The pedigree to the right shows a pattern of inheritance for galactosemia, a rare autosomal recessive trait in humans where the enzymes that digest and utilize galactose are missing or defective. Shaded individuals have galactosemia. If individuals B and D want to get married, what is the probability that their first offspring will have galactosemia? A D а. 1/24 b. 1/16 с. 1/2 d. 1/12 е. 1/48 f. 1/4

The pedigree to the right shows a pattern of inheritance for galactosemia, a rare autosomal recessive trait in humans where the enzymes that digest and utilize galactose are missing or defective. Shaded individuals have galactosemia. If individuals B and D want to get married, what is the probability that their first offspring will have galactosemia? A D а. 1/24 b. 1/16 с. 1/2 d. 1/12 е. 1/48 f. 1/4

Name: 5. The pedigree to the right shows a pattern of inheritance forgalact
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Description: 5. The pedigree to the right shows a pattern of inheritance forgalactosemia, a rare autosomal recessive trait in humanswhere the enzymes that digest and utilize galactose aremissing or defective. Shaded individuals have galactosemia.If individuals B

Biology: The Unity and Diversity of Life (MindTap Course List)

14th Edition

ISBN:9781305073951

Author:Cecie Starr, Ralph Taggart, Christine Evers, Lisa Starr

Publisher:Cecie Starr, Ralph Taggart, Christine Evers, Lisa Starr

Chapter13: Observing Patterns In Inherited Traits

Section: Chapter Questions

Problem 1SQ: A heterozygous individual has a _______ for a trait being studied. a. pair of identical alleles b....

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In individuals affected by cystic fibrosis, salt crystals may appear afterperspiration dries up. In addition, the disease causes respiratory disorderswhich can be both debilitating and lethal. It occurs in individuals homozygousfor the recessive gene. Two normal parents had a daughter with thesymptoms of this disease, and a normal son who marries a normal womanwith an afflicted A test (salt concentration in perspiration of heterozygotes ishigher than normal) disclosed that both are indeed carriers of the gene. If thefirst child born to the mating in (b) was defective, what is the probability thatthe 2nd child would also be defective?Express answer in fraction form
5) Humans who have an abnormally high level of cholesterol are said to suffer from familialhypercholesterolemia.  The gene for this disorder is dominant (C). A man who isheterozygous for familial hypercholesterolemia marries a woman who is homozygousfor the recessive allele. What is the probability that they will have children that sufferfrom this disorder? please draw the puutnet square and also write the parents and gemtaics top the puutnet square thanks
1. The allele G for yellow stigma is completely dominant to green (g). Supposingtwo strains of autotetraploid plants are available and their genotypes are as follows:GGgg – in this plant the gene is close to the centromereGggg – in this plant the gene is far from the centromere If these two plants are crossed:a) provide the gametes that can be obtained from the two plants;b) provide the genotypic and phenotypic ratios of the offspring. 2. Consider the illustration below. Diagram the configuration you would observe at Anaphase I if crossing-over happens within the inversion. (IMAGE ATTACHED)
9. Make a pedigree for each of the following situations. For each individual, write the individual's genotype (when possible) next to the individual's symbol (e.g. O xty, I Gg): a. Two parents do not have cystic fibrosis and they have a daughter with cystic fibrosis and a son who does not have cystic fibrosis. The daughter grows up and she mates with a male who does not have cystic fibrosis. Their only child is a boy and he has cystic fibrosis. b. A man with hemophilia mates with a female without hemophilia. They have one son and one daughter. The daughter has hemophilia and the son does not have hemophilia. The son grows up, and he marries and mates with a female. Their only child is a boy, and he has hemophilia.
Galactosemia is a recessive human disease that istreatable by restricting lactose and glucose in the diet.Susan Smithers and her husband are both heterozygous for the galactosemia gene.a. Susan is pregnant with twins. If she hasfraternal (nonidentical) twins, what is theprobability both of the twins will be girls whohave galactosemia?b. If the twins are identical, what is the probabilitythat both will be girls and have galactosemia?For parts (c–g), assume that none of the children isa twin.c. If Susan and her husband have four children, whatis the probability that none of the four will havegalactosemia?d. If the couple has four children, what is the probability that at least one child will have galactosemia?e. If the couple has four children, what is the probability that the first two will have galactosemia andthe second two will not?f. If the couple has three children, what is the probability that two of the children will have galactosemia and one will not, regardless of order?g. If…
The maternal-effect mutation bicoid (bcd) is recessive. Inthe absence of the bicoid protein product, embryogenesis isnot completed. Consider a cross between a female heterozygousfor the bicoid mutation (bcd+/ bcd-) and a homozygousmale(bcd-/ bcd-). How is it possible for a male homozygous for the mutationto exist?
6. A person is simultaneously heterozygous for two autosomal genetic traits. One is a recessive condition foralbinism (alleles A and a); this albinism gene is foundnear the centromere on the long arm of an acrocentricautosome. The other trait is the dominantly inheritedHuntington disease (alleles HD and HD+). TheHuntington gene is located near the telomere of oneof the arms of a metacentric autosome. Draw all copies of the two relevant chromosomes in this person asthey would appear during metaphase of (a) mitosis,(b) meiosis I, and (c) meiosis II. In each figure, labelthe location on every chromatid of the alleles forthese two genes, assuming that no recombinationtakes place.
FAlpQLSfiOhfAvlhxzCSiUll_6rt-nU5b0WI73UmWOxkOw8OCwk01ng/formResponse B 1 2 Bb x Bb b 4 The fur in both parents in this cross is * 1 B B Bb x Bb b 3 4 brown black O homozygous dominant homozygous recessive 3. 近

The maternal-effect mutation bicoid (bcd) is recessive. Inthe absence of the bicoid protein product, embryogenesis isnot completed. Consider a cross between a female heterozygousfor the bicoid mutation (bcd+/ bcd-) and a homozygousmale(bcd-/ bcd-). Predict the outcome (normal vs. failed embryogenesis) inthe F1 and F2 generations of the cross described.
age 105 of 114 3 5 € BUD > D Q Autosomal Genes 1. Parental Cross: It is possible that your mutation is autosomal (dominant or recessive). In this case, you only need one Punnett square to make predictions for the F1 generation that will result from the parental cross, as it does not matter if the female or male in the parental generation is the true breeding mutant. a. Provide a Punnett square to predict the genotype of the F1 generation. b. Q Search If the mutant allele is dominant, predict the phenotype of the F1 generation. Make sure to also summarize the prediction as a ratio, comparing Wild-type : Mutant. c. If your mutant allele is recessive, predict the phenotype of the F1 generation. Make sure to also summarize the prediction as a ratio, comparing Wild-type : Mutant. 2. F1 Self-Cross: Take the F1 results from question #1 to create a Punnett square predicting the F2 generation genotypes. 104 a. What are the predicted F2 phenotypes if the mutant allele is dominant? Make sure to…
1. In the pedigree below, Use "A" for the allele associated with the dominant phenotype, and lowercase "a" for the allele associated with the recessive phenotype. Determine what is the most likely mode of inheritance of this disease (whether it is inherited as the result of an autosomal dominant or autosomal recessive trait). Write the most probable genotype for each individual based on the mode of inheritance you have determined. Show how all the partners are crossed and the expected offspring produced (You may use Punnett Square) 1 dró || IV
1 pts The pedigree below shows the expression of Huntington's disease in a family. Huntington's disease is a fatal genetic disorder that causes the progressive breakdown of nerve cells in the brain. It deteriorates a person's physical and mental abilities usually during their prime working years and has no cure. Huntington's is inherited as a dominant allele (H). Examine the pedigree below and determine the genotype for individual "2" DODO0O 2. O HH O Hh О h O either HH or Hh