m having trouble with my genetics study guide, and am stuck on this question. If someone could explain it with the work it would help me so much. Thank you! In humans, the inheritance of the ABO blood
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I'm having trouble with my genetics study guide, and am stuck on this question. If someone could explain it with the work it would help me so much. Thank you!
In humans, the inheritance of the ABO blood group system exhibits both complete dominance (alleles IA and IB are dominant to allele i) and codominance (alleles IA and IB). In addition, thalassemia shows incomplete dominance; heterozygotes (Tt) exhibit a mild form of the disease (thalassemia minor) and homozygotes (tt) have a more severe form (thalassemia major).
A man has blood type A and his mother has blood type O. His wife has blood type AB. Both members of the couple have thalassemia minor. What is the probability that they will have a child with:
a. thalassemia minor and blood type A?
b. no anemia and blood type AB?
c. thalassemia major and blood type B?
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- 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?Hi, I'm having trouble with my study guide for my upcoming genetics exam. If someone could please help with work shown and an explanation it would help so much! Thank you!! 2a. The pedigree below represents inheritance of rare condition. What pattern of inheritance is most consistent with the data? Assign alleles to all individuals to support your answer. If an allele is unknown, assign it a ? symbol. NOTE: Individuals whose phenotype or genotype cannot be determined are assumed to be unaffected and homozygous, unless otherwise indicated. 2b. In addition to the alleles you’ve indicated, describe 2 overall features of the pedigree that make it consistent with your chosen form of inheritance. 2c. Based on your mode of inheritance, what is the probability that the child of couple IV-4 x IV-5 will be affected? Show your work. attached is the pedigree
- I need some help with this fill-in-the-blank problem. The answer choices are bolded and bracketed. Please see the attached photo to complete. 1. In the D2S441 locus, Sophie's allele [10, 12] is maternal and her allele 10 is paternal from [Sam or Bill only, Sam or Harry only, Bill or Harry only, Sam Bill or Harry (either 3)]. 2. Sophie's allele 13 for D19S433 is [maternal, paternal] and her allele 14 could have come from [Sam and Bill only, Sam and Harry only, Bill and Harry only, Sam Bill or Harry (either 3)]. 3. In the FGA locus, Sophie's allele [21, 22] is maternal and her other allele could have come only from [Sam, Bill, Harry]. 4. Based on all STRs in the 3 panels we studied, it is clear that [Sam, Bill, Harry] is Sophie's father. He has one allele of Sophie's alleles for all STR loci 5. Would these results stand in court as proof paternity: [Yes or No] 6. This type of DNA profiling can also be used to determine maternity. Is there any doubt that Donna is Sophie's biological…I just want an explanation of what is in bold please. Lab Introduction: A dihybrid cross is a cross between individuals that involves two pairs of contrasting traits. To Predict the results of a dihybrid, cross all possible combinations of the four alleles from each parent must be considered. You will examine a dihybrid cross involving both color and texture. Purple (P), is dominate to yellow (p), and smooth texture (S) is dominant to wrinkled (s). Both parent plants are heterozygous for both traits. Review genetics and the use of Punnett squares in a biology text before doing this experiment.MATERIALS: Assume you have ear of Corn. You need a heterozygous X heterozygous 9:3:3:1, purple/yellow, starchy/sweet. PROCEDURE: From above please write out: The crop The parental (P) cross phenotype, genotype, gametes The F1 progeny Genotype and Phenotype Cross between two F1 Selfed testcross The F1 gametes The expected F2 results, genotype, phenotype, genotypic ratio, phenotypic ratio. 1.…A couple enters your genetic counseling clinic for some family planning advice. The woman’s father was color blind, but her own vision is normal. The man has no family history of color blindness. Neither the man nor woman have any known history of hemophilia, but their first child (a boy) has hemophilia. They ask you to calculate the chance that their nextchild will be affected by one or both conditions. You remember from your genetics training that these are both X-linked recessive conditions and that they are closely linked: in fact, their genetic loci are separated by only 10cM! During the interview with this couple, you draw the following pedigree to represent their information. Given what you know, determine for this couple what chance they have of each of the following (in the table).
- Alberta is phenotypically normal, but her brother (Rodrigo) has albinism, which is caused by an autosomal recessive mutation. The probability that Alberta is a carrier (i.e., heterozygous for albinism) is [express your answer as a fraction]My Question is what is the probability their first child will have hemophilia and drawn pedigrees for family members with genotypes. My explantion so far: A man has both X and Y chromosomes as sex chromosomes in his body. Here, though the brother of the man is hemophiliac, a man can’t be a carrier of hemophilia. So, it can be said that his chromosome is “XnY”.Here, the “n” stands for “normal”.Though the paternal uncle is hemophiliac, a man cannot be a carrier of hemophilia, his niece will not be a career. So it can be said that the woman is also not a carrier and has the “XnX” chromosome.So, as the mother is not a carrier, their first child does not have a chance of having hemophilia. This can be determined as it is known that there is no hidden carrier of hemophilia in the family.I am having trouble answering these problem sets about sex-linked inheritance. Background: Hemophilia is a recessive X linked genetic disorder (refer to BIOL 40 B), a medical condition in which theability of the blood to clot is severely reduced, causing the sufferer to bleed severely from even a slightinjury. The condition is typically caused by a hereditary lack of a coagulation factor, most often factor VIII orfactor IX (refer to Bio 40 B Textbook chapter 19 on Blood). Based on a recent study that used data collectedon patients receiving care in federally funded hemophilia treatment centers during the period 2012-2018, asmany as 33,000 males in the United States are living with the disorder.Males inherit the X chromosome from their mothers and the Y chromosome from their fathers. Femalesinherit one X chromosome from each parent. Thus, males can have a disease like hemophilia if they inheritan affected X chromosome that has a mutation in either the factor VIII or factor IX gene.…
- Please answer all of them, they are all connected. PEDIGREE ANALYSIS and SYMBOLOGY: Examine the pedigree which has X linked Dominant inheritance of disorder. Use letter X* (asterisk denotes disorder) as genotype of the individuals which can be XX, XY, X*X*, X*X and X*Y. a. What is the genotype of IV-6? b. What is the genotype of III-6? c. What is the genotype of II-3? d. What is the genotype of III-8? e. If couple I-1 and I-2 will have a son, what is the probability of having the disorder? f. If couple III-8 and III-9 will have another child, what is the probability of having the disorder? g. Theoretically, if individual IV-3 and individual IV-5 will marry and will have a child, what is the probability of having a child without the X-linked disorder?You are a genetic counselor. How would you advise a couple who wants to know their chances of having a baby with Duchenne-muscular dystrophy. Hannah is not a carrier, but her husband Barry has a brother, James, who died from the disease. Duchenne-muscular dystrophy is an x-linked condition, the allele responsible is recessive and people who have the disease die in childhood. a)What are Hannah’s and Barry’s chances of having a baby with Duchenne-muscular dystrophy? b)What are the chances of Barry’s sister having a baby with the disease?Hi, I'm having some trouble with this practice problem from my study guide. If anyone can explain it it would be very helpful! Brindle coloration is a black and brown striping pattern in some dogs. This fur coloration is controlled by different alleles at a single autosomal locus. There are three alleles, KB, kbr and ky. The KB allele is dominant over the other two alleles and produces a solid black color. The allele kbr produces the brindle color pattern and is dominant over the ky allele, which produces yellow fur. Give the genotypes of the parents and offspring in each cross. P: black X yellow; F1: ½ brindle and ½ black P: black X brindle; F1: ½ black, ¼ brindle and ¼ yellow P: brindle X yellow; F1: 100% brindle