Draw two replicated, condensed, homologous chromosomes that have the genes E and F on them. This individual is heterozygous for gene E, and homozygous recessive for gene F. Be sure to label your chromosomes for full credit.
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Draw two replicated, condensed, homologous chromosomes that have the genes E and F on them. This individual is heterozygous for gene E, and homozygous recessive for gene F. Be sure to label your chromosomes for full credit.
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- Draw two replicated, un-condensed, homologous chromosomes that have the genes A and E on them. This individual is homozygous recessive for A, and heterozygous for gene E. Be sure to label your chromosomesMy 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.A woman who sought genetic counseling is found to be heterozygousfor a chromosomal rearrangement between the second andthird chromosomes. Her chromosomes, compared to those in anormal karyotype, are diagrammed on the next page:(a) What kind of chromosomal aberration is shown?
- tate. 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?Both hemophilia (h) and favism (gd) are inherited as X-linked recessive traits. Hemophilia is an inherited disorder of blood clotting, and favism is an inherited hemolytic anemia caused by absence of the enzyme glucose-6-phosphate dehydrogenase. A phenotypically normal woman is known to have the X chromosome genotype h + / + gd. The frequency of recombination between h and gd is 16%. What proportion of sons born to this woman are expected to be phenotypically normal with respect to both hemophilia and favism?A tomato geneticist attempts to assign five recessivemutations to specific chromosomes by using trisomics.She crosses each homozygous mutant (2n) with each ofthree trisomics, in which chromosomes 1, 7, and 10 takepart. From these crosses, the geneticist selects trisomicprogeny (which are less vigorous) and backcrosses themto the appropriate homozygous recessive. The diploidprogeny from these crosses are examined. Her results, inwhich the ratios are wild type:mutant, are as follows:Which of the mutations can the geneticist assign towhich chromosomes? (Explain your answer fully.)
- Pedigree attached shows an autosomal recessive genetic disease. G is the normal allele and g is the disease-causing allele. Individual 1’s father is heterozygous (*) and his mother is homozygous dominant. Other individuals in the pedigree may be carriers, but are not marked. The question mark (?) indicates that you do not yet know anything about this individual’s phenotype with regard to the disease. part a) What is the probability that individuals 1 and 2 will have a child (5) who is a boy with the disease (the child is unborn and the sex is not yet known)? a)1/8 b)1/4 c)0 d)1/16 part b) What is the probability that the daughter (6) that individual 3 and 4 just had will have the disease? a)1/8 b)1/6 c)1/4 d)1/12The genes F and G are on the same chromosome in a eukaryote. Using a microscope, you can see that a chiasma occurs between these two loci in 24% of the meioses. A double heterozygote could have genotype FG//fg, where the // represents the pair of homologous chromosomes that contain the F and G loci: one homolog contains F and G alleles and the other contains f and g. You cross this FG//fg individual to an fg//fg individual and examine their offspring. What proportion of the offspring do you expect to be Fg//fg? Group of answer choices 48% 6% None of these 12% 24%Consider the following two nonhomologous wildtype chromosomes, where letters or numbers represent genes, the "-" represents the centromere of each chromosome, and chromosomes are shown on separate lines. ABCDE-FGHIJK 123-45678 Identify the type of rearrangement shown in each of the following (A-C) and then identify whether it is balanced or unbalanced. Assume that the individual is diploid and heterozygous for the rearrangement. A. ABCDE-FGHIJKGH 123-45678 Rearrangement: [Select] • Balanced or Unbalanced: [Select] B. ABCDGF-EHIJK 123-45678 Rearrangement: [Select] Balanced or Unbalanced: [Select]
- Is this pedigree autosomal dominant, autosomal recessive, or X-linked recressive? Can you please label each square or circle with the correct genotypes?How many total chromosomes were on your karyotype sheets? How many of the chromosomes are autosomes? Is this child a male or a female? How do you know? Does this child appear to have a genetic disorder? If so, what genetic disorder does this child have? ( DID I DO THIS CORRECTLY I THINK ITS ) 1. 47 CHROMOSOMES 2. 44 ARE AUTOSOMES 3. MALE HAS A Y CHROMOSOME 4. YES, HAS EXTRA Y CHROMOSOME AND THE GENETIC DISORDER IS XYY SYNDROMEShown 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?