Question

please answer these papers, they are related to each other. thank you!

Student name and UNID Class 18 Mendelian Genetics Class 18 Mandalas The gene involved in the disease Sickle Cell Anemia (SCA) is on human chromosome 11. Allele "A" is the normal form of the gene and codes for a part of the protein complex called hemoglobin. Hemoglobin is required for your blood cells to carry oxygen. Allele "a is an abnormal form of the gene. The hemoglobin protein made from the "a" allele is defective. Red blood cells containing the defective protein are very fragile. This disease is recessive meaning that you have to inherit two copies of the recessive allele in order to be affected with the disease SCA. The gene involved in the disease Cystic Fibrosis (CF) is on human chromosome 7. Allole "B" is the normal form of the gene and codes for a channel protein that allows chloride to pass through a cell. Allele "b" is the abnormal form of the gene. The channel protein made from the 'b' allele is defective and doesn't insert into the cell's plasma membrane, resulting in an inability of sodium and chloride to cross the membrane. CF is also recessive therefore only people with two defective copies of the gene ("b") are affected with the disease. 1. Chris and Emma are phenotypically normal but carriers for SCA (gene A). a. What are their genotypes? b. With respect to gene A, what are the possible genotypes of the gametes that they can produce. Chris: 2 Emma: Dobro In front of each gamete genotype above write probability of producing that gamete. c. What is the probability that Chris and Emma will have a child affected with SCA? (affected=has the disease="aa homozygous recessive) t omonidan d. Write the cross and the Punnett square that demonstrates this. o X e. The phenotypic ratio is The genotypic ratio is normal : AA: affected Aa: _aa 2. Chris and Emma are also carriers for CF (gene B is on chromosome 7). Both are genotype Bb. a. What is the probability that they will have a child affected with CF? (affected = has the disease = "bb" shomozygous recessive) b. Write the cross and the Punnett square that demonstrates this.

Answer 1

1a. Chris and Emma are phenotypically normal but are carriers. Hence their genotypes are composed of one copy of normal allele and one copy of abnormal allele: Aa and Aa

1b. Possible gametic genotypes are:

Chris - A and a Emma- A and a and the probabililty of producing the gametes is 50% each for both

1c. If the gametes of Chris and Emma are crossed, the probable offspring phenotypes are:

AA, Aa, Aa and aa . The probability of producing a child with SCA is 1/4

1d.

1e. The phenotypic ratio : 3(normal) : 1(affected)

The genotypic ratio : 1 (AA) : 2(Aa) : 1(aa)

2a. If we cross carriers(Bb) Chris and Emma for cystic fibrosis, the genotyoes of offsprings will be:

BB, Bb, Bb and bb . So probability of having an affected child is 1/4.

2b.

3a. Genotypes of possible gametes: AB, Ab, aB and ab for each Chris and Emma

a.b Aь Aь "АВ о. В A BB Ao Bь АВ AABB AABB abb A Abb AaBb AЬ AABB aaBb aa BB ав AaBB AaBb aabb. AaBb | at aaBb Aabb Habb с 3dy Probability of each oft sprung AABB = I AaBB = 2 ААВь. 2 да въя AAbb= + A Abb= aasis аав aaBb = 2 2 2 Aabb = Aabba 2 садь - 2 Bel Probability of each phenotypes ABA Ab = 4 ов: 114 aь - 4

4. Combined probabilities-

AB: 1/4*1/4 = 1/16

Ab: 2/4*2/4 = 1/4

ab: 1/4*1/4 = 1/16

aB: 1/4*1/4 = 1/16

4a. The inheritance of the A or a alleles are not dependant on the inheritence of the alleles B or b. This is the Mendel's law of Independent assortment. The two different alleles of two different genes get assorted independantly and do not influence each other's expression.

4b. Chance that their child will be phenotyically normal is : 12/16

4c.Genotype aabb : 1/16

4d. Boy with AbBb : 4/16