Single nucleotide polymorphisms do not show mendelian segregation through meiosis. Is this true or false? Please explain briefly why and answer the question as quickly as possible.
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- 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.Complete each of the following statements or questions as directed: 1. In the case of inheritance of capsule shape in shepherd's purse, what was the phenotype ratio of the F2 progeny? Use the format #:# or #:#:# or #:#:#:#. 2. What genetic phenomenon is demonstrated by the results of this cross (2 words):152 Phenylketonuria (PKU) is a disorder caused by a recessive allele. Two carrier individuals have progeny. Answer the following questions in order and show solutions whenever relevant. If they have a normal child, what is the probability that he or she will be heterozygous? If they have three children, what is the probability of having 2 affected children and one normal child?
- Shown above is a family pedigree tree in which family members afflictedwith the disease Haemophilia are shown with filled-in squares (male) or circles (females). A couple is trying to determine the likelihood of passingon the disease to their future children (represented by the ? symbolabove) because the hemophilia runs in the woman’s family. Turner syndrome is a disease in which an individual is bornwith only a single X chromosome. Suppose the woman in thecouple is a carrier for hemophilia and has a child with Turnersyndrome. Would this child have the disease?Assume that there is complete dominance and complete penetrance at each locus and that epistasis does not occur. (These are the same conditions we used in class). Write your answers in numeric form (not in words). Referring to the cross: Dd EE GG hh xMxm x Dd ee Gg HH xMY How many different gametes are produced by the individual of genotype DdEEGGhhXMXm? What is the probability that the first offspring from this cross will be genotype DDEeGgHhXMxm? our answer rounded properly to 5 decimal digits. What is the probability that the first offspring from this cross will be a son that shows the dominant phenotype for all loci? (Type in your answer rounded properly to 5 decimal digits.)what is the probability that if the couple has three children, none of them will be affected by the syndrome. refer to the attached picture and show solution. is it 107/144? can you verify my answer. please refer to the picture below.
- From the powerpoint presentation, choose five (5) terms or concepts pertaining to chromosomal abnormality in the number. Describe each in a sentence. https://www.slideshare.net/farhanali911/chromosomal-abnormalities-33461290The following pedigree shows the inheritance of a rare genetic disorder. Determine the most likely mode of inheritance, and for exam practice (unmarked on the assignment) write a clear justification for your conclusion. Remember that some pedigrees don't provide sufficient data/information to discriminate between possibilities. If that is the case, select all the possible answers. I III Tb 2 3 1 1 3 autosomal dominant autosomal recessive Osex-linked dominant Osex-linked recessive O 2 HDraw a Punnett square for the dihybrid cross described below (it is the same story as given for question 8, above) and use it to fill in the blanks correctly in the text that follows. NOTE: Please type in whole numbers, no symbols. There are two known alleles of gene occupying a specific locus in the X chromosome. The gene in question codes for a transcription factor involved in digit development. The mutant allele is dominant and gives rise to an additional but non-functioning little finger (polydactyly) on both hands. A couple have had their DNA sequenced at the region of interest, the male exhibits polydactyly because of the mutation, the female is homozygous wild type at the same locus and therefore has the wild type phenotype. Both have green eyes. In this story; eye colour shows a monogenic autosomal inheritance pattern and the allele for brown eyes shows incomplete dominance with that for blue eyes, the heterozygote phenotype is green eyes. The genes for eye colour and…
- The genotype AABBCcDD will have the following gametestate. edu/d21/le/content/5003190/viewContent/44248878/View 10. The genes Stubble (Sb), Lyra (Ly), and bright (br) are all linked on chromosome 3 in Drosophila fruit flies. An organism that was heterozygous for all three genes was mated to an organism that was homozygous recessive for all three. The following phenotypes were seen in the offspring: All wild-type 422 Lyra and Stubble 4. Stubble and bright 16 Lyrà and bright 75 Lyra only 18 Stubble only 59 Bright only Lyra, Stubble, and bright 404. f. What are the alleles in the parental gametes?Determining a karyotype is an important clinical method for diagnosing genetic disorders. Explain why it is useful for diagnosing monosomies and trisomies.