A high-interest rate means the opportunity cost to built or buy a house is high it is a possible channel. So when the interest rate is higher, there may be fewer new housing starts. The abstract point-scoring task requires a minimal model that illustrates a plausible causal channel.
I think Oxygen is the best answer.
Answer:
6,25%
Explanation:
Considering that the couple has a trait of sickle cell anemia, we know that both are heterozygous for the disease (Aa) and therefore can have children with the following genotypes:
Parents: Aa X Aa
Children: AA(A x A), Aa(A x a), Aa (a x A) and aa(a x a)
Knowing that sickle cell anemia only occurs in homozygous individuals, the probability for children to have the disease according to each crossing is:
A x A = 1/4 = 25%
A x a = 1/4 = 25%
a x A = 1/4 = 25%
a x a = 1/4 = 25%
The probability of forming each homozygous child (aa) is 1/4 or 25%. Since they are two children, the probability of both having sickle cell anemia is calculated by multiplying the probability of each, so:
1/4 × 1/4 = 1/16 = 0.0625 = 6.25%
It is concluded that the probability of a heterozygous couple for sickle cell anemia to have two children with the disease is 6.25%.
The answer is c. Anaphase I separates homologous chromosomes and anaphase II separates sister chromatids into daughter cells.
Meiosis is a cell division which results in the reduction of chromosome number by half (from diploid to haploid) in daughter cells. It consists of meiosis I and meiosis II.
In anaphase I, the sister chromatids separate from each other to the opposite sides of the cells. In meiosis I there are 46 chromosomes in duplicates which are present as pairs of sister chromatids. When comes to separation, homologous chromosomes separates only, but not sister chromatids. Homologous chromosomes are present only in meiosis I.
In anaphase II, since the cell is haploid, there are 23 chromosomes in duplicates, which are present as sister chromatids. So, in this phase, sister chromatids are those who separates.
9:3: 3:1 is the phenotypic ratio showing traits as black and long hair : black and short hair: chestnut and long hair: chestnut and short hair when a chestnut horse heterozygous for pacing and hair length with a hybrid horse.
Explanation:
Dominant trait = black hair colour (BB,Bb), trotting (TT,Tt) , long hair (LL,Ll)
recessive trait = chesnut hair colour (bb), pacing gait (tt), short hair(ll)
cross between chestnut horse heterozygous for pacing and hair length will have alleles as BbLl
alleles for hybrid horse will also be heterozygous Bb, Ll
Punnett square to show the cross:
BL Bl bL bl
BL BBLL BBLl BbLL BbLl
Bl BBLl BBll BbLl Bbll
bL BbLL BbLl bbLL bbLl
bl BbLl Bbll bblL bbll
phenotype ratio
black and long hair : black and short hair: chestnut and long hair: chestnut and short hair
9:3: 3:1 is the phenotype ratio.
