A man and woman have decided to have children but are worried about possible genetic diseases. The couple visits geneticist to determine what the possible risks are for their future children. The couple discovers that the mom is a carrier, or heterozygous with the diabetes allele and the dad is homozygous dominant, so neither of them have the disease but the mom has one recessive allele. In order to have Diabetes, you have to have two recessive alleles. Th following year, the parents are blessed with a bouncing baby boy! What is the correct genotype of the mom? Aa B AA aa
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- Question:- Ben and Elise are pregnant with their 3rd child. The two prior pregnancies resulted in healthy boys but they are still concerned because Elise's first cousin was just diagnosed with DMD (Duchene muscular dystrophy – x-linked). Ben has three healthy brothers and no aunts or uncles. Elise has two healthy brothers and only the one maternal aunt with the son with DMD. What is the chance that Ben and Elise will have a child with DMD? a.1/20 b.1/140 c.1/209Please answer with reason: 1) The nurse is explaining the inheritance of Huntington disease (autosomal dominant) to a newly diagnosed patient whose partner does not have the gene mutation. Which statement should the nurse make regarding family planning? a. There is a 100% chance with each pregnancy that the child will inherit the gene for Huntington disease. b. There is a 50% chance with each pregnancy that the child will inherit the gene for Huntington disease c. There is a 25% chance with each pregnancy that the child will inherit the gene for Huntington disease d. There is a 0% chance with each pregnancy that the child will inherit the gene for Huntington disease.mentum.com/courseware-delivery/ua/49006400/724263801/aHR0cHM6Ly9mMi5hcH. Savvas EasyBridge Google Slides Infinite Campus M Inbox-francis.a@s... mosomal Inheritance: Tutorial What offspring will result from the mating of a non-carrier female and a ? hemophilic male? Half of the sons are hemophillacs and half are not. All of the daughter carriers. None of the sons are hemophiliacs, and all of the daughters are carrie None of the daughters are hemophiliacs, and all of the sons are carrier O Half of the daughters are hemophiliacs and half of the daughters are c None of the sons are hemophiliacs.
- XPY XDXD XdY Carrier? (Yes or No) Has the Trait? (Yes or No) Male or Female Phenotype or Appearance of the Individual Guided Questions: 1. What chromosomes will result in a normal phenotype (normal spongy pore trait)? 2. Do all individuals who have the spongy tiny pore allele show it in their physical appearance? Why or why not?Mrs. Xexy Lucero, G1P1, postpartum mother brought her child Adriana to a Neonatologist for a consultation. Physical examination were done. Diagnostic tests were conducted. Karyotyping revealed Turner's syndrome. Which of the following responses are appropriate for Nurse Zasha to make? (Select all that apply). a. It is a result of deletion of chromosome X b. lt is a result of nondisjunction chromosomal aberration c. lt contains 47 chromosomes d. It contains 45 chromosomes e. The child with the defect can have children through in vitro fertilization.Mr. Steve W. genetics question Albinism is a recessive autosomal trait for skin pigmentation. Hemophilia is a sex-linked recessive disorder of the blood. assign alleles to the traits A – normal skin pigmentation XH – normal blood a – albino Xh - hemophilia A double heterozygous woman marries a non-hemophilic man and is heterozygous for skin pigmentation. Double heterozygous means heterozygous for both traits. Aa for skin pigmentation and XHXh for blood traits. Therefore, the genotype of the woman is AaXHXh. Non-hemophilic man is XHY and heterozygous for skin pigmentation is Aa. The genotype, therefore, of the man is AaXHY. What is the probability that they will have: a child with normal skin? _____________________ a child with normal blood? _____________________ an albino girl? _____________________ A hemophilic boy? _____________________
- SUBJECT: GENETICS Topic: Polygenic Inheritance Question: Discuss possible presentation why a given family with 12 children, has 7 variations/differences in terms of height (contributing alleles like AABBCCDDEEFF, non-contributing alleles, aabbccddeeff)Question:- 8. Describe how Morgan’s experiment with the white-eyed mutant fly provided evidence for the chromosomal basis for inheritance. Be sure to include the significance of the x-linked gene./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…
- Directions: Analyze the given situation and answer the questions. Hemophilia is a disease caused by a gene found on the X chromosome. Therefore, it is referred to as a sex-linked disease. The recessive allele causes the disease. A man with hemophilia (XnY) marries a woman who is homozygous dominant (XHXH). A. Illustrate using a Punnett square the probability that their children will have the disease. B. Will any of their children have the disease? C. Predict the probabilities of their children having the disease.Kindly review my answers on the genotypes part. Please give me the correct answers if incorrect. Thank you.Question For Research 1. The M, N, and MN blood groups are determined by two alleles, IM and IN. The ABO blood types are determined by three alleles, I^, I®, and Iº. In a court case concerning a paternity dispute, each of two men claimed three children to be his own. The blood groups of the men, the children and their mother were as follows: Husband M 8 | Page The wife's lover АВ MN Wife A N Child 1 MN Child 2 A N Child 3 A MN From this evidence, can the paternity of the children be established? Show expected crosses.