Which condition is most consistent with this karyotype finding, and what is the underlying mechanism?

Answer: Down syndrome (trisomy 21), most often caused by nondisjunction during meiosis I in the mother. A third copy of chromosome 21 (trisomy 21) causes Down syndrome and most commonly results from maternal meiosis I nondisjunction.

In humans, sex is determined by the

Answer: Presence of the Y chromosome (XY = male, XX = female, in the typical case) The SRY gene on Y triggers male development. XX females lack Y; XY individuals have Y → male development.

What is the probability that a son from this cross will be affected with hemophilia?

Answer: 1/2 Among male offspring, half inherit X^H (unaffected) and half inherit X^h (affected). So 50% of sons are predicted to be hemophiliacs.

AP Biology · Topic 5.6

Chromosomal Inheritance Practice

Part of Heredity.(IST-1.H)

Practice questions

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Sample questions

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Sample 1difficulty 1/5
$Which condition is most consistent with this karyotype finding, and what is the underlying mechanism?$
- A
  $Klinefelter syndrome, caused by an extra X chromosome in males.$
- B
  $Down syndrome (trisomy 21), most often caused by nondisjunction during meiosis I in the mother.$
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- C
  $Cri-du-chat, caused by chromosome 5 deletion.$
- D
  $Turner syndrome, caused by a missing X chromosome.$
Why
$A third copy of chromosome 21 (trisomy 21) causes Down syndrome and most commonly results from maternal meiosis I nondisjunction.$
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Sample 2difficulty 1/5
$In humans, sex is determined by the$
- A
  $The total number of autosomal chromosomes (46 = male, 44 = female, in the typical case)$
- B
  $The ratio of X chromosomes to autosomes (1:1 = male, 2:1 = female, in the typical case)$
- C
  $Presence of the Y chromosome (XY = male, XX = female, in the typical case)$
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- D
  $The presence of mitochondrial DNA (maternal mtDNA = female, paternal mtDNA = male)$
Why
$The SRY gene on Y triggers male development. XX females lack Y; XY individuals have Y \to male development.$
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Sample 3difficulty 2/5
$A female heterozygous carrier of the X-linked recessive hemophilia allele (X^H X^h) is crossed with an unaffected male (X^H Y).$
$What is the probability that a son from this cross will be affected with hemophilia?$
- A
  $1/4$
- B
  $0$
- C
  $1$
- D
  $1/2$
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Why
$Among male offspring, half inherit X^H (unaffected) and half inherit X^h (affected). So 50% of sons are predicted to be hemophiliacs.$
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Sample 4difficulty 2/5
$A heterozygous AaBb fruit fly is testcrossed to a homozygous recessive aabb partner. The phenotypic counts of 1000 progeny are listed.$
$What is the recombination frequency between the A and B loci?$
- A
  $Approximately 20%, indicating linked genes about 20 mu apart$
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- B
  $Cannot be calculated from the data$
- C
  $Approximately 1%, indicating tight linkage$
- D
  $Exactly 50%, indicating unlinked genes$
Why
$Recombinants = 95 + 105 = 200 of 1000 total = 20%. Recombination below 50% indicates linkage; the value (~20 mu) estimates map distance between A and B.$
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Sample 5difficulty 2/5
$From this cross, the probability of a child being male is$
- A
  $0%$
- B
  $50%$
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- C
  $100%$
- D
  $25%$
Why
$Two of four boxes are XY (male). Sex is determined by which sex chromosome the father contributes — X or Y — at 50:50.$
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