6. In So laut, a gene for tal plants( S) ?s dominant over its allele for short p

Question

6. In So laut, a gene for tal plants( S) ?s dominant over its allele for short plants (s). The gene for rod fruits·7) is dominant over its allele for yellow fruits (). The genes are not linked. Calculate both phenotypic and genotypic ratios for the results of each of the following crosses: (4 marks) 7. In the following human pedigree, assume that the individual marked with an asterisk) does not carry any allele associated with the affected phenotype and that no other mutation spontancously occurs. Use "R" or X* for the allele associated with the dominant phenotype or "Y" or"x" for the aBele associated with the recessive phenotype 4 marks) What is the most likely mode of inheritance of this disease b. List all the possble genotypes of following individuals in the pedigree #1, #3 What is the probability of individual A being affected? d. What is the probability of individual B being affected? 8. In a plant having three alleles for the locus of leaf shade, the following relationship exists Green (G), Yelow(Y) and G is dominant to Y and Y is dominant to B. Obsve the following rasses and answer the questions Brown (B), w with proper justification: a. Cross between GB and YB what fraction of the progeny will be brown? b. Cross between GY and YB what fraction of the progeny will be yellow? c. Cross between GG and YB what fraction of the pro geny will be Brown? d. Cross between BB and YB, what fraction of the (2 marks) will be yellow? 9. In the given pedigree chart, the individuals from outside the family are all homozygous for the given trait. Using this information, answer the following inheritance? a. What is the most probable mode of Justify. (1 mark) b. If the individual 10and 1I mated, what would be the probability of the offspring being affected? Explain I mark) c. What is the genotype ofthe individuals #2.4. #8 and 10? Mention all the possibilides on the basis of the inheritance pattem.(1 mark each)

Explanation / Answer

9. a. This is the case of X linked recessive mode of inhetritence. Because it can be seen that when both mother is diseased then it passes the allele to daughter who is just carrier but do not shows symptoms and when both parents are diseased then only all the offsprings are diseased.

9. b. If the individual 10 and 11 will be mated:

11 is normal who could be carrier of the disease while the male is diseased

So the probability of offspring being diseased = ½*1/2 = ¼

c. Let say R denotes the dominant allele while r is ressivem,

Then genotype of individual 2 = XRY, which do not express recessive and hence do not show disease symptoms.

Individual 2 is XrXr

individual 8 is XRY which do not express recessive and hence do not show disease symptoms.

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