Question

1. sickle-cell anemia is an autosomal recessive genetic disorder that causes red blood cells to change shape. this can cause the red blood cells to become stuck in blood vessels. this blocking can deprive tissues of oxygen and cause organ damage like strokes. one benefit of this is that people who have one or two alleles of the sickle cell disease are resistant to malaria since the red blood cells are not conducive to the parasites. sickle-cell anemia also exhibits incomplete dominance at which the individual who does not have the disease retains immunity to malaria. fill in the punnett square and determine the expected genotypic ratios from crossing homozygous recessive and homozygous dominant parents.

genotypes: ____________________ genotypic ratio: _____
circle all phenotype(s): normal, carrier, has sickle-cell anemia
% of kids with disorder: ___ % carriers of the disorder: ___

1. hemophilia a is a recessive sex-linked genetic disorder that prevents the blood from clotting. the effects of this x chromosome disorder develops almost entirely in males even though the gene is inherited from one of the mother’s x chromosome. fill in the punnett square and determine the expected genotypes and phenotypes from crossing a male with hemophilia a and a female who is a carrier for hemophilia.

genotypes: ______________________
circle all phenotype(s): normal male, male with hemophilia a, normal female, carrier female, female with hemophilia a

1. red-green colorblindness is a recessive sex-linked (x chromosome) genetic disorder where the middle (green) or long (red-yellow) wavelength cones in the eyes have a partial or complete loss of function. fill in the punnett square and determine the expected genotypes and phenotypes from crossing a normal male and a female who is a carrier of color-blindness.

genotypes: ______________________
circle all phenotype(s): normal male, male with colorblindness, normal female, carrier female, female with colorblindness

1. pku is a recessive autosomal genetic disorder that increases the amount of a type of amino acid in the blood. fill in the punnett square and determine the expected genotypes and phenotypes from crossing a heterozygous and a heterozygous.

genotypes: ____________________ genotypic ratio: _____
circle all phenotype(s): normal, carrier, has pku
% of kids with disorder: ___ % carriers of the disorder: ___

Answer:

Problem 4: Sickle - cell anemia (autosomal recessive)

Step 1: Determine parental genotypes

Let the dominant allele be \(A\) and the recessive allele be \(a\). The homozygous dominant parent has the genotype \(AA\), and the homozygous recessive parent has the genotype \(aa\).

Step 2: Fill the Punnett square

When we cross \(AA\) (parent 1) and \(aa\) (parent 2), each gamete from parent 1 is \(A\), and each gamete from parent 2 is \(a\). So the Punnett square will have all cells with the genotype \(Aa\).

	\(A\)	\(A\)
\(a\)	\(Aa\)	\(Aa\)

Step 3: Determine genotypes and genotypic ratio

The genotypes of the offspring are all \(Aa\). The genotypic ratio is \(4:0:0\) (or simply all \(Aa\), so the ratio of \(AA:Aa:aa = 0:4:0\)).

Step 4: Determine phenotypes

The phenotype of \(Aa\) individuals (in the context of sickle - cell anemia with incomplete dominance) is a carrier (and also has malaria resistance). Since there are no \(aa\) genotypes, the percentage of kids with the disorder (sickle - cell anemia, which requires \(aa\)) is \(0\%\). The percentage of carriers (genotype \(Aa\)) is \(100\%\) (because all offspring are \(Aa\)). The phenotypes to circle are normal (no, because \(Aa\) is a carrier, but in terms of sickle - cell anemia, the non - diseased (no \(aa\)) and carrier. Wait, the problem says "the individual who does not have the disease retains immunity to malaria". The \(Aa\) individuals are carriers (and have malaria immunity), and do not have sickle - cell anemia (since sickle - cell anemia is \(aa\)). So the phenotypes: normal (no, \(Aa\) is a carrier), carrier (yes), has sickle - cell anemia (no). So we circle carrier.

Problem 5: Hemophilia A (sex - linked recessive)

The Punnett square is already filled. The genotypes are \(X^{H}X^{h}\), \(X^{h}X^{h}\), \(X^{H}Y\), \(X^{h}Y\). Let's verify:

• The male with hemophilia has genotype \(X^{h}Y\), so his gametes are \(X^{h}\) and \(Y\). The female carrier has genotype \(X^{H}X^{h}\), so her gametes are \(X^{H}\) and \(X^{h}\).
• When we fill the Punnett square:
• Top - left cell: \(X^{H}\) (from female) and \(X^{h}\) (from male) → \(X^{H}X^{h}\)
• Top - right cell: \(X^{h}\) (from female) and \(X^{h}\) (from male) → \(X^{h}X^{h}\)
• Bottom - left cell: \(X^{H}\) (from female) and \(Y\) (from male) → \(X^{H}Y\)
• Bottom - right cell: \(X^{h}\) (from female) and \(Y\) (from male) → \(X^{h}Y\)

The phenotypes:

• \(X^{H}X^{h}\): carrier female
• \(X^{h}X^{h}\): female with hemophilia A
• \(X^{H}Y\): normal male
• \(X^{h}Y\): male with hemophilia A

So the circled phenotypes should be: normal male, male with hemophilia A, carrier female, female with hemophilia A (wait, the original circling was a bit off. \(X^{H}X^{h}\) is a carrier female, \(X^{h}X^{h}\) is female with hemophilia, \(X^{H}Y\) is normal male, \(X^{h}Y\) is male with hemophilia. So all of these phenotypes are present.

Problem 6: Red - Green colorblindness (sex - linked recessive)

Step 1: Determine parental genotypes

Let the dominant allele (normal vision) be \(X^{B}\) and the recessive allele (colorblindness) be \(X^{b}\). A normal male has genotype \(X^{B}Y\), and a carrier female has genotype \(X^{B}X^{b}\).

Step 2: Determine gametes

The male (\(X^{B}Y\)) produces gametes \(X^{B}\) and \(Y\). The female (\(X^{B}X^{b}\)) produces gametes \(X^{B}\) and \(X^{b}\).

Step 3: Fill the Punnett square

	\(X^{B}\)	\(Y\)
\(X^{b}\)	\(X^{B}X^{b}\)	\(X^{b}Y\)

Step 4: Determine genotypes and phenotypes

• Genotypes: \(X^{B}X^{B}\), \(X^{B}X^{b}\), \(X^{B}Y\), \(X^{b}Y\)
• Phenotypes:
• \(X^{B}X^{B}\): normal female
• \(X^{B}X^{b}\): carrier female
• \(X^{B}Y\): normal male
• \(X^{b}Y\): male with colorblindness

So the circled phenotypes should be normal male, male with colorblindness, normal female, carrier female (we do not circle female with colorblindness because there are no \(X^{b}X^{b}\) genotypes here).

Problem 7: PKU (autosomal recessive)

Step 1: Determine parental genotypes

Let the dominant allele (normal) be \(P\) and the recessive allele (PKU) be \(p\). Both parents are heterozygous, so their genotype is \(Pp\).

Step 2: Determine gametes

Each parent produces gametes \(P\) and \(p\).

Step 3: Fill the Punnett square

	\(P\)	\(p\)
\(p\)	\(Pp\)	\(pp\)

Step 4: Determine genotypes and genotypic ratio

The genotypes are \(PP\), \(Pp\), \(Pp\), \(pp\). The genotypic ratio is \(PP:Pp:pp=1:2:1\).

Step 5: Determine phenotypes and percentages

• Phenotypes:
• \(PP\): normal
• \(Pp\): carrier
• \(pp\): has PKU
• Percentage of kids with disorder (\(pp\)): \(\frac{1}{4}\times100 = 25\%\)
• Percentage of carriers (\(Pp\)): \(\frac{2}{4}\times100 = 50\%\)

Final Answers (for each problem)

Problem 4

• Genotypes: \(Aa\) (all offspring)
• Genotypic Ratio: \(AA:Aa:aa = 0:4:0\) (or \(0:1:0\) when simplified per type, but since all are \(Aa\), the ratio is \(4\) (Aa) : \(0\) (AA) : \(0\) (aa))
• Circled phenotype: carrier
• % of kids with disorder: \(0\%\)
• % carriers of the disorder: \(100\%\)

Problem 5

• Genotypes: \(X^{H}X^{h}\), \(X^{h}X^{h}\), \(X^{H}Y\), \(X^{h}Y\)
• Circled phenotypes: normal male, male with hemophilia A, carrier female, female with hemophilia A

Problem 6

• Genotypes: \(X^{B}X^{B}\), \(X^{B}X^{b}\), \(X^{B}Y\), \(X^{b}Y\)
• Circled phenotypes: normal male, male with colorblindness, normal female, carrier female

Problem 7

• Genotypes: \(PP\), \(Pp\), \(Pp\), \(pp\)
• Genotypic Ratio: \(PP:Pp:pp = 1:2:1\)
• Circled phenotypes: normal, carrier, has PKU
• % of kids with disorder: \(25\%\)
• % carriers of the disorder: \(50\%\)