B30m6l06assignment

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School

Archbishop MacDonald**We aren't endorsed by this school

Course

BIOLOGY 30

Subject

Biology

Date

Jan 2, 2025

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Uploaded by ColonelFreedomPelican35

Biology 30: Module 6: Lesson 61Assignment BookletMODULE6: LESSON6 ASSIGNMENTThis Module 6: Lesson 6 Assignment is worth 12 marks. The value of each assignment and each question is stated in the left margin.(12 marks)Lesson 6 Assignment—Chromosomal Theory and Sex-Linked Inheritance(6 marks)1.Perform an internet search on sex-linked inherited conditions. Verify that these genes would be located on the X-chromosome. Create a sex-linked cross. Include the alleles and the solution to the Punnett square. Describe the phenotypes and genotypes of the parents and the resulting outcome of the offspring.Grading:●1 mark for trait on the X-chromosome●1 mark for description of the alleles●1 mark for the correct genotype of each parent●1 mark for the Punnett square (with each parent listed correctly) ●1 mark for correct genotype of offspring●0.5 marks for correct ratio of genotypes of offspring●0.5 marks for correct ratio of phenotype of offspringAnswer:Color blindness is a common example of a sex-linked inherited condition that is located on the X chromosome. It is caused by a recessive allele on the X chromosome.Alleles:●Xᴄ = X chromosome with the allele for color blindness (recessive)●Xᴺ = X chromosome with the normal vision allele (dominant)●Y = Y chromosome (does not carry the color blindness allele)Parents:●Mother (carrier): XᴺXᴄ (She has one normal vision allele and one color blindness allele, but since color blindness is recessive, she does not express it.)●Father (affected): XᴄY (He has the color blindness allele on his X chromosome and will express color blindness because males only have one X chromosome.)Punnett Square:

Biology 30: Module 6: Lesson 62Assignment BookletXᴺ (Mother)Xᴄ (Mother)Xᴄ (Father)XᴺXᴄ (Female, carrier)XᴄXᴄ (Female, color blind)Y (Father)XᴺY (Male, normal vision)XᴄY (Male, color blind)Genotype ratio:●1 XᴺXᴄ : 1 XᴄXᴄ : 1 XᴺY : 1 XᴄYPhenotype ratio:●1 female carrier with normal vision : 1 female color blind : 1 male normal vision : 1 male color blind(2 marks)2.Hemophilia, a blood disorder in humans, results from a sex-linked recessive allele. Suppose the daughter of a mother without the allele and a father with the allele marries a man with hemophilia. What is the probability that the daughter's children will develop the disease? Describe how you determined the probability. Answer:The probability that the daughter’s children will develop hemophilia is 50%. Figure out the daughter genotype by crossing her parents genotypes, then do the same with the daughters genotype and her husbands to figure out the probability. Both are found by using a punnett square. (2 marks)3.Colour-blindness results from a sex-linked recessive allele. Determine the genotypes of the offspring that would result from a cross between a colour-blind male and a homozygous female who has normal vision. Describe how you determined the genotypes of the offspring.Answer:The genotype for a female offspring would be XX normal(carrier), and the genotype for a male offspring would be XY colourblind. This could be determined through a punnett square and crossing the genotypes of both parents. (2 marks)4.Explain why X-linked traits appear more often in males than in females.Answer:This happens because males have XY sex chromosomes and women have XX chromosomes. Since men only have one X chromosomes, they express X linked traits with just one recessive allele, whereas females need two recessive alleles to express the trait, making their chances lower. Once you have completed all of the questions, submit your work to dropbox C7.

Biology 30: Module 6: Lesson 63Assignment Booklet