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AURORKA [14]
4 years ago
14

disorders can occur as a result of a person missing chromosomes or having too many chromosomes. which of the following cells wou

ld not have a disorder? a) sex cell with 23 chromosomes: 22 autosomes and one sex chromosome b) sex cell with 22 chromosomes: 22 autosomes and no sex chromosomes c) sex cell with 24 chromosomes: 22 autosomes and two sex chromosomes d) sex cell with 24 chromosomes: 23 autosomes and one sex chromosome
Biology
2 answers:
Oxana [17]4 years ago
8 0

A)  sex cell with 23 chromosomes: 22 autosomes and one sex chromosome

lesya692 [45]4 years ago
7 0

Answer:

a) sex cell with 23 chromosomes: 22 autosomes and one sex chromosome

Explanation:

Sex cells are the gametes and are products of meiosis. These cells have haploid chromosome number. Human sex cells are sperms and egg which serve as male and female gamete respectively. Human sex cells have n = 23 chromosomes. Out of total 23 chromosomes, 22 chromosomes are autosomes while one is the sex chromosome (either X or Y chromosome). Deviation from this pattern can cause any genetic disorder.

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Genetic linkage mapping for a large number of families identifies 4% recombination between the genes for Rh blood type and ellip
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0.2404

Explanation:

The genes R/r and E/e are linked and there is 4% recombination between them.

<u>The possible genotypes and phenotypes are:</u>

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  • rr: Rh- blood type
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Tom and Terri each have elliptocytosis (they are E_), and each is Rh+ (they are R_).

Tom's mother has elliptocytosis (E_) and is Rh- (rr), so she has the genotype Er/_r. His father is healthy (ee) and has Rh+ (R_), so he has the genotype eR/e_. Tom must have inherited his E allele from his mother and his R allele  from his father, so he has the genotype eR/Er.

Terri's father is Rh+ (R_) and has elliptocytosis (E_), while Terri's mother is Rh- (rr) and is healthy (ee) with the genotype er/er. Terry could only receive the chromosome <em>er </em>from her mother, and because she is heterozygous for both genes the dominant alleles were both received from her father. Terri's genotype is ER/er.

The frequency of recombination is 4%, so 4% of the produced gametes will be recombinant. There are two possible recombinant gametes, so each will appear 2% of the times (a frequency of 0.02).

<u />

<u>Tom will produce the following gametes:</u>

  • eR, parental (0.48)
  • Er, parental (0.48)
  • er, recombinant (0.02)
  • ER (recombinant (0.02)

<u>Terri will produce the following gametes:</u>

  • ER, parental (0.48)
  • er, parental (0.48)
  • Er, recombinant (0.02)
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A child Rh- with elliptocytosis has the genotype rrE_. This can happen from the independent combination of the following gametes from Tom and Terri respectively:

  • Er (0.48) × er (0.48) = 0.2304 Er/er
  • Er (0.48) × Er (0.02) = 0.0096 Er/Er
  • er (0.02) × Er (0.02) = 0.0004 er/Er

And the total probability of having a rrE_ child will be 0.2304 + 0.0096 + 0.0004 = 0.2404

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