Imagine that a couple is planning to have children. The male is heterozygous for Huntington’s disease and homozygous dominant for Tay-Sachs. The female is homozygous recessive for Huntington’s disease and heterozygous for Tay-Sachs. The couple is curious about the possibility and probability of their offspring inheriting Tay-Sachs and/or Huntington’s. For humans, Huntington’s disease is dominant (H) over the “normal” condition (h), and the “normal” condition is dominant (T) over Tay-Sachs (t). Complete a Punnett square for this cross and record the probabilities for genotypes and phenotypes of the offspring as ratios.

Imagine that a couple is planning to have children. The male is heterozygous for Huntington’s disease and homozygous dominant for Tay-Sachs. The female is homozygous recessive for Huntington’s disease and heterozygous for Tay-Sachs. The couple is curious about the possibility and probability of their offspring inheriting Tay-Sachs and/or Huntington’s. For humans, Huntington’s disease is dominant (H) over the “normal” condition (h), and the “normal” condition is dominant (T) over Tay-Sachs (t). Complete a Punnett square for this cross and record the probabilities for genotypes and phenotypes of the offspring as ratios.

Name: Imagine that a couple is planning to have children. The male is hetero
Uploaded: 2024-03-20T06:43:04.000Z
Channel: Bartleby
Description: Imagine that a couple is planning to have children. The male is heterozygous for Huntington’s disease and homozygous dominant for Tay-Sachs. The female is homozygous recessive for Huntington’s disease and heterozygous for Tay-Sachs. The couple is cur

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter18: Genetics Of Behavior

Section: Chapter Questions

Problem 14QP: A pedigree analysis was performed on the family of a man with schizophrenia. Based on the known...

See similar textbooks

Similar questions

young couple is planning to have children. The male is heterozygous for Huntington’s disease and homozygous dominant for Tay-Sachs. The female is homozygous recessive for Huntington’s disease and heterozygous for Tay-Sachs. The couple is curious about the possibility and probability of their offspring inheriting Tay-Sachs and/or Huntington’s. For humans, Huntington’s disease is dominant (H) over the “normal” condition (h), and the “normal” condition is dominant (T) over Tay-Sachs (t). Complete a Punnett square for this cross and record the percent probabilities for genotypes and phenotypes of the offspring.
Hemophilia is a sex-linked recessive trait. A male hemophiliac and phenotypically normal female have a son with hemophilia. They would like to have one more child. What is the probability of having a child without hemophilia? Explain using a Punnett square. Is it possible for a girl to be born with hemophilia? Explain.
In man, Huntington’s Chorea is autosomal dominant progressive brain disorder and often lethal (H), the individuals usually dying in middle age. Red-green color blindness is recessive and sex-linked (X c ). A double heterozygous woman marries a heterozygous Huntington’s Chorea colorblind man. What are the genotypes of the man and woman? What are the genotypes and phenotypes of the all the remaining offspring after all deaths caused by Huntington’s Chorea? Show COMPLETE cross. TIP: cancel out gametes with H
Hemophilia, disease in which the blood lacks a clotting factor, is caused by an X linked recessive gene. Joe doe not have hemophilia and Lucille is heterozygous for the condition. What is the chance that their MALE child will have hemophilia? (Note: you are calculating the probability for their MALE children only, in other words if they have 1 male child, what is the probability that he will be born with the disease?) 0% 1/4 or 25% 2/4 or 50% 3/4 or 75% 4/4 or 100%
A newly-married couple is thinking of having children. They are worried about their child having cystic fibrosis, a recessive disease that shows up in children with the genotype ff. Jake, the father is homozygous dominant with a genotype FF. Amy, the mother, is heterozygous with a genotype Ff. Should they be worried about their children having this disease? For full credit, discuss the probability of having each genotype (FF, Ff, and ff) in your explanation. You may first write and then RECORD YOUR ANSWER.
A WOMAN IS HETEROZYGOUS FOR TWO HARMFUL RECESSIVE ALLELES IN DIFFERENT CHROMOSOMES, ONE FOR PHENYLKETONURIA (PKU) AND THE OTHER FOR CYSTIC FIBROSIS (CF). SHE MARRIES AN UNAFFECTED MAN WHO IS A CARRIER FOR NEITHER DISEASE. IF SHE HAS A DAUGHTER, WHAT IS THE PROBABILITY THAT THE CHILD WILL CARRY NEITHER OF THE RECESSIVE ALLELES? EXACTLY ONE? BOTH?
Sickle cell anemia is inherited as an autosomal recessive condition. It also exhibits incomplete dominance in that the heterozygous genotype displays a mild form of the disease known as sickle cell trait while individuals with the homozygous recessive genotype have a severe form of SCA. A man who has severe sickle cell anemia marries a woman who suffers from a mild trait. What is the probabilitu they will have a child with severe SCA?What is the probability they will have a child with mild SCA? What is the probability they will have a normal child? Show ALL work using punnett squares.
Duchenne muscular dystrophy is a sex-linked recessive disorder characterized by a progressive loss of muscle tissue. Neither Carla nor Rudy has Duchenne muscular dystrophy, but their first son does have it. What can you say about Rudy's or Carla's genotype in relation with the disorder? If the couple has a second child, what is the probability that he or she will also have the disease? Show your answers with a Punnet square.
hemophilia is a sex linked recessive disease. Dick does not have hemophilia. His wife Debi is a heterozygote. They are having fraternal twins, one of each sex, who they have named Dick Jr. and Debi Jr. What are the chances of Dick Jr. having hemophilia? Debi Jr.?
Pompe disease is a glycogen storage disorder caused by a lack of ⍺-glucosidase, the enzyme that converts glycogen to glucose in the muscles. Babies born with this disorder die by the age of two. Pompe disease is a genetic disorder caused by the presence of two recessive autosomal alleles. A man and a woman heterozygous for the condition have two female unaffected children. Determine the probability, expressed as a whole number percentage, of a third baby being born a male with Pompe disease. Your answer must include the use of a Punnett square as well as a legend indicating the allele symbols used. (2 marks)
Albinism and sickle cell anemia are both autosomal recessive disorders that are expressed only in the homozygous state. Two parents who are carriers for both albinism and sickle cell anemia plan to have a child. What is the probability that the child will have exactly one of the conditions? 3/8 7/16 O 1/16 O 1/4 9/16
If both parents are carriers for an autosomal recessive disorder (i.e. both Aa) and they have 2 children, what is the probability that both will be affected by the disorder? (Give your answer to 4 decimal places)"