In humans, haemophilia is a sex-linked recessive trait. What are the possible genotypes and phenotypes of offspring from a haemophiliac female and a normal male?

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Asked on April 1, 2026Biology

This biology question covers important biological concepts and processes. The step-by-step explanation below helps you understand the underlying mechanisms and reasoning.

Accepted Answer · 2026-04-01T05:05:07.199Z

Here are the solutions to the questions: 3.3 Step 1: Determine the genotypes of the parents. A haemophiliac female has the genotype X^hX^h. A normal male has the genotype X^HY. Step 2: Perform a genetic cross. Parents: X^hX^h × X^HY Gametes from female: X^h Gametes from male: X^H, Y Punnett Square: |c|c|c| Gametes & X^H & Y \\ X^h & X^HX^h & X^hY \\ X^h & X^HX^h & X^hY \\ Step 3: Determine the genotypes and phenotypes of the offspring. Possible genotypes: X^HX^h, X^hY Possible phenotypes: Normal female (carrier), Haemophiliac male Explanation: All sons inherit an X chromosome from their mother and a Y chromosome from their father. Since the mother is haemophiliac (X^hX^h), she can only pass on an X^h allele to her sons. The father provides the Y chromosome. Therefore, all sons will have the genotype X^hY, which means they will be haemophiliac. 2.4.1 Step 1: Determine the genotypes of the parents. A man suffering from haemophilia has the genotype X^hY. A woman who is a carrier has the genotype X^HX^h. Step 2: Perform a genetic cross. Parents: X^hY × X^HX^h Gametes from male: X^h, Y Gametes from female: X^H, X^h Punnett Square: |c|c|c| Gametes & X^H & X^h \\ X^h & X^HX^h & X^hX^h \\ Y & X^HY & X^hY \\ Step 3: Determine the possible genotypes and phenotypes of the children. Possible genotypes: X^HX^h, X^hX^h, X^HY, X^hY Possible phenotypes: • X^HX^h: Normal female (carrier) • X^hX^h: Haemophiliac female • X^HY: Normal male • X^hY: Haemophiliac male 2.4.2 Haemophilia affects mostly males because it is a sex-linked recessive disorder on the X chromosome. Males have only one X chromosome (XY). If a male inherits the recessive allele (X^h) on his single X chromosome, he will express the disorder (X^hY) because there is no dominant allele on a second X chromosome to mask it. Females have two X chromosomes (XX), so they must inherit two copies of the recessive allele (X^hX^h) to express the disorder. If a female inherits only one recessive allele (X^HX^h), she will be a carrier but will not show symptoms due to the presence of a dominant normal allele (X^H). 2.1.1 Step 1: Determine the genotypes of the parents. A normal father has the genotype X^HY. A heterozygous mother (carrier) has the genotype X^HX^h. Step 2: Represent the genetic cross. Parents: X^HY × X^HX^h Gametes from father: X^H, Y Gametes from mother: X^H, X^h Punnett Square: |c|c|c| Gametes & X^H & X^h \\ X^H & X^HX^H & X^HX^h \\ Y & X^HY & X^hY \\ Step 3: Determine the possible genotypes and phenotypes of the children. Possible genotypes: X^HX^H, X^HX^h, X^HY, X^hY Possible phenotypes: • X^HX^H: Normal female • X^HX^h: Normal female (carrier) • X^HY: Normal male • X^hY: Haemophiliac male 2.1.2 Step 1: Identify the genotype for a haemophiliac male from the Punnett square in 2.1.1. The genotype for a haemophiliac male is X^hY. Step 2: Calculate the chances. From the Punnett square, there is 1 out of 4 possible offspring genotypes that results in a haemophiliac male. The chances are (1)/(4) or 25%. The chances of the parents having a child that will be a haemophiliac male are 25%. 2.1.3 Males have only one X chromosome (XY). Haemophilia is a recessive disorder on the X chromosome. If a male inherits the recessive allele (X^h) on his single X chromosome, he will express the disorder (X^hY) because there is no second X chromosome with a dominant normal allele to mask it. A carrier is an individual who possesses a recessive allele but does not express the trait. Therefore, a male cannot be a carrier for haemophilia; he either has the disorder or he does not.

Accepted Answer · 2026-04-01T05:05:07.199Z

Here are the solutions to the questions: 3.3 Step 1: Determine the genotypes of the parents. A haemophiliac female has the genotype X^hX^h. A normal male has the genotype X^HY. Step 2: Perform a genetic cross. Parents: X^hX^h × X^HY Gametes from female: X^h Gametes from male: X^H, Y Punnett Square: |c|c|c| Gametes & X^H & Y \\ X^h & X^HX^h & X^hY \\ X^h & X^HX^h & X^hY \\ Step 3: Determine the genotypes and phenotypes of the offspring. Possible genotypes: X^HX^h, X^hY Possible phenotypes: Normal female (carrier), Haemophiliac male Explanation: All sons inherit an X chromosome from their mother and a Y chromosome from their father. Since the mother is haemophiliac (X^hX^h), she can only pass on an X^h allele to her sons. The father provides the Y chromosome. Therefore, all sons will have the genotype X^hY, which means they will be haemophiliac. 2.4.1 Step 1: Determine the genotypes of the parents. A man suffering from haemophilia has the genotype X^hY. A woman who is a carrier has the genotype X^HX^h. Step 2: Perform a genetic cross. Parents: X^hY × X^HX^h Gametes from male: X^h, Y Gametes from female: X^H, X^h Punnett Square: |c|c|c| Gametes & X^H & X^h \\ X^h & X^HX^h & X^hX^h \\ Y & X^HY & X^hY \\ Step 3: Determine the possible genotypes and phenotypes of the children. Possible genotypes: X^HX^h, X^hX^h, X^HY, X^hY Possible phenotypes: • X^HX^h: Normal female (carrier) • X^hX^h: Haemophiliac female • X^HY: Normal male • X^hY: Haemophiliac male 2.4.2 Haemophilia affects mostly males because it is a sex-linked recessive disorder on the X chromosome. Males have only one X chromosome (XY). If a male inherits the recessive allele (X^h) on his single X chromosome, he will express the disorder (X^hY) because there is no dominant allele on a second X chromosome to mask it. Females have two X chromosomes (XX), so they must inherit two copies of the recessive allele (X^hX^h) to express the disorder. If a female inherits only one recessive allele (X^HX^h), she will be a carrier but will not show symptoms due to the presence of a dominant normal allele (X^H). 2.1.1 Step 1: Determine the genotypes of the parents. A normal father has the genotype X^HY. A heterozygous mother (carrier) has the genotype X^HX^h. Step 2: Represent the genetic cross. Parents: X^HY × X^HX^h Gametes from father: X^H, Y Gametes from mother: X^H, X^h Punnett Square: |c|c|c| Gametes & X^H & X^h \\ X^H & X^HX^H & X^HX^h \\ Y & X^HY & X^hY \\ Step 3: Determine the possible genotypes and phenotypes of the children. Possible genotypes: X^HX^H, X^HX^h, X^HY, X^hY Possible phenotypes: • X^HX^H: Normal female • X^HX^h: Normal female (carrier) • X^HY: Normal male • X^hY: Haemophiliac male 2.1.2 Step 1: Identify the genotype for a haemophiliac male from the Punnett square in 2.1.1. The genotype for a haemophiliac male is X^hY. Step 2: Calculate the chances. From the Punnett square, there is 1 out of 4 possible offspring genotypes that results in a haemophiliac male. The chances are (1)/(4) or 25%. The chances of the parents having a child that will be a haemophiliac male are 25%. 2.1.3 Males have only one X chromosome (XY). Haemophilia is a recessive disorder on the X chromosome. If a male inherits the recessive allele (X^h) on his single X chromosome, he will express the disorder (X^hY) because there is no second X chromosome with a dominant normal allele to mask it. A carrier is an individual who possesses a recessive allele but does not express the trait. Therefore, a male cannot be a carrier for haemophilia; he either has the disorder or he does not.

In humans, haemophilia is a sex-linked recessive trait. What are the possible genotypes and phenotypes of offspring from a haemophiliac female and a normal male?

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In humans, haemophilia is a sex-linked recessive trait. What are the possible genotypes and phenotypes of offspring from a haemophiliac female and a normal male?

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