Short term effects include memory loss, hangovers, and blackouts. Long term problems associated with heavy drinking include stomach ailments, heart problems, cancer, brain damage, serious memory loss and liver cirrhosis.
Four cells and 10 chromosomes each. Before going through meiosis ii it has to go through meiosis i (homologous chromosomes separate) first to form two cells. Then both cells goes through meiosis ii (sister chromatids separate) to form two cells each. Therefore 4 cells.<span />
Answer/Explanation:
Two unaffected parents can have affected children, that means cystic fibrosis must be a recessive trait.
The two parents must each have one copy of the cystic fibrosis allele, but are themselves unaffected. By chance, the son (shaded) inherited both of these copies and is affected. The daughter either inherited one or zero copies of the cystic fibrosis allele, as she is unaffected.
Answer:
Mintu should have chosen a material with a lower specific heat.
Explanation:
When we talk about specific heat, we refer to a specific quantity of heat that is needed for a substance or material to rise its temperature. In more concrete terms, is the amount of heat required for 1 gram of substance/material to elevate its temperature by 1 degree Celsius.
This would mean that a material with a high specific heat would require a higher quantity of heat to increase its temperature, so in this case, if Mintu wanted to make the heat transfer easier between the metal iron and the other piece of metal, he should have chosen a material that could elevate its temperature by receiving a lower quantity of heat, that is a material with a low specific heat. But instead he chose a metal with a higher specific heat, which will now make the heat transfer more difficult,and that was the error.
Answer:
<em>The genotypic ratio of the F1 generation is: </em>
1:1 OR 50% heterozygous wire-haired, 50% homozygous smooth-haired.
<em>The phenotypic ratio of the F1 generation is:</em>
1:1 OR 50% wire-haired, 50% smooth haired.
Explanation:
A cross between the parents with genotype:
Parent 1; Ss and Parent 2; ss, will yield two types of progeny. Half the offspring in the F1 generation will be heterozygous wire-haired just as in parent 1, while the other half will be homozygous short-haired as in parent 2.
Demonstrating it with a punnet square:
S s
s Ss ss
s Ss ss 2/4 offspring: Ss,
2/4 offspring: ss
Since the allele for wire hair is dominant over that of short hair, the presence of one dominant allele, translates into the offspring having wire hair phenotypically or by physical assessment. The offspring who are homozygous for the recessive allele however, express short hair because they carry both alleles for short hair.