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Good Day! Can u help me answer number 5-7 ?thank youu
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- Dihybrid crosses I only need the phenotypic ratio. Secretors (genotypes SS and Ss) secrete their A and B blood group antigens into their saliva and other body fluids, while “non-secretors” (genotype ss) do not. An ABSs woman and an OOSs man have several kids. How many of the kids could be secretors? _______________________________________ How many of the kids could be non-secretors? _________________________________ Of the kids who could be secretors – what is the phenotypic ratio in terms of their blood types? ______________________________Write your ratio like this – e.g. 1AB:1A Of the kids who could be non- secretors – what is the phenotypic ratio in terms of their blood types? ____________________________Write your ratio like this – e.g. 1AB:1A i need help finding the answer please explain and show a punnet sqaure please to explain the answer pleaseInterspecification crosses are rare in nature and intergenic crosses almost unknown.why?
- I don't really understand how to solve this question. For part a, I think my approach would be to find the genotypes of the parents first, and find the genotype of the F1. For part B, I would like to get some clarifications on the difference between linked genes with no crossing over vs. unlinked genes when approaching Punnett square type questions. Thank you also answer part a and bThe images attached show a parental cross that is homozygous wild female x white male. F1 were intercrossed to produce the F2 generation as indicated below: Wild Female: 416 Wild Male: 192 White Female: 0 White Male: 192 A chi-square analysis was done and an image was also attached. Please discus what conclusions can be made based on the data/findings.A single allele gives rise to the Hbs form of hemoglobin. Individuals who are homozygous for the allele (HbS/HbS) develop sickle-cell anemia (Section 9.6). Heterozygous individuals (HbA/HbS) have few symptoms. A couple who are both heterozygous for the HbS allele plan to have children. For each of the pregnancies, state the probability that they will have a child who is: a. homozygous for the HbS allele b. homozygous for the normal allele (HbA) c. heterozygous: HbA/HbS
- Drosohpila Punnet Square of Crosses. I need results of F1 & F2 generation using Punnett Squares for: Make Punnet Squares of the following crosses •Drosophila Female wildtype cross Male White-eye •Drosophila Male wildtype cross Female White-eye •Drosophila Female Wild Type cross Male Scarlet Eye •Drosophila Male Wild Type cross Female Scarlet Eye Also, Which allele is heterozygous and which is homozygous, & which is dominant and which is recessive?What is the diploid genotype after mating these two strains? Parent 1: MATα msh6Δ::kanMX leu2-3,112 ura3-1 trp1-1 his3-11,15 + pRS415 (msh6- K337T and LEU2 gene) Parent 2: Mata his3∆ leu2∆ lys2∆ ura3∆ TRP1 MSH6In com, male sterility is controlled by maternal cytoplasmic elements. However, the presence ofa nuclear fertility restorer gene (F_) restores fertility to male sterile lines. Draw your simulated crosses male sterile female x FF male and give the genotypes and phenotypes of the offspring in each cross.
- 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!! 1a. The pedigree below represents inheritance of rare condition (filled symbols used for affected individuals). Test the hypothesis of X-linked dominant inheritance by assigning alleles (A or a) to sex chromosomes of all individuals in generations I and II. Does the X-linked dominant hypothesis agree with the data? It not, indicate all at least 2 individuals by generation and number (e.g. II-8) that are not consistent with the genotype you’ve proposed for the individuals in generation I. 1b. Test the hypothesis of autosomal dominant inheritance by assigning alleles (A or a) to autosomes of all individuals in the pedigree (generations I – IV). Does the autosomal dominant hypothesis agree with the data? It not, indicate all individuals by generation and number (e.g. II-8) that are not consistent with the genotype…4 offspring, 3 normal and 1 (with disease) What is the probability for this if the father (has the disease) and mother (does not carry disease). BOTH are heterozygous3 offspring in the following order: normal, with disease, normal What would be the probability for this if the father (with the disease) and mother (doesn't carry the disease) had offspring. Both are heterozygous