Imagine you have a blood group of "X" which is recessive and expressed by xx. The dominant blood groups are Y and Z, where homozygous of these alleles are expressed as YY and ZZ, respectively. What will be the genotype of your parents blood group? Why? Please explain in your own words. [Max 200 words]

Imagine you have a blood group of "X" which is recessive and expressed by xx. The dominant blood groups are Y and Z, where homozygous of these alleles are expressed as YY and ZZ, respectively. What will be the genotype of your parents blood group? Why? Please explain in your own words. [Max 200 words]

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter16: Reproductive Technology, Genetic Testing, And Gene Therapy

Section: Chapter Questions

Problem 18QP

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Mike was referred for genetic counseling because he was concerned about his extensive family history of colon cancer. That family history was highly suggestive of hereditary nonpolyposis colon cancer (HNPCC). This predisposition is inherited as an autosomal dominant trait, and those who carry the mutant allele have a 75% chance of developing colon cancer by age 65. Mike was counseled about the inheritance of this condition, the associated cancers, and the possibility of genetic testing (on an affected family member). Mikes aunt elected to be tested for one of the genes that may be altered in this condition and discovered that she did have an altered MSH2 gene. Other family members are in the process of being tested for this mutation. Is colon cancer treatable? What are the common treatments, and how effective are they?
Mike was referred for genetic counseling because he was concerned about his extensive family history of colon cancer. That family history was highly suggestive of hereditary nonpolyposis colon cancer (HNPCC). This predisposition is inherited as an autosomal dominant trait, and those who carry the mutant allele have a 75% chance of developing colon cancer by age 65. Mike was counseled about the inheritance of this condition, the associated cancers, and the possibility of genetic testing (on an affected family member). Mikes aunt elected to be tested for one of the genes that may be altered in this condition and discovered that she did have an altered MSH2 gene. Other family members are in the process of being tested for this mutation. Seventy-five percent of people who carry the mutant allele will get colon cancer by age 65. This is an example of incomplete penetrance. What could cause this?
A couple was referred for genetic counseling because they wanted to know the chances of having a child with dwarfism. Both the man and the woman had achondroplasia (MIM 100800), the most common form of short-limbed dwarfism. The couple knew that this condition is inherited as an autosomal dominant trait, but they were unsure what kind of physical manifestations a child would have if it inherited both mutant alleles. They were each heterozygous for the FGFR3 (MIM 134934) allele that causes achondroplasia. Normally, the protein encoded by this gene interacts with growth factors outside the cell and receives signals that control growth and development. In achrodroplasia, a mutation alters the activity of the receptor, resulting in a characteristic form of dwarfism. Because both the normal and mutant forms of the FGFR3 protein act before birth, no treatment for achrondroplasia is available. The parents each carry one normal allele and one mutant allele of FGRF3, and they wanted information on their chances of having a homozygous child. The counsellor briefly reviewed the phenotypic features of individuals with achondroplasia. These include facial features (large head with prominent forehead; small, flat nasal bridge; and prominent jaw), very short stature, and shortening of the arms and legs. Physical examination and skeletal X-ray films are used to diagnose this condition. Final adult height is approximately 4 feet. Because achondroplasia is an autosomal dominant condition, a heterozygote has a 1-in-2, or 50%, chance of passing this trait to his or her offspring. However, about 75% of those with achondroplasia have parents of average size who do not carry the mutant allele. In these cases, achondroplasia is due to a new mutation. In the couple being counseled, each individual is heterozygous, and they are at risk for having a homozygous child with two copies of the mutated gene. Infants with homozygous achondroplasia are either stillborn or die shortly after birth. The counselor recommended prenatal diagnosis via ultrasounds at various stages of development. In addition, a DNA test is available to detect the homozygous condition prenatally. What if the couple wanted prenatal testing so that a normal fetus could be aborted?
A couple was referred for genetic counseling because they wanted to know the chances of having a child with dwarfism. Both the man and the woman had achondroplasia (MIM 100800), the most common form of short-limbed dwarfism. The couple knew that this condition is inherited as an autosomal dominant trait, but they were unsure what kind of physical manifestations a child would have if it inherited both mutant alleles. They were each heterozygous for the FGFR3 (MIM 134934) allele that causes achondroplasia. Normally, the protein encoded by this gene interacts with growth factors outside the cell and receives signals that control growth and development. In achrodroplasia, a mutation alters the activity of the receptor, resulting in a characteristic form of dwarfism. Because both the normal and mutant forms of the FGFR3 protein act before birth, no treatment for achrondroplasia is available. The parents each carry one normal allele and one mutant allele of FGRF3, and they wanted information on their chances of having a homozygous child. The counsellor briefly reviewed the phenotypic features of individuals with achondroplasia. These include facial features (large head with prominent forehead; small, flat nasal bridge; and prominent jaw), very short stature, and shortening of the arms and legs. Physical examination and skeletal X-ray films are used to diagnose this condition. Final adult height is approximately 4 feet. Because achondroplasia is an autosomal dominant condition, a heterozygote has a 1-in-2, or 50%, chance of passing this trait to his or her offspring. However, about 75% of those with achondroplasia have parents of average size who do not carry the mutant allele. In these cases, achondroplasia is due to a new mutation. In the couple being counseled, each individual is heterozygous, and they are at risk for having a homozygous child with two copies of the mutated gene. Infants with homozygous achondroplasia are either stillborn or die shortly after birth. The counselor recommended prenatal diagnosis via ultrasounds at various stages of development. In addition, a DNA test is available to detect the homozygous condition prenatally. What is the chance that this couple will have a child with two copies of the dominant mutant gene? What is the chance that the child will have normal height?
A couple was referred for genetic counseling because they wanted to know the chances of having a child with dwarfism. Both the man and the woman had achondroplasia (MIM 100800), the most common form of short-limbed dwarfism. The couple knew that this condition is inherited as an autosomal dominant trait, but they were unsure what kind of physical manifestations a child would have if it inherited both mutant alleles. They were each heterozygous for the FGFR3 (MIM 134934) allele that causes achondroplasia. Normally, the protein encoded by this gene interacts with growth factors outside the cell and receives signals that control growth and development. In achrodroplasia, a mutation alters the activity of the receptor, resulting in a characteristic form of dwarfism. Because both the normal and mutant forms of the FGFR3 protein act before birth, no treatment for achrondroplasia is available. The parents each carry one normal allele and one mutant allele of FGRF3, and they wanted information on their chances of having a homozygous child. The counsellor briefly reviewed the phenotypic features of individuals with achondroplasia. These include facial features (large head with prominent forehead; small, flat nasal bridge; and prominent jaw), very short stature, and shortening of the arms and legs. Physical examination and skeletal X-ray films are used to diagnose this condition. Final adult height is approximately 4 feet. Because achondroplasia is an autosomal dominant condition, a heterozygote has a 1-in-2, or 50%, chance of passing this trait to his or her offspring. However, about 75% of those with achondroplasia have parents of average size who do not carry the mutant allele. In these cases, achondroplasia is due to a new mutation. In the couple being counseled, each individual is heterozygous, and they are at risk for having a homozygous child with two copies of the mutated gene. Infants with homozygous achondroplasia are either stillborn or die shortly after birth. The counselor recommended prenatal diagnosis via ultrasounds at various stages of development. In addition, a DNA test is available to detect the homozygous condition prenatally. Should the parents be concerned about the heterozygous condition as well as the homozygous mutant condition?
tion 8: below is the pedigree of inheritance of phenylketonuria (PKU). We will designate the letter Caven for the dominant allele and "p" for the recessive allele. 4 The pedigree shows that the pattern of inheritance for the allele for phenylk ylketonuria is: I. II. 1 III. IV. Autosomal dominant Autosomal recessive X-linked dominant X-linked recessive b. The parents in generation I have how many children: I. 3 Boys II. 3 Girls III. IV. 3 Boys and 1 Girl 3 Girls and 1 Boy c. What is the genotype of individual 1 in generation III: I. PP II. pp III. Pp " O 1 III. 50% E III 1 ▬ 2 2 IV. 25% 1 3 IV. Can be PP or Pp ii. Suppose that a man having type AB blood marries a woman having type O blood. What is the probability that their child will have type A blood? I. 100% II. 75% 2 4 3
Trivla Game Show _Make Your Own Tri ngston.schoology.com/common-assessment-delivery/start/4789189591?action=onresume&submissionld=463322566 Dillon WF g Aa v Done In guinea pigs, black hair (B) is dominant to white hair (b) and rough hair (R) is dominant to smooth hair (r). What are all the possible genotypes of a guinea pig that has black, rough hair? (Select all that apply.) O BBRR BBRr BBrr BBRR BbRr O bbRR O bbRr O bbrr O Black O White O Rough OSmooth O Rough O Smooth
9 Genet x K Kami 9 Point x K point- x 9 Band, X S Week x 9 Geog x My Qu X STI In X > (2 x R Read X Discu X A web.kamihq.com/web/viewer.html?file=https%3A%2F%2Ftuscaloosacity.schoology.com%2Fattachment%2F1716642933%2Fsource%2F775235e0ba17ab191f792e5f.. K * : = 3+Branches+of+Go. O STI InformationNo. E Schubert = Pod 2 7th Master S. Other bookmarks Каmi Student Upgrade O O A My Drive Kami Export - Scan Feb 26, 2021.pdf A Turn In 100% JT Use the following information for questions 10-12: In dogs, the gene for fur color has two alleles. The dominant allele (F) codes for grey fur and the recessive allele (f) codes for black fur. 10) The female dog is heterozygous. The male dog is homozygous recessive. What is the chance their offspring have grey fur? T 14 - 11) The female dog has black fur. The male dog has black fur. What is the chance their offspring has a heterozygous genotype? 12) The female dog is heterozygous. The male dog is heterozygous. What is the chance their offspring is…
Let us practice it again! Analyze the pedigree below to answer the questions that follow. Huntington's disease a disorder in which nerve cells waste away, or disintegrate, is passed down through families. certain parts of the brain Huntington's diseate llustration ereated in htps://pregenygenetion.com/ 1. What members of the family above are affected with the Huntington's disease? 2. Tnere are no carriers ior Huntungton's disease you either have it or you do not. Is Huntington's disease caused-by a dominant or recessive trait? 3. Identify the genotypes of the following individuals using the pedigree above. (homozygous dominant, homozygous recessive, heterozygous). I- 1 II -1: II -3: III - 4 : 4. How many children did individuals I-1 and I-2 have? 5. How many girls did II-1 and II-2 have? How many have Huntington's Disease? 6. How are individuals III-2 and II-4 related? I-2 and III-5?
/d/1n5NtidRwTwUzcDkDPi5Z9P_SHPZ91A-XH-pfftLbhNc/edit (1) O pols Add-ons Help Last edit was seconds ago BIUA ミ: 12 + ext Calibri I|1 6 I 2 Section 5: Trihybrid cross and Laws of probability For a trihybrid cross, in which inheritance of alleles for three genes is tracked, drawing a Punnett square that combines all three genes may not be practical. Instead the laws of probability may be used. The product law of probabilities says that when alleles for separate genes segregate independently, we can figure out the probability of a particular combined genotype by multiplying the probability of the alleles for each gene. 13. We cross a homozygous tall pea plant with yellow, round seeds to a homozygous dwarf pea plant with green, wrinkled seeds. All the F1 offspring are all tall plants with yellow, round seeds. a. What are the expected F2 ratios (use fractions) of tall and dwarf plants? b. What are the expected F2 ratios (use fractions) of yellow and green seeds? C. What are the expected F2…
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The Meeting Sarah stared blankly at the blue paisley wallpaper. Her husband Mike sat by her side, bending and unbending a small paper clip. “Sarah and Michael, it’s good to meet you,” welcomed the genetic counselor, as she entered the room. “I apologize for being late, but I was just meeting with another couple. Let’s see, you’d like to have a child, but you’re concerned because of your family history of cystic fibrosis.” “Yes,” Sarah replied softly. “Mike and I met at a CF support group meeting a few years ago. He had a younger brother who died of cystic fibrosis, and I had a younger sister. We saw the painful lives they had—difficulty breathing, the constant respiratory infections. Although the treatments for CF are better now, we just don’t know if we can…” she trailed off. “I can certainly understand your concern,” the genetic counselor responded sympathetically. “That’s where I hope to help, by providing as much information and advice as I can. I’m glad that you came to see me…