parents are heterozygous for albinism, what is the probability that four are normal and two are albinos?

0 of 6 completed

Concept #1: Probability

Practice #1: Product Law

Practice #2: Binomial

Practice #3: Binomial

Practice #4: Binomial

Practice #5: Binomial

Adjusting probabilities based on new informationIndividual IV-3 is born as a male affected with condition B but not condition A. His parents are bred again, and an ultrasound shows that the fetus is another male. The relevant portion of the pedigree is shown below. Use this new information to determine the parents.

In mice, an allele for apricot eyes (a) is recessive to an allele for brown eyes (a+). At an independently assorting locus, an allele for tan color (t) is recessive to an allele for black coat color (t+). A mouse that is homozygous for brown eyes and black coat color is crossed with a mouse having apricot eyes and a tan coat. The resulting F1 are intercrossed to produce the F2. In a litter of eight F2 mice, what is the probability that two will have apricot eyes and tan coats?

In the F2 generation of a homozygous round (AA) × homozygous wrinkled (aa) cross in peas, two round seeds are chosen at random. What is the probability that one is AA and the other Aa? The key here is the stipulation that the chosen seeds are round (i.e., AA or Aa).A) 2(2/3)(1/3)B) (1/3)2C) (2/3)2

Sickle cell anemia and albinism are both recessive traits in humans. Imagine that a couple, already pregnant with twins, has just learned that they are both heterozygous for both of these traits.As the couple's genetic counselor, the couple asks you the following questions about how their carrier status will affect their offspring.A. If the couple has fraternal twins, what is the probability that both children will be unaffected by both conditions?B. If the couple has fraternal twins, what is the probability that both of the couple's children will have both sickle cell anemia and albinism?C. What is the probability that one of the fraternal twins is a carrier of either, but not both, of the conditions?(Hint: You will need to use both the product law and the sum law to answer this question.)D. If the couple has fraternal twins, what is the probability of having two phenotypically normal children, one being a carrier of only the sickle cell anemia recessive allele, and the other being a carrier of only the recessive allele for albinism?(Hint: You will need to use both the product law and the sum law to answer this question.)

Use the completed Punnett square in Part B to answer the questions below about the F2 generation.Answer each question in the table by dragging the correct label to the appropriate location. Labels can be used once, more than once, or not at all.

Huntington's disease is an inherited autosomal dominant disorder that can affect both men and women. Imagine a couple has had seven children, and later in life, the husband develops Huntington's disease. He is tested and it is discovered he is heterozygous for the disease allele, Hh. The wife is also genetically tested for the Huntington's disease allele, and her test results show she is unaffected, hh. What is the percent probability that the first child of this couple will have Huntington's disease?also.What is the probability that any two of the seven children will have Huntington's disease?

In the accompanying figure, the chance that individual IV-7 is a heterozygous carrier is ________.a) 3/4 b) 1/4 c) 2/3d) 1/2 e) 1/3

In the accompanying figure, the chance that individual III-2 is a heterozygous carrier is ________.A) 100 %B) 50 %C) 25 %D) 0 %

Suppose that two parents are both heterozygous for sickle cell anemia, which is an autosomal recessive disease. They have seven children. Use the binomial theorem to determine the probability that three of the children have sickle cell anemia and four of the children are healthy. Round your answer to the nearest hundredth.

What is the probability that Mary and Justin will have an affected child?

Enter your friends' email addresses to invite them:

We invited your friends!

Join **thousands** of students and gain free access to **10 hours** of Genetics videos that follow the topics **your textbook** covers.