Sex Linkage and Meiosis (1 Viewer)

[bayan92]

Can someone please explain to me, how to answer questions:

• 18 a) ii)
• 18 b)
• 19 a)

http://mscw.catholic.edu.au/Assessm...xam Choice HSC/Bio Trial 2009 Exam Choice.pdf

Thanks :uhhuh:

 

[Prashant Sallan]

Can someone please explain to me, how to answer questions:

• 18 a) ii)
• 18 b)
• 19 a)

http://mscw.catholic.edu.au/Assessm...xam Choice HSC/Bio Trial 2009 Exam Choice.pdf

Thanks :uhhuh:

18)b

You work back from sue (towards the parents), and work out everyones genotype, by doing this you get the the conclusion that:
Sues genotype is either:
XX or XX' (this depends on what gene her mother gave to her (X' or X))
Since her child is a boy, her husband gives a y allele, therefore there is a 25% chance that her son will DMD, since the genotypes she can pass on from the pool of possibilities are: X, X, X, or X' therefore: 1/4 (25%)

19) a isn't this just..
AB, ab, Cd, CD :S

 

[hermine]

19) a isn't this just..
AB, ab, Cd, CD :S

no that can't be right, the two homologous chromosomes should be the same size, so the pairs are the two longer ones and the two shorter ones with the same letter representing corresponding alleles. The genotype should then be Aa, Bb, CC and Dd

 

[hermine]

as for 18 a) (ii), a female would need to have 2 copies of the recessive allele (one on each X chromosome) to show the sex-linked conditions of DMD. One of these alleles would thus need to be inherited from a father who also suffers from DMD. Males with DMD are weelchairbound and rarely live beyond twenty as stated in the question. This then means it would be very unlikely for such males to be able to engage in sexual intercourse to reproduce to in turn pass the allele to a possible female.