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Answer:
Please find the expected genotypes and phenotypes of the progenies of each cross below.
Explanation:
In the MN blood-group locus of the gene, alleles LM and LN exhibit codominance i.e one allele is not dominant or recessive to the other, hence, they are both expressed when they occur in a heterozygous state (MN).
Considering the following crosses (find the punnet square attached);
a)LMLM x LMLN - The progeny are LMLM and LMLN in the genotypic ratio 1:1. Phenotypic ratio is Blood type M (1) : blood type MN (1)
b) LNLN x LNLN - The progeny are all LNLN offsprings with a phenotypic and genotypic ratio 4:0. All offsprings will have a blood type N (4)
c) LMLN x LMLN - The progenies are LMLM, LMLN and LNLN in the genotypic ratio 1:2:1 respectively. The phenotypic ratio is Blood type M (1) : L
Blood type MN (2) : Blood type N (1)
d) LMLN x LNLN - The progeny are LMLN and LNLN with genotypic ratio 1:1 and phenotypic ratio blood type MN (1) : blood type N (1)
e) LMLM x LNLN - The progeny are all LMLN offsprings with penotypic ratio blood type MN (4)
B. They decrease. As more we use the sources of nature, the less is left.
Answer:
See the explanation
Explanation:
Answer 1.
As given that in F1 all are short and white then it can be said according to Mendel's law that short and white are dominant over tall and purple.
Let S for short s for tall and W for white and w for purple allele.
So the genotype of short purple will be Ssww or SSww. So In first case Ssww self crossed then resultant offsprings will be,
................ Sw .................... sw
Sw ........ SSww ............. Ssww
sw ........ Ssww .............. ssww (tall and purple)
So from this 1/4 will be tall and purple while 3/4 will be short and purple.
In second case SSww only short purple progeny will appear.
Answer 2.
2. a) The female progeny will not show any trait because there are two X chromosome in females , so female offspring can be carrier but not show any trait in case of X- linked trait.
2. b) Half of male offsprings show trait because X is inherited from mother. So the chance of having X-linked recessive allele is 1/2.
2. c) The chance of having X linked affect allele in daughter is 1/2. So the chance of inheriting that X to son will be 1/2 so in total there is chance of 1/4 that son will be affected.
2. d) the chance of first child show this trait will be 1/4 in case of male offspring while 0 in case of female offspring.
Hope this helps!
It is important that it has half because there is a half that comes from each parent. so to equal it all out human gametes have a half set of dna instead of a full set
